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1l6nnts5x | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let S be the set of all a $$\in R$$ for which the angle between the vectors $$
\vec{u}=a\left(\log _{e} b\right) \hat{i}-6 \hat{j}+3 \hat{k}$$ and $$\vec{v}=\left(\log _{e} b\right) \hat{i}+2 \hat{j}+2 a\left(\log _{e} b\right) \hat{k}$$, $$(b>1)$$ is acute. Then S is equal to :</p> | [{"identifier": "A", "content": "$$\\left(-\\infty,-\\frac{4}{3}\\right)$$"}, {"identifier": "B", "content": "$$\\Phi $$"}, {"identifier": "C", "content": "$$\\left(-\\frac{4}{3}, 0\\right)$$"}, {"identifier": "D", "content": "$$\\left(\\frac{12}{7}, \\infty\\right)$$"}] | ["B"] | null | <p>$$\overrightarrow u = a({\log _e}b)\widehat i - 6\widehat j + 3\widehat k$$</p>
<p>$$\overrightarrow v = ({\log _e}b)\widehat i + 2\widehat j + 2a({\log _e}b)\widehat k$$</p>
<p>For acute angle $$\overrightarrow u \,.\,\overrightarrow v > 0$$</p>
<p>$$ \Rightarrow a{({\log _e}b)^2} - 12 + 6a({\log _e}b) > 0$$</p>
<p>$$\because$$ $$b > 1$$</p>
<p>Let $${\log _e}b = t \Rightarrow t > 0$$ as $$b > 1$$</p>
<p>$$a{t^2} + 6at - 12 > 0\,\,\,\,\,\,\,\forall t > 0$$</p>
<p>$$ \Rightarrow a \in \phi $$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldo522uv | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$$ and $$\vec{b}=\hat{i}+3 \hat{j}+5 \hat{k}$$ be two vectors. Then which one of the following statements is TRUE ?</p> | [{"identifier": "A", "content": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$."}, {"identifier": "B", "content": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{13}{\\sqrt{35}}$$ and the direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$."}, {"identifier": "C", "content": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{13}{\\sqrt{35}}$$ and the direction of the projection vector is same as of $$\\vec{b}$$."}, {"identifier": "D", "content": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vector is same as of $$\\vec{b}$$."}] | ["A"] | null | $\begin{aligned} & \text { Projection of }\vec{a} \text { on } \vec{b} =\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\\\ & = \frac{(5 \hat{i}-\hat{j}-3 \hat{k}) \cdot(\hat{i}+3 \hat{j}+5 \hat{k})}{\sqrt{1^2+3^2+5^2}}=\frac{5-3-15}{\sqrt{35}} \\\\ & = \frac{-13}{\sqrt{35}}\end{aligned}$
<br/><br/>Negative sign indicates that direction of the projection vector is opposite to the direction
of $$\vec{b}$$. | mcq | jee-main-2023-online-1st-february-evening-shift |
1ldo6n5a3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}$$ and $$\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}$$ be three given vectors. If $$\overrightarrow{\mathrm{r}}$$ is a vector such that $$\vec{r} \times \vec{a}=\vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b}=0$$, then $$|\vec{r}|$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{11}{7}$$"}, {"identifier": "B", "content": "$$\\frac{11}{5} \\sqrt{2}$$"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{914}}{7}$$"}, {"identifier": "D", "content": "$$\\frac{11}{7} \\sqrt{2}$$"}] | ["D"] | null | $\begin{aligned} & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \Rightarrow \vec{r}-\vec{c}=\lambda \vec{a}((\vec{r}-\vec{c} ) \text{and} \overrightarrow{a} \text { are parallel }) \\\\ & \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a} \\\\ & \Rightarrow \vec{r} \cdot \vec{b}=\vec{c} \cdot \vec{b}+\lambda \vec{a} \cdot \vec{b} \\\\ & 0=(1-3)+\lambda(2+5) \Rightarrow \lambda=\frac{2}{7} \\\\ & \text { Hence, } \vec{r}=\vec{c}+\frac{2 \vec{a}}{7} \\\\ & \vec{r} \Rightarrow \frac{11}{7} \hat{i}-\frac{11}{7} \hat{k} \\\\ & |\vec{r}|=\sqrt{\left(\frac{11}{7}\right)^2+\left(-\frac{11}{7}\right)^2} \Rightarrow r=\frac{11 \sqrt{2}}{7}\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
1ldpsrr3l | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$ and $$\vec{b} \cdot \vec{c}=0$$. Consider the following two statements:</p>
<p>(A) $$|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$$ for all $$\lambda \in \mathbb{R}$$.</p>
<p>(B) $$\vec{a}$$ and $$\vec{c}$$ are always parallel.</p>
<p>Then,</p> | [{"identifier": "A", "content": "only (B) is correct"}, {"identifier": "B", "content": "both (A) and (B) are correct"}, {"identifier": "C", "content": "only (A) is correct"}, {"identifier": "D", "content": "neither (A) nor (B) is correct"}] | ["C"] | null | $|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$
<br/><br/>$$ \Rightarrow $$ $|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}+\vec{b}-\vec{c}|^{2}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a}) \\\\
& =|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{a}) \\\\
& \Rightarrow \vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=0 \Rightarrow \vec{c} \cdot \vec{a}=0 \\\\
& |\vec{a}+\lambda \vec{c}|^{2}=|\vec{a}|^{2}+\lambda^{2}|\vec{c}|^{2}+0 \geq|\vec{a}|^{2}
\end{aligned}
$$
<br/><br/>So $\mathrm{A}$ is correct.
<br/><br/>$B$ is incorrect. | mcq | jee-main-2023-online-31st-january-morning-shift |
1ldswfhf7 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>If the vectors $$\overrightarrow a = \lambda \widehat i + \mu \widehat j + 4\widehat k$$, $$\overrightarrow b = - 2\widehat i + 4\widehat j - 2\widehat k$$ and $$\overrightarrow c = 2\widehat i + 3\widehat j + \widehat k$$ are coplanar and the projection of $$\overrightarrow a $$ on the vector $$\overrightarrow b $$ is $$\sqrt {54} $$ units, then the sum of all possible values of $$\lambda + \mu $$ is equal to :</p> | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "6"}] | ["A"] | null | $\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}, \vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$
<br/><br/>
Now, $\vec{a} \cdot \vec{b}=\sqrt{54} \Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$
<br/><br/>
$\Rightarrow-2 \lambda+4 \mu-8=36$
<br/><br/>
$\Rightarrow 2 \mu-\lambda=22\quad...(i)$
<br/><br/>
and $\left|\begin{array}{ccc}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{array}\right|=0$
<br/><br/>
$10 \lambda-2 \mu-56=0 \quad...(ii)$
<br/><br/>
By (i) & (ii) $\lambda=\frac{78}{9}, \mu=\frac{138}{9}$
<br/><br/>
$\therefore \mu+\lambda=24$ | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldv14lsr | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>The vector $$\overrightarrow a = - \widehat i + 2\widehat j + \widehat k$$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $$\overrightarrow b $$. Then the projection of $$3\overrightarrow a + \sqrt 2 \overrightarrow b $$ on $$\overrightarrow c = 5\widehat i + 4\widehat j + 3\widehat k$$ is :</p> | [{"identifier": "A", "content": "$$\\sqrt6$$"}, {"identifier": "B", "content": "2$$\\sqrt3$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3$$\\sqrt2$$"}] | ["D"] | null | <p>First, we write $\overrightarrow{b}$ as a linear combination of $\overrightarrow{a}$ and $\overrightarrow{j}$ since $\overrightarrow{b}$ is a rotation of $\overrightarrow{a}$ about the y-axis.</p>
<p>$\vec{b}=\lambda \vec{a}+\mu \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})+\mu \hat{j}=-\lambda \hat{i}+(2 \lambda+\mu \hat{j})+\lambda \hat{k}$</p>
<p>$\overrightarrow{b}$ is orthogonal to $\overrightarrow{a}$ due to the right angle rotation, so $\overrightarrow{b} \cdot \overrightarrow{a} = 0$. </p>
<p>This implies :</p>
<p>$\begin{aligned} & (-\hat{i}+2 \hat{j}+\hat{k}) \cdot(-\lambda \hat{i}+(2 \lambda+\mu) \hat{j}+\lambda \hat{k})=0 \\\\ & \lambda+2(2 \lambda+\mu)+\lambda=0 \Rightarrow 6 \lambda+2 \mu=0 \Rightarrow \mu+3 \lambda=0\end{aligned}$</p>
<p>$$ \therefore $$ $\vec{b}=\lambda \vec{a}-3 \lambda \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})-3 \lambda \hat{j}=\lambda(-\hat{i}-\hat{j}+\hat{k})$</p>
<p>The magnitude of $\overrightarrow{b}$ is the same as the magnitude of $\overrightarrow{a}$ because a rotation doesn't change the magnitude of a vector. This gives us :</p>
<p>$$
|\vec{b}|=\sqrt{3}|\lambda|=\sqrt{6}[\because|a|=\sqrt{6}] \Rightarrow|\lambda|=\sqrt{2} \Rightarrow \lambda \neq \sqrt{2}
$$
<br/><br/>as for this value of $\lambda$ angle between $b$ and $y$-axis is not acute.
<br/><br/>Therefore $\lambda=-\sqrt{2}$</p>
<p>Thus, we have :</p>
<p>$\overrightarrow{b} = -\sqrt{2}(-\hat{i} - \hat{j} + \hat{k}) = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$</p>
<p>Then, we find the vector $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ :</p>
<p>$3\overrightarrow{a} + \sqrt{2}\overrightarrow{b} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}) $
<br/><br/>$= -3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} - 2\hat{k} = -\hat{i} + 8\hat{j} + \hat{k}$</p>
<p>The projection of this vector onto $\overrightarrow{c}$ is given by the dot product divided by the magnitude of $\overrightarrow{c}$ :</p>
<p>$\frac{(-\hat{i} + 8\hat{j} + \hat{k}) \cdot (5\hat{i} + 4\hat{j} + 3\hat{k})}{\sqrt{5^2 + 4^2 + 3^2}} = \frac{-5 + 32 + 3}{\sqrt{50}} = \frac{30}{5\sqrt{2}} = 3\sqrt{2}$</p>
<p>So the projection of $3\overrightarrow{a} + \sqrt{2}\overrightarrow{b}$ onto $\overrightarrow{c}$ is $3\sqrt{2}$ which is option D.</p>
| mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwwcfsd | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow \alpha = 4\widehat i + 3\widehat j + 5\widehat k$$ and $$\overrightarrow \beta = \widehat i + 2\widehat j - 4\widehat k$$. Let $${\overrightarrow \beta _1}$$ be parallel to $$\overrightarrow \alpha $$ and $${\overrightarrow \beta _2}$$ be perpendicular to $$\overrightarrow \alpha $$. If $$\overrightarrow \beta = {\overrightarrow \beta _1} + {\overrightarrow \beta _2}$$, then the value of $$5{\overrightarrow \beta _2}\,.\left( {\widehat i + \widehat j + \widehat k} \right)$$ is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "11"}] | ["B"] | null | Let $\vec{\beta}_1=\lambda \vec{\alpha}$<br/><br/>
Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$<br/><br/>
$$
\begin{aligned}
& =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})-\lambda(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \\\\
& =(1-4 \lambda) \hat{\mathrm{i}}+(2-3 \lambda) \hat{\mathrm{j}}-(5 \lambda+4) \hat{\mathrm{k}} \\\\
& \vec{\beta}_2 \cdot \vec{\alpha}=0 \\\\
& \Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0 \\\\
& \Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0 \\\\
& \Rightarrow 50 \lambda=-10 \\\\
& \Rightarrow \lambda=\frac{-1}{5} \\\\
& \vec{\beta}_2=\left(1+\frac{4}{5}\right) \hat{\mathrm{i}}+\left(2+\frac{3}{5}\right) \hat{\mathrm{j}}-(-1+4) \hat{\mathrm{k}} \\\\
& \vec{\beta}_2=\frac{9}{5} \hat{\mathrm{i}}+\frac{13}{5} \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\\\
& 5 \vec{\beta}_2=9 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\\\
& 5 \vec{\beta}_2 \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=9+13-15=7
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift |
lsblj5uf | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2 \hat{j}+2 \hat{k}$ and $\alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k}$ is acute, is ___________. | [] | null | 5 | <p>$$\begin{aligned}
& \cos \theta=\frac{(\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\alpha \hat{\mathrm{i}}+2 \alpha \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{\alpha^2+4+4} \sqrt{\alpha^2+4 \alpha^2+4}} \\
& \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \alpha^2+4}} \\
& \Rightarrow \alpha^2-4 \alpha-4>0 \quad \Rightarrow(\alpha-2)^2>8 \\
& \Rightarrow \alpha^2-4 \alpha+4>8 \quad \alpha-2<-2 \sqrt{2} \\
& \Rightarrow \alpha-2>2 \sqrt{2} \text { or } \alpha-2<2 \sqrt{2} \\
& \alpha>2+2 \sqrt{2} \text { or } \alpha<2-2 \sqrt{2} \\
& \alpha \in(-\infty,-0.82) \cup(4.82, \infty)
\end{aligned}$$</p>
<p>Least positive integral value of $$\alpha \Rightarrow 5$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lsfkf8y5 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let a unit vector $$\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$$ make angles $$\frac{\pi}{2}, \frac{\pi}{3}$$ and $$\frac{2 \pi}{3}$$ with the vectors $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ and $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$$ respectively. If $$\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ then $$|\hat{u}-\vec{v}|^2$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{11}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{2}$$"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "9"}] | ["B"] | null | <p>Unit vector $$\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$$</p>
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}
\end{aligned}$$</p>
<p>Now angle between $$\hat{\mathrm{u}}$$ and $$\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}$$</p>
<p>$$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$$</p>
<p>$$\Rightarrow \mathrm{x}+\mathrm{z}=0$$ ...... (i)</p>
<p>Angle between $$\hat{\mathrm{u}}$$ and $$\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}$$</p>
<p>$$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}$$</p>
<p>$$\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \Rightarrow y+z=\frac{1}{\sqrt{2}}$$ ...... (ii)</p>
<p>Angle between $$\hat{\mathrm{u}}$$ and $$\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}$$</p>
<p>$$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3}$$</p>
<p>$$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}}$$ ..... (iii)</p>
<p>from equation (i), (ii) and (iii) we get</p>
<p>$$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$$</p>
<p>$$\begin{aligned}
& \text { Thus } \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\
& \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \\
& \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
lv0vxcgi | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let a unit vector which makes an angle of $$60^{\circ}$$ with $$2 \hat{i}+2 \hat{j}-\hat{k}$$ and an angle of $$45^{\circ}$$ with $$\hat{i}-\hat{k}$$ be $$\vec{C}$$. Then $$\vec{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$$ is:</p> | [{"identifier": "A", "content": "$$-\\frac{\\sqrt{2}}{3} \\hat{i}+\\frac{\\sqrt{2}}{3} \\hat{j}+\\left(\\frac{1}{2}+\\frac{2 \\sqrt{2}}{3}\\right) \\hat{k}$$\n"}, {"identifier": "B", "content": "$$\\left(\\frac{1}{\\sqrt{3}}+\\frac{1}{2}\\right) \\hat{i}+\\left(\\frac{1}{\\sqrt{3}}-\\frac{1}{3 \\sqrt{2}}\\right) \\hat{j}+\\left(\\frac{1}{\\sqrt{3}}+\\frac{\\sqrt{2}}{3}\\right) \\hat{k}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{2}}{3} \\hat{i}-\\frac{1}{2} \\hat{k}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{2}}{3} \\hat{i}+\\frac{1}{3 \\sqrt{2}} \\hat{j}-\\frac{1}{2} \\hat{k}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \text { Let } \vec{C}=a \hat{i}+b \hat{j}+c \hat{k} \\
& (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}+2 \hat{j}-\hat{k})=1 \times 3 \times \frac{1}{2} \\
& 2 a+2 b-c=\frac{3}{2} \qquad \text{... (1)}\\
& (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}-\hat{k})=1 \times \sqrt{2} \times \frac{1}{\sqrt{2}} \\
& a-c=1 \quad \text{... (2)}\\
& a^2+b^2+c^2=1 \quad \text{... (3)}\\
\end{aligned}$$</p>
<p>Solving (1), (2) and (3)</p>
<p>$$\begin{aligned}
& a+2 b=\frac{1}{2} \\
& a^2+b^2+(a-1)^2=1 \\
& 2 a^2-2 a+b^2=0 \\
& 2 a^2-2 a+\left(\frac{2 a-1}{4}\right)^2=0 \\
& 32 a^2-32 a+4 a^2-4 a+1=0 \\
& 36 a^2-36 a+1=0 \\
& a=\frac{36 \pm \sqrt{(36)^2-4(36)}}{2 \times 36} \\
& =\frac{1}{2} \pm \frac{\sqrt{2}}{3} \\
& b=\frac{1-2 a}{4} \Rightarrow b=\frac{1 \pm \frac{2 \sqrt{2}}{3}-1}{4} \\
& =\mp \frac{1}{3 \sqrt{2}} \\
& C=-\frac{1}{2} \pm \frac{\sqrt{2}}{3}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& C+\left(\frac{-1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right) \\
& =\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2erzmn | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>For $$\lambda>0$$, let $$\theta$$ be the angle between the vectors $$\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. If the vectors $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ are mutually perpendicular, then the value of (14 cos $$\theta)^2$$ is equal to</p> | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "40"}] | ["A"] | null | <p>$$\begin{aligned}
& \text { Given } \vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k} \\
& \vec{b}=3 \hat{i}-\hat{j}+2 \hat{k} \\
& \vec{a}+\vec{b}=4 \hat{i}+(\lambda-1) \hat{j}-\hat{k} \\
& \vec{a}-\vec{b}=-2 \hat{i}+(\lambda+1) \hat{j}-5 \hat{k} \\
& (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 \\
& -8+\lambda^2-1+5=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \lambda^2=4 \\
& \lambda= \pm 2 \because \lambda>0 \text { (Given) } \\
& \therefore \lambda=2 \\
& \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\
& \cos \theta=\frac{3-2-6}{\sqrt{14} \times \sqrt{14}}=\frac{-5}{14} \\
& (14 \cos \theta)^2=\left(14 \times \frac{-5}{14}\right)^2=25
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv3veez6 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$$ and $$\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$$ be three vectors. Let $$\overrightarrow{\mathrm{r}}$$ be a unit vector along $$\vec{b}+\vec{c}$$. If $$\vec{r} \cdot \vec{a}=3$$, then $$3 \lambda$$ is equal to:</p> | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "30"}] | ["B"] | null | <p>$$\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\
& \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \\
& \vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k} \\
& \vec{b}+\vec{c}=5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}
\end{aligned}$$</p>
<p>$$\vec{r}$$ is a unit vector along $$\vec{b}+\vec{c}$$</p>
<p>$$\therefore \quad \vec{r}=\frac{5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}}{\sqrt{25+4+(\lambda-5)^2}}$$</p>
<p>Now, $$\vec{r} \cdot \vec{a}=3$$</p>
<p>$$\frac{1}{\sqrt{29+(\lambda-5)^2}}[5+4+3(\lambda-5)]=3$$</p>
<p>Squaring both sides</p>
<p>$$\begin{aligned}
& \Rightarrow \quad \frac{1}{29+(\lambda-5)^2}\left[9+3(\lambda-5)^2\right]=9 \\
& \Rightarrow \quad[3+(\lambda-5)]^2=29+(\lambda-5)^2 \\
& \Rightarrow \quad 9+(\lambda-5)^2+6(\lambda-5)=29+(\lambda-5)^2 \\
& \Rightarrow \quad 9+6(\lambda-5)=29 \\
& \Rightarrow \quad \lambda=\frac{20}{6}+5=\frac{25}{3} \\
& \therefore \quad 3 \lambda=3 \times \frac{25}{3}=25
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv5grw8k | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>The set of all $$\alpha$$, for which the vectors $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$ and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$ are inclined at an obtuse angle for all $$t \in \mathbb{R}$$, is</p> | [{"identifier": "A", "content": "$$[0,1)$$\n"}, {"identifier": "B", "content": "$$\\left(-\\frac{4}{3}, 0\\right]$$\n"}, {"identifier": "C", "content": "$$(-2,0]$$\n"}, {"identifier": "D", "content": "$$\\left(-\\frac{4}{3}, 1\\right)$$"}] | ["B"] | null | <p>Given $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$</p>
<p>and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$</p>
<p>angle between $$\vec{a}$$ and $$\vec{b}$$ is given by</p>
<p>$$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$</p>
<p>We have, $$\cos \theta < 0(\because$$ angle between $$\vec{a}$$ and $$\vec{b}$$ is obtuse)</p>
<p>$$\begin{aligned}
& \Rightarrow \quad \vec{a} \cdot \vec{b}<0 \\
& \Rightarrow \alpha t^2-12+6 \alpha t<0 \forall t \in \mathbb{R}
\end{aligned}$$</p>
<p>If $$\alpha=0$$, then $$-12<0$$ (condition holds)</p>
<p>If $$\alpha \neq 0 \Rightarrow \alpha<0\quad \text{.... (i)}$$</p>
<p>And maximum value of $$\alpha t^2+6 \alpha t-12<0$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{-D}{4 a}<0 \text { (where } D \text { is discriminant and } a=\alpha) \\
& \Rightarrow \frac{36 \alpha^2+48 \alpha}{4 \alpha}>0 \\
& \Rightarrow \alpha>\frac{-4}{3} \\
& \therefore \alpha \in\left(\frac{-4}{3}, 0\right]
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
TRVqTQF2Low17LyL | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If the vectors $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ from the sides $B C, C A$ and $A B$ respectively of a triangle $A B C$, then : | [{"identifier": "A", "content": "$\\overrightarrow{\\mathbf{a}} \\cdot \\overrightarrow{\\mathbf{b}}=\\overrightarrow{\\mathbf{b}} \\cdot \\overrightarrow{\\mathbf{c}}=\\overrightarrow{\\mathbf{c}} \\cdot \\overrightarrow{\\mathbf{b}}=0$"}, {"identifier": "B", "content": "$\\overrightarrow{\\mathbf{a}} \\times \\overrightarrow{\\mathbf{b}}=\\overrightarrow{\\mathbf{b}} \\times \\overrightarrow{\\mathbf{c}}=\\overrightarrow{\\mathbf{c}} \\times \\overrightarrow{\\mathbf{a}}$"}, {"identifier": "C", "content": "$\\overrightarrow{\\mathbf{a}} \\cdot \\overrightarrow{\\mathbf{b}}=\\overrightarrow{\\mathbf{b}} \\cdot \\overrightarrow{\\mathbf{c}}=\\overrightarrow{\\mathbf{c}} \\cdot \\overrightarrow{\\mathbf{a}}=0$"}, {"identifier": "D", "content": "$\\overrightarrow{\\mathbf{a}} \\times \\overrightarrow{\\mathbf{a}}+\\overrightarrow{\\mathbf{a}} \\times \\overrightarrow{\\mathbf{c}}+\\overrightarrow{\\mathbf{c}} \\times \\overrightarrow{\\mathbf{a}}=\\overrightarrow{\\mathbf{0}}$"}] | ["B"] | null | If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are the sides of $\mathbf{a}$ triangle, then <br/><br/>$\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$
<br/><br/>Since,
<br/><br/>$$
\begin{aligned}
\vec{a}+\vec{b}+\vec{c} & =\overrightarrow{0} \\\\
\vec{a}+\vec{b} & =-\vec{c}
\end{aligned}
$$
<br/><br/>$$
\begin{array}{ll}
\Rightarrow & (\vec{a}+\vec{b}) \times \vec{c}=-\vec{c} \times \vec{c} \\\\
\Rightarrow & \vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\overrightarrow{0} \\\\
\Rightarrow & \vec{b} \times \vec{c}=\vec{c} \times \vec{a} \\\\
\text { Similarly, } & \vec{a} \times \vec{b}=\vec{b} \times \vec{c} \\\\
\text { Hence, } & \vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}
\end{array}
$$ | mcq | aieee-2002 |
EmeP0gA9DJPiHDRu | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\left| {\overrightarrow a } \right| = 4,\left| {\overrightarrow b } \right| = 2$$ and the angle between $${\overrightarrow a }$$ and $${\overrightarrow b }$$ is $$\pi /6$$ then $${\left( {\overrightarrow a \times \overrightarrow b } \right)^2}$$ is equal to : | [{"identifier": "A", "content": "$$48$$ "}, {"identifier": "B", "content": "$$16$$"}, {"identifier": "C", "content": "$$\\overrightarrow a $$ "}, {"identifier": "D", "content": "none of these "}] | ["B"] | null | $${\left( {\overrightarrow a \times \overrightarrow b } \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}\,\,{\sin ^2}{\pi \over 6}$$
<br><br>$$ = 16 \times 4 \times {1 \over 4} = 16$$ | mcq | aieee-2002 |
SUoKDdZSEYkrRjwJ | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If the vectors $$\overrightarrow c ,\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$$ and $$\widehat b = \widehat j$$ are such that $$\overrightarrow a ,\overrightarrow c $$ and $$\overrightarrow b $$ form a right handed system then $${\overrightarrow c }$$ is : | [{"identifier": "A", "content": "$$z\\widehat i - x\\widehat k$$ "}, {"identifier": "B", "content": "$$\\overrightarrow 0 $$ "}, {"identifier": "C", "content": "$$y\\widehat j$$ "}, {"identifier": "D", "content": "$$ - z\\widehat i + x\\widehat k$$ "}] | ["A"] | null | Since $$\overrightarrow a ,\overrightarrow c ,\overrightarrow b $$ form a right handed system,
<br><br>$$\therefore$$ $$\overrightarrow c = \overrightarrow b \times \overrightarrow a = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 1 & 0 \cr
x & y & z \cr
} } \right|$$
<br><br>$$ = z\widehat i - x\widehat k$$ | mcq | aieee-2002 |
cgkZdqQPALB8vGYw | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | $$\overrightarrow a = 3\widehat i - 5\widehat j$$ and $$\overrightarrow b = 6\widehat i + 3\widehat j$$ are two vectors and $$\overrightarrow c $$ is a vector such that $$\overrightarrow c = \overrightarrow a \times \overrightarrow b $$ then $$\left| {\overrightarrow a } \right|:\left| {\overrightarrow b } \right|:\left| {\overrightarrow c } \right|$$ = | [{"identifier": "A", "content": "$$\\sqrt {34} :\\sqrt {45} :\\sqrt {39} $$ "}, {"identifier": "B", "content": "$$\\sqrt {34} :\\sqrt {45} :39$$ "}, {"identifier": "C", "content": "$$34:39:45$$ "}, {"identifier": "D", "content": "$$\\,39:35:34$$ "}] | ["B"] | null | <p>To solve this problem, let's take it step by step, beginning with calculating each of the vector magnitudes (or norms) and then finding the magnitude of the cross product vector $\overrightarrow{c}$.</p>
<p>Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are:</p>
<p>$\overrightarrow{a} = 3\widehat{i} - 5\widehat{j}$</p>
<p>$\overrightarrow{b} = 6\widehat{i} + 3\widehat{j}$</p>
<ol>
<li><strong>Magnitude of $\overrightarrow{a}$</strong></li>
</ol>
<p>The magnitude of vector $\overrightarrow{a}$ is calculated using the formula:</p>
<p>$\left| \overrightarrow{a} \right| = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}.$</p>
<ol start="2">
<li><strong>Magnitude of $\overrightarrow{b}$</strong></li>
</ol>
<p>Similarly, the magnitude of vector $\overrightarrow{b}$ is calculated as:</p>
<p>$\left| \overrightarrow{b} \right| = \sqrt{(6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45}.$</p>
<ol start="3">
<li><strong>Calculating $\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b}$</strong></li>
</ol>
<p>The cross product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in 3-dimensional space is given by:</p>
<p>$\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b} = \left| \begin{array}{ccc} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{array} \right|$</p>
<p>For our provided vectors, the $k$ component of both vectors is $0$, thus their cross product will be purely in the $k$ direction. The determinant simplifies to:</p>
<p>$\overrightarrow{c} = (3 \times 3 - (-5) \times 6)\widehat{k} = (9 + 30)\widehat{k} = 39\widehat{k}.$</p>
<ol start="4">
<li><strong>Magnitude of $\overrightarrow{c}$</strong></li>
</ol>
<p>The magnitude of vector $\overrightarrow{c}$ is:</p>
<p>$\left| \overrightarrow{c} \right| = \sqrt{39^2} = 39.$</p>
<p>Finally, the ratio of the magnitudes of vectors $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$ is thus:</p>
<p>$\sqrt{34}:\sqrt{45}:39.$</p>
<p>Therefore, the correct answer is:</p>
<p>Option B) $\sqrt{34}:\sqrt{45}:39.$</p> | mcq | aieee-2002 |
FRP6dd3QwqsKilpn | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a $$ then $$\overrightarrow a + \overrightarrow b + \overrightarrow c = $$ | [{"identifier": "A", "content": "$$abc$$ "}, {"identifier": "B", "content": "$$-1$$"}, {"identifier": "C", "content": "$$0$$"}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Let $$\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow r .$$ Then
<br><br>$$\overrightarrow a \times \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = \overrightarrow a \times \overrightarrow r $$
<br><br>$$ \Rightarrow 0 + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c = \overrightarrow a \times \overrightarrow r $$
<br><br>$$ \Rightarrow \overrightarrow a \times \overrightarrow b - \overrightarrow c \times \overrightarrow a = \overrightarrow a \times \overrightarrow r $$
<br><br>$$ \Rightarrow \overrightarrow a \times \overrightarrow r = \overrightarrow 0 $$
<br><br>Similarly $$\overrightarrow b \times \overrightarrow r = \overrightarrow 0 \,\,\,\& \,\,\,\overrightarrow c \times \overrightarrow r = \overrightarrow 0 $$
<br><br>Above three conditions will be satisfied for non-zero vectors if and only if $$\overrightarrow r = \overrightarrow 0 $$ | mcq | aieee-2003 |
WpRf7gPaDyKnEwUr | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | A tetrahedron has vertices at $$O(0,0,0), A(1,2,1) B(2,1,3)$$ and $$C(-1,1,2).$$ Then the angle between the faces $$OAB$$ and $$ABC$$ will be : | [{"identifier": "A", "content": "$${90^ \\circ }$$ "}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {{{19} \\over {35}}} \\right)$$ "}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{{17} \\over {31}}} \\right)$$"}, {"identifier": "D", "content": "$${30^ \\circ }$$"}] | ["B"] | null | Vector perpendicular to the face $$OAB$$
<br><br>$$ = \overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 1 \cr
2 & 1 & 3 \cr
} } \right| = 5\widehat i - \widehat j - 3\widehat k$$
<br><br>Vector perpendicular to the face $$ABC$$
<br><br>$$ = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 1} & 2 \cr
{ - 2} & { - 1} & 1 \cr
} } \right| = \widehat i - 5\widehat j - 3\widehat k$$
<br><br>Angle between the faces $$=$$ angle between their normals
<br><br>$$\cos \theta = \left| {{{5 + 5 + 9} \over {\sqrt {35} \sqrt {35} }}} \right| = {{19} \over {35}}$$
<br><br>or $$\theta = {\cos ^{ - 1}}\left( {{{19} \over {35}}} \right)$$ | mcq | aieee-2003 |
mKOt6iTT5rj0k5Bl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow u = \widehat i + \widehat j,\,\overrightarrow v = \widehat i - \widehat j$$ and $$\overrightarrow w = \widehat i + 2\widehat j + 3\widehat k\,\,.$$ If $$\widehat n$$ is a unit vector such that $$\overrightarrow u .\widehat n = 0$$ and $$\overrightarrow v .\widehat n = 0\,\,,$$ then $$\left| {\overrightarrow w .\widehat n} \right|$$ is equal to : | [{"identifier": "A", "content": "$$3$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["A"] | null | Since, $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{u}}$ and $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{v}}$
<br/><br/>$$
\begin{aligned}
\Rightarrow \hat{\mathbf{n}} & =\frac{\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \\\\
\therefore|\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}| & =\left|\frac{\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|}\right| \\\\
& =\frac{|\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, } \overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})=\left|\begin{array}{rrr}
1 & 2 & 3 \\
1 & 1 & 0 \\
1 & -1 & 0
\end{array}\right|=-6 \\\\
& \begin{array}{ll}
\Rightarrow |\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|=6 \\\\
\text { and } \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-2 \hat{\mathbf{k}} \\\\
\Rightarrow |\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|=2 \\\\
\therefore |\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}|=\frac{6}{2}=3
\end{array}
\end{aligned}
$$ | mcq | aieee-2003 |
8fOb3BJc4CXuZI26 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be non-zero vectors such that $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a \,\,.$$ If $$\theta $$ is the acute angle between the vectors $${\overrightarrow b }$$ and $${\overrightarrow c },$$ then $$sin\theta $$ equals : | [{"identifier": "A", "content": "$${{2\\sqrt 2 } \\over 3}$$ "}, {"identifier": "B", "content": "$${{\\sqrt 2 } \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$ "}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["A"] | null | Given $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>Clearly $$\overrightarrow a $$ and $$\overrightarrow b $$ are noncollinear
<br><br>$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>$$\therefore$$ $$\overrightarrow a .\overrightarrow c = 0$$
<br><br>and $$ - \overrightarrow b .\overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| \Rightarrow \cos \theta = {{ - 1} \over 3}$$
<br><br>$$\therefore$$ $$\sin \theta = \sqrt {1 - {1 \over 9}} = {{2\sqrt 2 } \over 3}$$
<br><br>$$\,\,\,\,\,\,\,\left[ {} \right.$$ $$\theta $$ is acute angle between $$\overrightarrow b $$ and $$\overrightarrow c $$ $$\left. {} \right]$$ | mcq | aieee-2004 |
bnFroekS8xy90KW5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | For any vector $${\overrightarrow a }$$ , the value of $${\left( {\overrightarrow a \times \widehat i} \right)^2} + {\left( {\overrightarrow a \times \widehat j} \right)^2} + {\left( {\overrightarrow a \times \widehat k} \right)^2}$$ is equal to : | [{"identifier": "A", "content": "$$3{\\overrightarrow a ^2}$$ "}, {"identifier": "B", "content": "$${\\overrightarrow a ^2}$$"}, {"identifier": "C", "content": "$$2{\\overrightarrow a ^2}$$"}, {"identifier": "D", "content": "$$4{\\overrightarrow a ^2}$$"}] | ["C"] | null | Let $$\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $$
<br><br>$$\overrightarrow a \times \overrightarrow i = z\overrightarrow j - y\overrightarrow k $$
<br><br>$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = {y^2} + {z^2}$$
<br><br>Similarly, $${\left( {\overrightarrow a \times \overrightarrow j } \right)^2} = {x^2} + {z^2}\,\,$$
<br><br>and $${\left( {\overrightarrow a \times \overrightarrow k } \right)^2} = {x^2} + {y^2}$$
<br><br>$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow j } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow k } \right)^2}$$
<br><br>$$ = 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2\overrightarrow a 2$$ | mcq | aieee-2005 |
KRG2QzjpRTUqQrKh | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\widehat u$$ and $$\widehat v$$ are unit vectors and $$\theta $$ is the acute angle between them, then $$2\widehat u \times 3\widehat v$$ is a unit vector for : | [{"identifier": "A", "content": "no value of $$\\theta $$ "}, {"identifier": "B", "content": "exactly one value of $$\\theta $$ "}, {"identifier": "C", "content": "exactly two values of $$\\theta $$ "}, {"identifier": "D", "content": "more than two values of $$\\theta $$ "}] | ["B"] | null | Given $$\left| {2\widehat u \times 3\widehat v} \right| = 1$$
<br><br>and $$\theta $$ is acute angle between $$\widehat u$$
<br><br>and $$\widehat v,\,\,\left| {\widehat u} \right| = 1,\,\,\left| {\widehat v} \right| = 1\,\,\,$$
<br><br>$$ \Rightarrow \,\,\,6\left| {\widehat u} \right|\left| {\widehat v} \right|\left| {\sin \theta } \right| = 1$$
<br><br>$$ \Rightarrow 6\left| {\sin \theta } \right| = 1 \Rightarrow \sin \theta = {1 \over 6}$$
<br><br>Hence, there is exactly one value of $$\theta $$
<br><br>for which $$2\widehat u \times 3\widehat v$$ is a unit vector. | mcq | aieee-2007 |
GzoMaNVU4DTaCF86 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | The vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ are not perpendicular and $$\overrightarrow c $$ and $$\overrightarrow d $$ are two vectors satisfying $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $$ and $$\overrightarrow a .\overrightarrow d = 0\,\,.$$ Then the vector $$\overrightarrow d $$ is equal to : | [{"identifier": "A", "content": "$$\\overrightarrow c + \\left( {{{\\overrightarrow a .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow b $$ "}, {"identifier": "B", "content": "$$\\overrightarrow b + \\left( {{{\\overrightarrow b .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow c $$ "}, {"identifier": "C", "content": "$$\\overrightarrow c - \\left( {{{\\overrightarrow a .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow b $$ "}, {"identifier": "D", "content": "$$\\overrightarrow b - \\left( {{{\\overrightarrow b .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow c $$ "}] | ["C"] | null | $$\overrightarrow a .\overrightarrow b \ne 0,\overrightarrow a .\overrightarrow d = 0$$
<br><br>Now, $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $$
<br><br>$$ \Rightarrow \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow d } \right)$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c = \left( {\overrightarrow a .\overrightarrow d } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow d $$
<br><br>$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow d = - \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b + \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c $$
<br><br>$$\overrightarrow d = \overrightarrow c - \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow b $$ | mcq | aieee-2011 |
c5QQOLp2BK9AMeWV | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$.
<br/><br>Let $$\overrightarrow c $$ be a vector such that $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$,
<br/><br>$$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3$$ and the angle between $$\overrightarrow c $$ and $\overrightarrow a \times \overrightarrow b$ is $$30^\circ $$.
<br/><br>Then $$\overrightarrow a .\overrightarrow c $$ is equal to :</br></br></br> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${{25} \\over 8}$$"}] | ["A"] | null | Given:
<br><br>$$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k,\,\,\overrightarrow b = \widehat i + \widehat j$$
<br><br>$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right| = 3$$
<br><br>$$ \therefore $$ $$\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$$
<br><br>$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3$$
<br><br>We have $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin 30^\circ$$
<br><br>$$ \Rightarrow $$ $$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3\left| {\overrightarrow c } \right|.{1 \over 2}$$
<br><br>$$ \Rightarrow $$ $$3 = 3\left| {\overrightarrow c } \right|.{1 \over 2}$$
<br><br>$$ \therefore $$ $$\left| {\overrightarrow c } \right| = 2$$
<br><br>Now $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$
<br><br>On squaring, we get
<br><br>$$ \Rightarrow $$ $${c^2} + {a^2} - 2 - \overrightarrow c .\overrightarrow a = 9$$
<br><br>$$ \Rightarrow $$ $$4 + 9 - 2 - \overrightarrow a .\overrightarrow c = 9$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow a .\overrightarrow c = 2$$ [$$ \because $$ $$\overrightarrow c .\overrightarrow a \,\, = \,\,\overrightarrow a .\overrightarrow c $$] | mcq | jee-main-2017-offline |
Pf8T6ywBp4CA8ZLZEdLkh | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | The area (in sq. units) of the parallelogram whose diagonals are along the vectors $$8\widehat i - 6\widehat j$$ and $$3\widehat i + 4\widehat j - 12\widehat k,$$ is : | [{"identifier": "A", "content": "26"}, {"identifier": "B", "content": "65"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "52"}] | ["B"] | null | When diagonal $${\overrightarrow {{d_1}} }$$ and $${\overrightarrow {{d_2}} }$$ are given of a parallelogram then the area of parallelogram = $${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$
<br><br>Given, $${\overrightarrow {{d_1}} }$$ = 8$$\widehat i$$ $$-$$ 6$$\widehat j$$ + 0$$\widehat k$$
<br><br>and $${\overrightarrow {{d_2}} }$$ = 3$$\widehat i$$ + 4$$\widehat j$$ $$-$$ 12$$\widehat k$$
<br><br>$$\therefore\,\,\,$$ $${\overrightarrow {{d_1}} }$$ $$ \times $$ $${\overrightarrow {{d_2}} }$$ = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
8 & { - 6} & 0 \cr
3 & 4 & { - 12} \cr
} } \right|$$
<br><br>= 72 $$\widehat i$$ $$-$$ ($$-$$ 96) $$\widehat j$$ + 50$$\widehat k$$
<br><br>= 72 $$\widehat i$$ + 96 $$\widehat j$$ + 50 $$\widehat k$$
<br><br>$$\therefore\,\,\,$$ $$\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$ = $$\sqrt {{{72}^2} + {{96}^2} + {{50}^2}} $$
<br><br>= $$\sqrt {16900} $$
<br><br>= 130
<br><br>$$\therefore\,\,\,$$ Area of parallelogram = $${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$ = $${1 \over 2}$$ $$ \times $$ 130 = 65 | mcq | jee-main-2017-online-8th-april-morning-slot |
vNlAtSKfwwTlji9dN29Ja | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If the vector $$\overrightarrow b = 3\widehat j + 4\widehat k$$ is written as the
sum of a vector $$\overrightarrow {{b_1}} ,$$ paralel to $$\overrightarrow a = \widehat i + \widehat j$$ and a vector $$\overrightarrow {{b_2}} ,$$ perpendicular to $$\overrightarrow a ,$$ then $$\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} $$ is equal to : | [{"identifier": "A", "content": "$$ - 3\\widehat i + 3\\widehat j - 9\\widehat k$$"}, {"identifier": "B", "content": "$$6\\widehat i - 6\\widehat j + {9 \\over 2}\\widehat k$$ "}, {"identifier": "C", "content": "$$ - 6\\widehat i + 6\\widehat j - {9 \\over 2}\\widehat k$$"}, {"identifier": "D", "content": "$$3\\widehat i - 3\\widehat j + 9\\widehat k$$ "}] | ["B"] | null | $$\overrightarrow {{b_1}} = {{\left( {\overrightarrow {{b_1}} .\overrightarrow a } \right)\widehat a} \over 1}$$
<br><br>= $$\left\{ {{{\left( {3\widehat j + 4\widehat k} \right).\left( {\widehat i + \widehat j} \right)} \over {\sqrt 2 }}} \right\}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
<br><br>= $${{3\left( {\widehat i + \widehat j} \right)} \over {\sqrt 2 \times \sqrt 2 }} = {{3\left( {\widehat i + \widehat j} \right)} \over 2}$$
<br><br>$$\overrightarrow {{b_1}} + \overrightarrow {{b_2}} = \overrightarrow b $$
<br><br>$$ \Rightarrow $$ $$\overrightarrow {{b_2}} = \overrightarrow b - \overrightarrow {{b_1}} $$
<br><br>= $$\left( {3\widehat j + 4\widehat k} \right) - {3 \over 2}\left( {\widehat i + \widehat j} \right)$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow {{b_2}} $$ = $$ - {3 \over 2}\widehat i + {3 \over 2}\widehat j + 4\widehat k$$
<br><br>& $$\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{{3 \over 2}} & {{3 \over 2}} & 0 \cr
{ - {3 \over 2}} & {{3 \over 2}} & 4 \cr
} } \right|$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \widehat i\left( 6 \right) - \widehat j\left( 6 \right) + \widehat k\left( { - {9 \over 4} + {9 \over 4}} \right)$$
<br><br>$$ \Rightarrow $$ $$6\widehat i - 6\widehat j + {9 \over 2}\widehat k$$ | mcq | jee-main-2017-online-9th-april-morning-slot |
MeeggVCHIAFtgLpDiij1E | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\overrightarrow a ,\,\,\overrightarrow b ,$$ and $$\overrightarrow C $$ are unit vectors such that $$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,$$ then $$\left| {\overrightarrow a \times \overrightarrow c } \right|$$ is equal to : | [{"identifier": "A", "content": "$${{\\sqrt {15} } \\over 4}$$ "}, {"identifier": "B", "content": "$${{1} \\over {4}}$$"}, {"identifier": "C", "content": "$${{15} \\over {16}}$$"}, {"identifier": "D", "content": "$${{\\sqrt {15} } \\over 16}$$"}] | ["A"] | null | Given, <br><br>
$$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 $$<br><br/>
$$ \Rightarrow $$ $$\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b $$<br>
<br>
Squaring both sides,<br><br/>
$${\left| {\overrightarrow a } \right|^2} + 4\overrightarrow a .\overrightarrow c + 4{\left| {\overrightarrow c } \right|^2} = 4{\left| {\overrightarrow b } \right|^2}$$<br><br/>
$$ \Rightarrow $$ 1 + $$4\overrightarrow a .\overrightarrow c $$ + 4 = 4 [as $$\left| {\overrightarrow a } \right|^2$$ = $$\left| {\overrightarrow b } \right|^2$$ = $$\left| {\overrightarrow c } \right|^2$$ = 1]<br><br/>
$$ \Rightarrow $$ $$\overrightarrow a .\overrightarrow c = - {1 \over 4}$$<br><br/>
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \theta = - {1 \over 4}$$<br><br/>
$$ \therefore $$ $$\cos \theta $$ = $$ - {1 \over 4}$$<br><br/>
$$ \therefore $$ $$\sin ^2 \theta $$ = 1 - $$\cos ^2 \theta $$<br><br/>
= 1 - $$1 \over 16$$<br><br/>
= $$15\over 16$$<br><br/>
$$ \therefore $$ sin$$\theta $$ = $${{\sqrt {15} } \over 4}$$<br><br/>
$$ \therefore $$ $$\left| {\overrightarrow a \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta $$<br><br/>
= 1 . 1 . $${{\sqrt {15} } \over 4}$$<br><br/>
= $${{\sqrt {15} } \over 4}$$
| mcq | jee-main-2018-online-15th-april-morning-slot |
t3jFwvsaHi92dFMbV9vrP | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$$ and a vector $$\overrightarrow b $$ be such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow a .\overrightarrow b = 3.$$ Then $$\left| {\overrightarrow b } \right|$$ equals : | [{"identifier": "A", "content": "$${{11} \\over 3}$$"}, {"identifier": "B", "content": "$${{11} \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$$\\sqrt {{{11} \\over 3}} $$"}, {"identifier": "D", "content": "$${{\\sqrt {11} } \\over 3}$$"}] | ["C"] | null | $$ \because $$ $$\overrightarrow a $$ $$=$$ $$\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3 $$
<br><br>& $$\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2 $$
<br><br>Now, $$\overrightarrow a $$ $$ \times $$ $$\overrightarrow b $$ = $$\overrightarrow c $$ (Given)
<br><br>$$ \Rightarrow $$ $$\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta = \left| {\overrightarrow c } \right|$$
<br><br>$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \sqrt 2 $$
<br><br>also $$\overrightarrow a .\overrightarrow b = 3$$
<br><br>$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3$$
<br><br>Dividing [i] by [iii], we get
<br><br>tan$$\theta $$ = $${{\sqrt 2 } \over 3}$$
<br><br>$$ \therefore $$ sin$$\theta $$ $$=$$ $${{\sqrt 2 } \over {\sqrt {11} }}$$
<br><br>Substituting value of sin$$\theta $$ in [i] we get
<br><br>$$\sqrt 3 \left| {\overrightarrow b } \right|{{\sqrt 2 } \over {\sqrt {11} }} = \sqrt 2 $$
<br><br>$$\left| {\overrightarrow b } \right| = {{\sqrt {11} } \over {\sqrt 3 }}$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
AsjpuvwmmzcqAZgjE5Xn1 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\mathop a\limits^ \to = 3\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + x\mathop k\limits^ \wedge $$ and $$\mathop b\limits^ \to = \mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $$
, for some real x. Then $$\left| {\mathop a\limits^ \to \times \mathop b\limits^ \to } \right|$$ = r
is possible if : | [{"identifier": "A", "content": "0 < r < $$\\sqrt {{3 \\over 2}} $$"}, {"identifier": "B", "content": "$$3\\sqrt {{3 \\over 2}} < r < 5\\sqrt {{3 \\over 2}} $$"}, {"identifier": "C", "content": "$$ r \\ge 5\\sqrt {{3 \\over 2}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{3 \\over 2}} < r \\le 3\\sqrt {{3 \\over 2}} $$"}] | ["C"] | null | $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 2 & x \cr
1 & { - 1} & 1 \cr
} } \right|$$
<br><br>= (2 + x)$${\widehat i}$$ + (3 - x)$${\widehat j}$$ - 5$${\widehat k}$$
<br><br>$$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = r
<br><br>= $$\sqrt {{{\left( {2 + x} \right)}^2} + {{\left( {x - 3} \right)}^2} + {{\left( { - 5} \right)}^2}} $$
<br><br>$$ \Rightarrow $$ r = $$\sqrt {4 + 4x + {x^2} + {x^2} + 9 - 6x + 25} $$
<br><br>= $$\sqrt {2{x^2} - 2x + 38} $$
<br><br>= $$\sqrt {2\left( {{x^2} - x + {1 \over 4}} \right) + 38 - {1 \over 2}} $$
<br><br>= $$\sqrt {2{{\left( {x - {1 \over 2}} \right)}^2} + {{75} \over 2}} $$
<br><br>$$ \Rightarrow $$ r $$ \ge $$ $$\sqrt {{{75} \over 2}} $$
<br><br>$$ \Rightarrow $$ $$ r \ge 5\sqrt {{3 \over 2}} $$ | mcq | jee-main-2019-online-8th-april-evening-slot |
kz4wkHygFPPLV412753rsa0w2w9jx65kz0q | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k$$ be two vectors. If a vector perpendicular to both the vectors
$$\overrightarrow a + \overrightarrow b $$ and $$\overrightarrow a - \overrightarrow b $$ has the magnitude 12 then one such vector is :
| [{"identifier": "A", "content": "$$4\\left( {2\\widehat i - 2\\widehat j - \\widehat k} \\right)$$"}, {"identifier": "B", "content": "$$4\\left( { - 2\\widehat i - 2\\widehat j + \\widehat k} \\right)$$"}, {"identifier": "C", "content": "$$4\\left( {2\\widehat i + 2\\widehat j + \\widehat k} \\right)$$"}, {"identifier": "D", "content": "$$4\\left( {2\\widehat i + 2\\widehat j - \\widehat k} \\right)$$"}] | ["A"] | null | Required vector is $\overrightarrow r$ = $$\lambda \left( {\left( {\overline a + \overline b } \right) \times \left( {\overline a - \overline b } \right)} \right)$$<br><br>
$$ \Rightarrow \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
4 & 4 & 0 \cr
2 & 0 & 4 \cr
} } \right| = \lambda \left( {16\widehat i - 16\widehat j - 8\widehat k} \right)$$<br><br>
$$ \Rightarrow \overrightarrow r = 8\lambda \left( {2\widehat i - 2\widehat j - \widehat k} \right) = \left| {\overrightarrow r } \right|$$<br><br>
$$ \Rightarrow \left| {8\lambda } \right|.3 \Rightarrow 8\lambda = \pm 4$$<br><br>
$$ \Rightarrow \overrightarrow r = \pm 4\left( {2\widehat i - 2\widehat j - \widehat k} \right)$$
| mcq | jee-main-2019-online-12th-april-morning-slot |
bOQOFNGP3tT7x6szeD3rsa0w2w9jx2gbzy3 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | The distance of the point having position vector $$ - \widehat i + 2\widehat j + 6\widehat k$$
from the straight line passing through the point
(2, 3, – 4) and parallel to the vector, $$6\widehat i + 3\widehat j - 4\widehat k$$ is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "$$2\\sqrt {13} $$"}, {"identifier": "D", "content": "$$4\\sqrt 3 $$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264693/exam_images/n0zytcbrpcd3kgpqtlsb.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263784/exam_images/axuiomv56i9jmm3qd8t0.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264289/exam_images/rwmz7tdkddlyrjqm0lzn.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264220/exam_images/v4qgp1kbwwxsznkf9iow.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266997/exam_images/abp8adaeqlzspavd0ck4.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265414/exam_images/voujeb24cnhzz9uutjy9.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266490/exam_images/uthfwx3rpj07ivgncl1h.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265482/exam_images/mcpi8ysmwwzqgrff8js2.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Evening Slot Mathematics - Vector Algebra Question 170 English Explanation"></picture>
<br>
$$AD = \left| {{{\overrightarrow {AP} .\overrightarrow n } \over {\left| {\overrightarrow n } \right|}}} \right| = \sqrt {61} $$<br><br>
$$ \Rightarrow PD = \sqrt {A{P^2} - A{D^2}} $$<br><br>
$$ = \sqrt {110 - 61} $$<br><br>
= 7 | mcq | jee-main-2019-online-10th-april-evening-slot |
3h3PJW4wWUVQxcoR3jRCO | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow \alpha = 3\widehat i + \widehat j$$ and $$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$
. If $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$,
where $${\overrightarrow \beta _1}$$
is parallel to $$\overrightarrow \alpha $$ and $$\overrightarrow {{\beta _2}} $$
is perpendicular
to $$\overrightarrow \alpha $$ , then $${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} $$
is equal to | [{"identifier": "A", "content": "$$ 3\\widehat i - 9\\widehat j - 5\\widehat k$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$($$ - 3\\widehat i + 9\\widehat j + 5\\widehat k$$)"}, {"identifier": "C", "content": "$$ - 3\\widehat i + 9\\widehat j + 5\\widehat k$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$($$ 3\\widehat i - 9\\widehat j + 5\\widehat k$$)"}] | ["B"] | null | Given $$\overrightarrow \alpha = 3\widehat i + \widehat j$$<br><br>$$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$
<br><br>$${\overrightarrow \beta _1}$$
is parallel to $$\overrightarrow \alpha $$
<br><br>$$ \therefore $$ $${\overrightarrow \beta _1}$$ = $$\lambda $$ $$\overrightarrow \alpha$$
<br><br>$$ \Rightarrow $$ $${\overrightarrow \beta _1}$$ = $$3\lambda \widehat i + \lambda \widehat j$$
<br><br>$$\overrightarrow {{\beta _2}} $$
is perpendicular
to $$\overrightarrow \alpha $$
<br><br>$$\overrightarrow {{\beta _2}} $$ . $$\overrightarrow \alpha$$ = 0
<br><br>Let $$\overrightarrow {{\beta _2}} $$ = $$x\widehat i + y\widehat j + z\widehat k$$
<br><br>$$ \therefore $$ ($$x\widehat i + y\widehat j + z\widehat k$$).($$3\widehat i + \widehat j$$) = 0
<br><br>$$ \Rightarrow $$ 3x + y = 0
<br><br>$$ \Rightarrow $$ y = -3x
<br><br>$$ \therefore $$ $$\overrightarrow {{\beta _2}} $$ = $$x\widehat i -3x\widehat j + z\widehat k$$
<br><br>Given $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$
<br><br>$$ \Rightarrow $$ $$(2\widehat i - \widehat j + 3 \widehat k$$) = ($$3\lambda \widehat i + \lambda \widehat j$$) - ($$x\widehat i -3x\widehat j + z\widehat k$$)
<br><br>= $$\left( {3\lambda - x} \right)\widehat i + \left( {\lambda + 3x} \right)\widehat j - z\widehat k$$
<br><br>$$ \therefore $$ 3$$\lambda $$ - x = 2 ...(1)
<br><br>$$\lambda $$ + 3x = -1.....(2)
<br><br>z = -3
<br><br>Solving (1) and (2), we get
<br><br>$$\lambda $$ = $${1 \over 2}$$ and x = $$-{1 \over 2}$$
<br><br>$$ \therefore $$ $${\overrightarrow \beta _1}$$ = $${3 \over 2}\widehat i + {1 \over 2}\widehat j$$
<br><br>and $$\overrightarrow {{\beta _2}} $$ = $$ - {1 \over 2}\widehat i + {3 \over 2}\widehat j - 3\widehat k$$
<br><br>$${\overrightarrow \beta _1} \times {\overrightarrow \beta _2} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{{3 \over 2}} & {{1 \over 2}} & 0 \cr
{ - {1 \over 2}} & {{3 \over 2}} & { - 3} \cr
} } \right|$$
<br><br>= $${1 \over 2}$$($$ - 3\widehat i + 9\widehat j + 5\widehat k$$) | mcq | jee-main-2019-online-9th-april-morning-slot |
Cc04Cv6xi4G4IkiaNfkHp | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$$ $$\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$$ and $$\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$$ be coplanar vectors. Then the non-zero vector $$\overrightarrow a \times \overrightarrow c $$ is : | [{"identifier": "A", "content": "$$ - 10\\widehat i - 5\\widehat j$$"}, {"identifier": "B", "content": "$$ - 10\\widehat i + 5\\widehat j$$"}, {"identifier": "C", "content": "$$ - 14\\widehat i + 5\\widehat j$$"}, {"identifier": "D", "content": "$$ - 14\\widehat i - 5\\widehat j$$"}] | ["B"] | null | $$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
1 & 2 & 4 \cr
1 & \lambda & 4 \cr
2 & 4 & {{\lambda ^2} - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$$
<br><br>$$ \Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$$
<br><br>$$ \Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$$
<br><br>$$ \Rightarrow \lambda = 2,3, - 3$$
<br><br>So, $$\lambda $$ = 2 (as $$\overrightarrow a $$ is parallel to $$\overrightarrow c $$ for $$\lambda $$ = $$ \pm $$3)
<br><br>Hence $$\overrightarrow a \times \overrightarrow c = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 4 \cr
2 & 4 & 3 \cr
} } \right|$$
<br><br>$$ = - 10\widehat i + 5\widehat j$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
HFYRDICglIKazRYErj7k9k2k5fl9rbl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$
, $$\overrightarrow b $$
and $$\overrightarrow c $$
be three unit vectors such that
<br/>$$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$. If $$\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a $$ and
<br/>$$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $$, then the ordered pair, $$\left( {\lambda ,\overrightarrow d } \right)$$ is equal to : | [{"identifier": "A", "content": "$$\\left( {{3 \\over 2},3\\overrightarrow a \\times \\overrightarrow c } \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {3 \\over 2},3\\overrightarrow c \\times \\overrightarrow b } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {3 \\over 2},3\\overrightarrow a \\times \\overrightarrow b } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{3 \\over 2},3\\overrightarrow b \\times \\overrightarrow c } \\right)$$"}] | ["C"] | null | $$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$
<br><br>$$ \Rightarrow $$ $${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$$ = 0
<br><br>$$ \Rightarrow $$ $${{{\left| {\overrightarrow a } \right|}^2}}$$ + $${{{\left| {\overrightarrow b } \right|}^2}}$$ + $${{{\left| {\overrightarrow c } \right|}^2}}$$ + <br>$${2\left( {\overrightarrow a .\overrightarrow b } \right)}$$ + $${2\left( {\overrightarrow b .\overrightarrow c } \right)}$$ + $${2\left( {\overrightarrow c .\overrightarrow {a} } \right)}$$ = 0
<br><br>$$ \Rightarrow $$ 3 + $${2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)}$$ = 0
<br><br>$$ \Rightarrow $$ 3 + 2$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $$ - {3 \over 2}$$
<br><br>$$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $$
<br><br>= $${\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \left( { - \overrightarrow a - \overrightarrow b } \right) + \left( { - \overrightarrow a - \overrightarrow b } \right) \times \overrightarrow a }$$
<br><br>= $${\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a + }$$ $${\overrightarrow b \times \overrightarrow b - \overrightarrow a \times \overrightarrow a - \overrightarrow b \times \overrightarrow a }$$
<br><br>= $${\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b }$$
<br><br>= $${3\left( {\overrightarrow a \times \overrightarrow b } \right)}$$ | mcq | jee-main-2020-online-7th-january-evening-slot |
CB4aJTXbdSnUgLtX7X7k9k2k5hiep0f | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$$ and $$\overrightarrow b = \widehat i - \widehat j + \widehat k$$ be two
vectors. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$ and $$\overrightarrow c .\overrightarrow a = 0$$, then $$\overrightarrow c .\overrightarrow b $$ is equal to | [{"identifier": "A", "content": "$$ - {1 \\over 2}$$"}, {"identifier": "B", "content": "$$ - {3 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "-1"}] | ["A"] | null | $$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$$
<br><br>$$\overrightarrow b = \widehat i - \widehat j + \widehat k$$
<br><br>$$\left| {\overrightarrow a } \right|$$ = $$\sqrt 6 $$, $$\left| {\overrightarrow b } \right|$$ = $$\sqrt 3 $$
<br><br>and $${\overrightarrow a .\overrightarrow b }$$ = 4
<br><br>Given $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$
<br><br>$$ \Rightarrow $$ $${\left( {\overrightarrow b \times \overrightarrow c } \right)}$$ - $${\left( {\overrightarrow b \times \overrightarrow a } \right)}$$ = 0
<br><br>$$ \Rightarrow $$ $${\overrightarrow b \times \left( {\overrightarrow c - \overrightarrow a } \right)}$$ = 0
<br><br>$$ \therefore $$ $${\overrightarrow b \parallel \left( {\overrightarrow c - \overrightarrow a } \right)}$$
<br><br>$$ \Rightarrow $$ $${\left( {\overrightarrow c - \overrightarrow a } \right) = \lambda \overrightarrow b }$$
<br><br>$$ \Rightarrow $$ $${\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b }$$
<br><br>$$ \Rightarrow $$ $${\overrightarrow c .\overrightarrow a = \overrightarrow a .\overrightarrow a + \lambda \overrightarrow a .\overrightarrow b }$$
<br><br>$$ \Rightarrow $$ 0 = $${{{\left| {\overrightarrow a } \right|}^2} + \lambda \left( {\overrightarrow a .\overrightarrow b } \right)}$$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${{{ - {{\left| {\overrightarrow a } \right|}^2}} \over {\overrightarrow a .\overrightarrow b }}}$$ = $${{ - 6} \over 4} = - {3 \over 2}$$
<br><br>$$ \therefore $$ $$\overrightarrow c $$ = $${\overrightarrow a - {3 \over 2}\overrightarrow b }$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow c $$ = ($$\widehat i - 2\widehat j + \widehat k$$) - $${3 \over 2}$$($$\widehat i - \widehat j + \widehat k$$)
<br><br>$$ \Rightarrow $$ $$\overrightarrow c $$ = $$ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$$
<br><br>$$ \therefore $$ $$\overrightarrow c .\overrightarrow b $$ = $$ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$$($$\widehat i - \widehat j + \widehat k$$)
<br><br>$$ \therefore $$ $$\overrightarrow c .\overrightarrow b $$ = $$ - {1 \over 2}$$ | mcq | jee-main-2020-online-8th-january-evening-slot |
uxkbzvwhQhZu358eew7k9k2k5ki4pum | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three vectors such that $$\left| {\overrightarrow a } \right| = \sqrt 3 $$,
$$\left| {\overrightarrow b } \right| = 5,\overrightarrow b .\overrightarrow c = 10$$ and the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$
is $${\pi \over 3}$$. If $${\overrightarrow a }$$ is perpendicular to the vector $$\overrightarrow b \times \overrightarrow c $$ , then $$\left| {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)} \right|$$ is equal to _____. | [] | null | 30 | Given $$\left| {\overrightarrow a } \right| = \sqrt 3 $$,
$$\left| {\overrightarrow b } \right| = 5$$
<br><br>Given $$\overrightarrow b .\overrightarrow c = 10$$
<br><br>And the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$
is $${\pi \over 3}$$
<br><br>$$ \therefore $$ $$bc\cos {\pi \over 3}$$ = 10
<br><br>$$ \Rightarrow $$ c = 4
<br><br>$${\overrightarrow a }$$ is perpendicular to the vector $$\overrightarrow b \times \overrightarrow c $$
<br><br>$$ \therefore $$ $$\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$$ = 0 and angle between them is $${\pi \over 2}$$
<br><br>Now $$\left| {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)} \right|$$
<br><br>= $$\left| {\overrightarrow a } \right|\left| {\overrightarrow b \times \overrightarrow c } \right|\sin {\pi \over 2}$$
<br><br>= $$\left| {\overrightarrow a } \right|$$.$${\left| {\overrightarrow b } \right|.\left| {\overrightarrow c } \right|}$$$$\sin {\pi \over 3}$$.1
<br><br>= $$\sqrt 3 \times 5 \times 4 \times {{\sqrt 3 } \over 2}$$
<br><br>= 30 | integer | jee-main-2020-online-9th-january-evening-slot |
I6tuy9NF399qWeyYVhjgy2xukezfmp9m | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let the position vectors of points 'A' and 'B' be
<br/>$$\widehat i + \widehat j + \widehat k$$ and $$2\widehat i + \widehat j + 3\widehat k$$, respectively. A point
'P' divides the line segment AB internally in the
ratio
$$\lambda $$ : 1 (
$$\lambda $$ > 0). If O is the origin and
<br/>$$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$$, then
$$\lambda $$ is equal
to______.
| [] | null | 0.8 | Let, $$\overrightarrow a $$ = $$\widehat i + \widehat j + \widehat k$$
<br><br>and $$\overrightarrow b $$ = $$2\widehat i + \widehat j + 3\widehat k$$
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265164/exam_images/hoezbwyywsjaggdoi7yh.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263940/exam_images/k8d9rkwhhmevap3yejqs.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263783/exam_images/sxqwwrfimbrsv1mu3ogd.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265171/exam_images/opa8dsj1xubasfvhffg0.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265154/exam_images/rqbu22xlaezgep227k51.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267454/exam_images/x1cszvcuyccpsznwi8cm.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264768/exam_images/u0s88dbojoevcxa28ijb.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267112/exam_images/jvoksih0q1foxtdpr9uw.webp"><source media="(max-width: 1760px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264476/exam_images/krelhhyhhnsrklvk49bw.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267524/exam_images/mzqj6yentbg3ruppcugm.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Evening Slot Mathematics - Vector Algebra Question 159 English Explanation"></picture>
<br>$$\overrightarrow {OB} = \overrightarrow b $$
<br><br>$$\overrightarrow {OP} = {{\overrightarrow a + \lambda \overrightarrow b } \over {1 + \lambda }}$$
<br><br>$$\overrightarrow {OA} = \overrightarrow a $$
<br><br>$$ \therefore $$ $$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow b .{{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {\overrightarrow a \times {{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$$ = 6
<br><br>$$ \Rightarrow $$ $${{\overrightarrow b .\overrightarrow a + \lambda \left( {\overrightarrow b .\overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {{{\overrightarrow a \times \overrightarrow a + \lambda \left( {\overrightarrow a \times \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$$ = 6
<br><br>[ $${\overrightarrow b .\overrightarrow a }$$ = ($$2\widehat i + \widehat j + 3\widehat k$$).($$\widehat i + \widehat j + \widehat k$$)
<br><br> = 2 + 1 + 3 = 6
<br><br>$${\overrightarrow b .\overrightarrow b }$$ = ($$2\widehat i + \widehat j + 3\widehat k$$).($$2\widehat i + \widehat j + 3\widehat k$$)
<br><br> = 4 + 1 + 9 = 14
<br><br>$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 1 & 1 \cr
2 & 1 & 3 \cr
} } \right|$$
<br><br>= (3 - 1)$${\widehat i}$$ - (3 - 2)$${\widehat j}$$ + (1 - 2)$${\widehat k}$$
<br><br>= 2$${\widehat i}$$ - $${\widehat j}$$ - $${\widehat k}$$
<br><br>$$ \therefore $$ $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = $$\sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} $$
= $$\sqrt 6 $$]
<br><br>$$ \Rightarrow $$ $${{6 + 14\lambda } \over {1 + \lambda }} - 3{\left| {{{\lambda \left( {\sqrt 6 } \right)} \over {1 + \lambda }}} \right|^2}$$ = 6
<br><br>$$ \Rightarrow $$ $${{6 + 14\lambda } \over {1 + \lambda }} - {{3{\lambda ^2} \times 6} \over {{{\left( {1 + \lambda } \right)}^2}}}$$ = 6
<br><br>$$ \Rightarrow $$ (14$$\lambda $$ + 6)($$\lambda $$ + 1)—18$$\lambda $$<sup>2</sup> = 6($$\lambda $$ + 1)<sup>2</sup>
<br><br>$$ \Rightarrow $$ —4$$\lambda $$<sup>2</sup>
+ 20$$\lambda $$ + 6 = 6$$\lambda $$<sup>2</sup>
+ 12$$\lambda $$ + 6
<br><br>$$ \Rightarrow $$ 10$$\lambda $$<sup>2</sup> — 8$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$(10$$\lambda $$ — 8) = 0
<br><br>As given $$\lambda $$ > 0
<br><br>$$ \therefore $$ $$\lambda $$ = $${8 \over {10}}$$ = 0.8
| integer | jee-main-2020-online-2nd-september-evening-slot |
XDKiZVSmcBrvn7Fb1njgy2xukfal01ga | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$$, then the value of<br/><br/>
$${\left| {\widehat i \times \left( {\overrightarrow a \times \widehat i} \right)} \right|^2} + {\left| {\widehat j \times \left( {\overrightarrow a \times \widehat j} \right)} \right|^2} + {\left| {\widehat k \times \left( {\overrightarrow a \times \widehat k} \right)} \right|^2}$$ is equal to____
| [] | null | 18 | Let $$\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$$<br><br>Now $$\widehat i \times \left( {\overrightarrow a \times \widehat i} \right) = \left( {\widehat i.\widehat i} \right)\overrightarrow a - \left( {\widehat i.\overrightarrow a } \right)\widehat i$$<br><br>= $$y\widehat j + z\widehat k$$<br><br>Similarly $$\widehat j \times \left( {\overrightarrow a \times \widehat j} \right) = x\widehat i + z\widehat k$$<br><br>$$\widehat k \times \left( {\overrightarrow a \times \widehat k} \right) = x\widehat i + y\widehat j$$<br><br>Now $${\left| {y\widehat j + z\widehat k} \right|^2} + {\left| {x\widehat i + z\widehat k} \right|^2} + {\left| {x\widehat i + y\widehat j} \right|^2}$$<br><br>= $$2({x^2} + {y^2} + {z^2}) $$
<br><br>Given $$\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$$
<br><br>$$ \therefore $$ x = 2, y = 1, z = 2
<br><br>= 2(4 + 1 + 4) = 18 | integer | jee-main-2020-online-4th-september-evening-slot |
RkTo4epEnfRC9GpYlw1kls5v75w | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + 2\widehat j - \widehat k$$, $$\overrightarrow b = \widehat i - \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j - \widehat k$$ be three given vectors. If $$\overrightarrow r $$ is a vector such that $$\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a $$ and $$\overrightarrow r .\,\overrightarrow b = 0$$, then $$\overrightarrow r .\,\overrightarrow a $$ is equal to __________. | [] | null | 12 | <p>Given, $$\overrightarrow a = \widehat i + 2\widehat j - \widehat k$$,</p>
<p>$$\overrightarrow b = \widehat i - \widehat j$$,</p>
<p>$$\overrightarrow c = \widehat i - \widehat j - \widehat k$$</p>
<p>$$\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a $$</p>
<p>$$ \Rightarrow \overrightarrow r \times \overrightarrow a - \overrightarrow c \times \overrightarrow a = 0$$</p>
<p>$$ \Rightarrow (\overrightarrow r - \overrightarrow c ) \times \overrightarrow a = 0$$</p>
<p>$$\therefore$$ $$\overrightarrow r - \overrightarrow c = \lambda \overrightarrow a $$</p>
<p>$$ \Rightarrow \overrightarrow r = \lambda \overrightarrow a + \overrightarrow c $$</p>
<p>$$ \Rightarrow \overrightarrow r \,.\,\overrightarrow b = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b $$ (taking dot with $$\overrightarrow b $$)</p>
<p>$$ \Rightarrow 0 = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b $$ [$$\because$$ $$\overrightarrow r \,.\,\overrightarrow b = 0$$]</p>
<p>$$ \Rightarrow \lambda (\widehat i + 2\widehat j - \widehat k)\,.\,(\widehat i - \widehat j) + (\widehat i - \widehat j - \widehat k)\,.\,(\widehat i - \widehat j) = 0$$</p>
<p>$$ \Rightarrow \lambda (1 - 2) + 2 = 0$$</p>
<p>$$ \Rightarrow \lambda = 2$$</p>
<p>$$\therefore$$ $$\overrightarrow r = 2\overrightarrow a + \overrightarrow c $$</p>
<p>$$ \Rightarrow \overrightarrow r \,.\,\overrightarrow a = 2\overrightarrow a \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a $$ [taking dot with $${\overrightarrow a }$$]</p>
<p>$$ = 2{\left| {\overrightarrow a } \right|^2} + \overrightarrow a \,.\,\overrightarrow c $$</p>
<p>$$ = 2(1 + 4 + 1) + (1 - 2 + 1)$$</p>
<p>$$ \Rightarrow \overrightarrow r \,.\,\overrightarrow a = 12$$</p> | integer | jee-main-2021-online-25th-february-morning-slot |
s7RlE6savu4SIIyjxr1klta1k13 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$ and $$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$8\sqrt 3 $$ square units, then $$\overrightarrow a $$ . $$\overrightarrow b $$ is equal to __________. | [] | null | 2 | $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$<br><br>$$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$<br><br>Area of parallelogram = $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$<br><br>$$ = \left| {(\widehat i + \alpha \widehat j + 3\widehat k) \times (3\widehat i - \alpha \widehat j + \widehat k)} \right|$$<br><br>$$8\sqrt 3 = \left| {(4\alpha )\widehat i + 8\widehat j - (4\alpha )\widehat k} \right|$$<br><br>$$(64)(3) = 16{\alpha ^2} + 64 + 16{\alpha ^2}$$<br><br>$$(64)(3) = 32{\alpha ^2} + 64$$<br><br>$$6 = {\alpha ^2} + 2$$<br><br>$${\alpha ^2} = 4$$<br><br>$$ \therefore $$ $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$<br><br>$$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$<br><br>$$\overrightarrow a \,.\,\overrightarrow b = 3 - {\alpha ^2} + 3$$<br><br>$$ = 6 - {\alpha ^2}$$<br><br>$$ = 6 - 4$$<br><br>$$ = 2$$ | integer | jee-main-2021-online-25th-february-evening-slot |
hkfGcy1FV2nIdZCA6X1kmiw7pzg | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$, <br/><br/>$$\overrightarrow r $$ . $$\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $$-$$1, $$\alpha$$ $$\in$$ R, then the <br/><br/>value of $$\alpha$$ + $${\left| {\overrightarrow r } \right|^2}$$ is equal to : | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "15"}] | ["D"] | null | Given $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$
<br><br>$$ \Rightarrow $$ $$\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $$<br><br>$$\overrightarrow r \times (\overrightarrow a + \overrightarrow b ) = 0$$<br><br>$$\overrightarrow r ||(\overrightarrow a + \overrightarrow b )$$<br><br>$$\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b )$$<br><br>$$(\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k)$$<br><br>$$ \because $$ $$\overrightarrow r \,.\,(2\widehat i + 5\widehat j - \alpha \widehat k) = - 1$$<br><br>$$\lambda \left[ {3\widehat i - \widehat j + 2\widehat k} \right]\,.\,\left[ {2\widehat i + 5\widehat j - \alpha \widehat k} \right] = - 1$$<br><br>$$ \Rightarrow \lambda (6 - 5 - 2\alpha ) = - 1$$<br><br>$$\lambda (1 - 2\alpha ) = - 1$$ .... (1)<br><br>$$\overrightarrow r \,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$$<br><br>$$\lambda (3\widehat i - \widehat j + 2\widehat k)\,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$$<br><br>$$ \Rightarrow \lambda [3\alpha - 2 + 2] = 3 \Rightarrow \lambda \alpha = 1$$ .... (2)<br><br>From (1) & (2)<br><br>$$\lambda \left[ {1 - {2 \over \lambda }} \right] = - 1$$<br><br>$$\lambda - 2 = - 1 \Rightarrow \lambda = 1\,\alpha = 1$$<br><br>$$\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k$$<br><br>$$\alpha + |\overrightarrow r {|^2} = 1 + 14 = 15$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
Y2uW5HKoyGplfqL1nL1kmm2snqv | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two non-zero vectors perpendicular to each other and $$|\overrightarrow a | = |\overrightarrow b |$$. If $$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$$, then the angle between the vectors $$\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)$$ and $${\overrightarrow a }$$ is equal to : | [{"identifier": "A", "content": "$${\\sin ^{ - 1}}\\left( {{1 \\over {\\sqrt 6 }}} \\right)$$"}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 2 }}} \\right)$$"}, {"identifier": "C", "content": "$${\\sin ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$"}, {"identifier": "D", "content": "$${\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$"}] | ["D"] | null | $$\overrightarrow a $$ is perpendicular to $$\overrightarrow b $$<br><br>$$ \therefore $$ $$\overrightarrow a $$ . $$\overrightarrow b $$ = 0<br><br>Given, | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ |<br><br>and | $$\overrightarrow a $$ | = | $$\overrightarrow b $$ |<br><br>$$ \therefore $$ | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ | = | $$\overrightarrow b $$ | = k(assume)<br><br>Now, angle between $$\overrightarrow a $$ and ($$\overrightarrow a $$ + $$\overrightarrow b $$ + ($$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$))<br><br>$$\cos \theta = {{\overrightarrow a ((\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ))} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$<br><br>$$ = {{|\overrightarrow a {|^2} + \,\overrightarrow a .\,\overrightarrow b + \overrightarrow a (\overrightarrow a \times \overrightarrow b )} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$<br><br>[Note $$\overrightarrow a .(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow a \overrightarrow a \overrightarrow b ] = 0$$]<br><br>$$ = {{|\overrightarrow a {|^2} + 0 + 0} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$<br><br>Now, $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2}$$<br><br>$$ = |\overrightarrow a {|^2} + |\overrightarrow a {|^2} + |\overrightarrow a \times \overrightarrow b {|^2} + $$<br><br>$$2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .(\overrightarrow a \times \overrightarrow b ) + 2\overrightarrow a .(\overrightarrow a \times \overrightarrow b )$$<br><br>$$ = {k^2} + {k^2} + {k^2}$$<br><br>$$ \therefore $$ $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2} = 3{k^2}$$<br><br>$$ \Rightarrow |\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )| = \sqrt 3 k$$<br><br>$$ \therefore $$ $$\cos \theta = {{{k^2}} \over {k(\sqrt 3 k)}} = {1 \over {\sqrt 3 }}$$<br><br>$$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1krq0xpcz | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If the shortest distance between the lines $$\overrightarrow {{r_1}} = \alpha \widehat i + 2\widehat j + 2\widehat k + \lambda (\widehat i - 2\widehat j + 2\widehat k)$$, $$\lambda$$ $$\in$$ R, $$\alpha$$ > 0 and $$\overrightarrow {{r_2}} = - 4\widehat i - \widehat k + \mu (3\widehat i - 2\widehat j - 2\widehat k)$$, $$\mu$$ $$\in$$ R is 9, then $$\alpha$$ is equal to ____________. | [] | null | 6 | If $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $$ and $$\overrightarrow r = \overrightarrow c + \lambda \overrightarrow d $$ then shortest distance between two lines is <br><br>$$L = {{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )} \over {|b \times d|}}$$<br><br>$$\therefore$$ $$\overrightarrow a - \overrightarrow c = ((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k)$$<br><br>$${{\overrightarrow b \times \overrightarrow d } \over {|b \times d|}} = {{(2\widehat i + 2\widehat j + \widehat k)} \over 3}$$<br><br>$$\therefore$$ $$((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k).{{(2\widehat i + 2\widehat j + \widehat k)} \over 3} = 9$$<br><br>or $$\alpha$$ = 6 | integer | jee-main-2021-online-20th-july-morning-shift |
1krw2kdmg | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$$ be two vectors. If a vector $$\overrightarrow r = (\alpha \widehat i + \beta \widehat j + \gamma \widehat k)$$ is perpendicular to each of the vectors ($$(\overrightarrow p + \overrightarrow q )$$ and $$(\overrightarrow p - \overrightarrow q )$$, and $$\left| {\overrightarrow r } \right| = \sqrt 3 $$, then $$\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$$ is equal to _______________. | [] | null | 3 | $$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$$ (Given )<br><br>$$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$$<br><br>Now, $$(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q ) = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 5 & 2 \cr
1 & 1 & 0 \cr
} } \right|$$<br><br>$$ = - 2\widehat i - 2\widehat j - 2\widehat k$$<br><br>$$ \Rightarrow \overrightarrow r = \pm \sqrt 3 {{\left( {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right)} \over {\left| {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right|}} = \pm {{\sqrt 3 \left( { - 2\widehat i - 2\widehat j - 2\widehat k} \right)} \over {\sqrt {{2^2} + {2^2} + {2^2}} }}$$<br><br>$$\overrightarrow r = \pm \left( { - \widehat i - \widehat j - \widehat k} \right)$$<br><br>According to question <br><br>$$\overrightarrow r = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$$<br><br>So, |$$\alpha$$| = 1, |$$\beta$$| = 1, |$$\gamma$$| = 1<br><br>$$\Rightarrow$$ $$\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$$ = 3 | integer | jee-main-2021-online-25th-july-morning-shift |
1krzmy5aw | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | If $$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$ and $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = 8, then $$\left| {\overrightarrow a .\,\overrightarrow b } \right|$$ is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["A"] | null | $$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$<br><br>$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$$<br><br>$$\sin \theta = \pm \,{4 \over 5}$$<br><br>$$\therefore$$ $$\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $$<br><br>$$ = 10.\left( { \pm \,{3 \over 5}} \right) = \pm 6$$<br><br>$$\left| {\overrightarrow a .\,\overrightarrow b } \right| = 6$$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1ks05h77r | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$ and $$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$$. Then the vector product $$\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$$ is equal to : | [{"identifier": "A", "content": "$$5(34\\widehat i - 5\\widehat j + 3\\widehat k)$$"}, {"identifier": "B", "content": "$$7(34\\widehat i - 5\\widehat j + 3\\widehat k)$$"}, {"identifier": "C", "content": "$$7(30\\widehat i - 5\\widehat j + 7\\widehat k)$$"}, {"identifier": "D", "content": "$$5(30\\widehat i - 5\\widehat j + 7\\widehat k)$$"}] | ["B"] | null | $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$<br><br>$$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$$<br><br>$$\overrightarrow a + \overrightarrow b = 3\widehat j + 5\widehat k;\overrightarrow a.\overrightarrow b = - 1 + 2 + 6 = 7$$<br><br>$$\left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$$<br><br>$$\left( {\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$$<br><br>$$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b - 0} \right)} \right) \times \overrightarrow b $$<br><br>$$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right) \times \overrightarrow b $$<br><br>$$\left( {\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow a .\overrightarrow a } \right)\overrightarrow b } \right) \times \overrightarrow b $$<br><br>$$\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a \times \overrightarrow b - \left( {\overrightarrow a .\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow b } \right)$$<br><br>$$\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow b } \right)$$<br><br>$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
i & j & k \cr
1 & 1 & 2 \cr
{ - 1} & 2 & 3 \cr
} } \right| = - \widehat i - 5\widehat j + 3\widehat k$$<br><br>$$\therefore$$ $$7\left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$$<br><br>$$\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {7\left( { - \widehat i - 5\widehat j + 3\widehat k} \right)} \right)$$<br><br>$$7\left( {0\widehat i + 3\widehat j + 5\widehat k} \right) \times \left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$$<br><br>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 3 & 5 \cr
{ - 1} & { - 5} & 3 \cr
} } \right|$$<br><br>$$ \Rightarrow 34\widehat i - (5)\widehat j + (3\widehat k)$$<br><br>$$ \Rightarrow 34\widehat i - 5\widehat j + 3\widehat k$$<br><br>$$\therefore$$
$$7\left( {0\widehat i + 3\widehat j + 5\widehat k} \right) \times \left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$$
<br><br>= $$7(34\widehat i - 5\widehat j + 3\widehat k)$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ks0bqpuy | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b $$ and $$\overrightarrow c = \widehat j - \widehat k$$ be three vectors such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow a \,.\,\overrightarrow b = 1$$. If the length of projection vector of the vector $$\overrightarrow b $$ on the vector $$\overrightarrow a \times \overrightarrow c $$ is l, then the value of 3l<sup>2</sup> is equal to _____________. | [] | null | 2 | $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$<br><br>Take Dot with $$\overrightarrow c $$<br><br>$$\left( {\overrightarrow a \times \overrightarrow b } \right).\,\overrightarrow c = {\left| {\overrightarrow c } \right|^2} = 2$$<br><br>Projection of $$\overrightarrow b $$ or $$\overrightarrow a \times \overrightarrow c = l$$<br><br>$${{\left| {\overrightarrow b \,.\,(\overrightarrow a \times \overrightarrow c )} \right|} \over {|\overrightarrow a \times \overrightarrow c |}} = l$$<br><br>$$\therefore$$ $$l = {2 \over {\sqrt 6 }} \Rightarrow {l^2} = {4 \over 6}$$<br><br>$$3{l^2} = 2$$ | integer | jee-main-2021-online-27th-july-morning-shift |
1kteolpch | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k$$, $$\overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k$$ and $$\overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k$$ be three vectors such that, $$\left| {\overrightarrow b \times \overrightarrow c } \right| = 5\sqrt 3 $$ and $${\overrightarrow a }$$ is perpendicular to $${\overrightarrow b }$$. Then the greatest amongst the values of $${\left| {\overrightarrow a } \right|^2}$$ is _____________. | [] | null | 90 | Since, $$\overrightarrow a .\,\overrightarrow b = 0$$<br><br>$$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$$ .... (1)<br><br>Also, <br><br>$${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$$<br><br>$$\Rightarrow$$ 5$$\beta$$<sup>2</sup> + 30$$\beta$$ + 40 = 0<br><br>$$\Rightarrow$$ $$\beta$$ = $$-$$4, $$-$$2<br><br>$$\Rightarrow$$ $$\alpha$$ = 4, 8<br><br>$$ \Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$$ | integer | jee-main-2021-online-27th-august-morning-shift |
1ktoc2aax | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$$. Let a vector $$\overrightarrow v $$ be in the plane containing $$\overrightarrow a $$ and $$\overrightarrow b $$. If $$\overrightarrow v $$ is perpendicular to the vector $$3\widehat i + 2\widehat j - \widehat k$$ and its projection on $$\overrightarrow a $$ is 19 units, then $${\left| {2\overrightarrow v } \right|^2}$$ is equal to _____________. | [] | null | 1494 | $$\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$$<br><br>$$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$$<br><br>$$\overrightarrow c = 3\widehat i + 2\widehat j - \widehat k$$<br><br>$$\overrightarrow v = x\overrightarrow a + y\overrightarrow b $$<br><br>$$\overrightarrow v \left( {3\widehat i + 2\widehat j - \widehat k} \right) = 0$$<br><br>$$\overrightarrow v .\widehat a = 19$$<br><br>$$\overrightarrow v = \lambda \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)$$<br><br>$$\overrightarrow v = \lambda \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right]$$<br><br>$$ = \lambda [(3 + 4 + 1)\left( {2\widehat i - \widehat j + 2\widehat k} \right) - \left( {{{6 - 2 - 2} \over 2}} \right)\left( {\widehat i + 2\widehat j + \widehat k} \right)$$<br><br>$$ = \lambda [16\widehat i - 8\widehat j + 16\widehat k - 2\widehat i - 4\widehat j + 2\widehat k]$$<br><br>$$\overrightarrow v = \lambda \left[ {14\widehat i - 12\widehat j + 18\widehat k} \right]$$<br><br>$$\lambda [14\widehat i - 12\widehat j + 18\widehat k].{{\left( {2\widehat i - \widehat j + 2\widehat k} \right)} \over {\sqrt {4 + 1 + 4} }} = 19$$<br><br>$$\lambda {{[28 + 12 + 36]} \over 3} = 19$$<br><br>$$\lambda \left( {{{76} \over 3}} \right) = 19$$<br><br>$$4\lambda = 3 \Rightarrow \lambda = {3 \over 4}$$<br><br>$$|2{v^2}| = {\left| {2 \times {3 \over 4}(14\widehat i - 12\widehat j + 18\widehat k)} \right|^2}$$<br><br>$${9 \over 4} \times 4{\left( {7\widehat i - 6\widehat j + 9\widehat k} \right)^2}$$<br><br>$$ = 9(49 + 36 + 81)$$<br><br>$$ = 9(166)$$<br><br>$$ = 1494$$ | integer | jee-main-2021-online-1st-september-evening-shift |
1l544t6p5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$, $$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$ and $$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$ where $$\alpha ,\,\beta \in R$$, be three vectors. If the projection of $$\overrightarrow a $$ on $$\overrightarrow c $$ is $${{10} \over 3}$$ and $$\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$$, then the value of $$\alpha + \beta $$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <p>$$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$</p>
<p>$$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$</p>
<p>$$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$</p>
<p>Projection of $$\overrightarrow a $$ on $$\overrightarrow c $$ is</p>
<p>$${{\overrightarrow a \,.\,\overrightarrow c } \over {|\overrightarrow b |}} = {{10} \over 3}$$</p>
<p>$${{\alpha + 6 + 2} \over {\sqrt {{1^2} + {2^2} + {{( - 2)}^2}} }} = {{\alpha + 8} \over 3} = {{10} \over 3}$$</p>
<p>$$\therefore$$ $$\alpha$$ = 2</p>
<p>$$\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$$</p>
<p>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & { - \beta } & 4 \cr
1 & 2 & { - 2} \cr
} } \right| = (2\beta - 8)\widehat i + 10\widehat j + (6 + \beta )\widehat k = - 6\widehat i + 10\widehat j + 7\widehat k$$</p>
<p>$$2\beta - 8 = - 6$$ & $$6 + \beta = 7$$</p>
<p>$$\therefore$$ $$\beta$$ = 1</p>
<p>$$\alpha + \beta = 2 + 1 = 3$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54tqgwd | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$$, $$\overrightarrow b = \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c $$ be a vector such that $$\overrightarrow a + \left( {\overrightarrow b \times \overrightarrow c } \right) = \overrightarrow 0 $$ and $$\overrightarrow b \,.\,\overrightarrow c = 5$$. Then the value of $$3\left( {\overrightarrow c \,.\,\overrightarrow a } \right)$$ is equal to _________.</p> | [] | null | BONUS | $$
\vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=2
$$ ........(i)
<br/><br/>Given: $\vec{a}+(\vec{b} \times \vec{c})=0$
<br/><br/>$$
\Rightarrow \vec{a} \cdot \vec{b}=0
$$ ........(ii)
<br/><br/>Equation (i) and equation (ii) are contradicting. | integer | jee-main-2022-online-29th-june-evening-shift |
1l55iksny | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$$ and $$\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$$, where $$\alpha \in R$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$\sqrt {15({\alpha ^2} + 4)} $$, then the value of $$2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a \,.\,\overrightarrow b } \right){\left| {\overrightarrow b } \right|^2}$$ is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$$ and $$\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$$</p>
<p>$$\therefore$$ $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\alpha & 2 & { - 1} \cr
{ - 2} & \alpha & 1 \cr
} } \right| = (2 + \alpha )\widehat i - (\alpha - 2)\widehat j + ({\alpha ^2} + 4)\widehat k$$</p>
<p>Now $$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {15({\alpha ^2} + 4)} $$</p>
<p>$$ \Rightarrow {(2 + \alpha )^2} + {(\alpha - 2)^2} + {({\alpha ^2} + 4)^2} = 15({\alpha ^2} + 4)$$</p>
<p>$$ \Rightarrow {\alpha ^4} - 5{\alpha ^2} - 36 = 0$$</p>
<p>$$\therefore$$ $$\alpha = \, \pm \,3$$</p>
<p>Now, $$2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a - \overrightarrow b } \right){\left| {\overrightarrow b } \right|^{ - 2}} = 2.14 - 14 = 14$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l55ithsn | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a $$ be a vector which is perpendicular to the vector $$3\widehat i + {1 \over 2}\widehat j + 2\widehat k$$. If $$\overrightarrow a \times \left( {2\widehat i + \widehat k} \right) = 2\widehat i - 13\widehat j - 4\widehat k$$, then the projection of the vector $$\overrightarrow a $$ on the vector $$2\widehat i + 2\widehat j + \widehat k$$ is :</p> | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${5 \\over 3}$$"}, {"identifier": "D", "content": "$${7 \\over 3}$$"}] | ["C"] | null | <p>Let $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$</p>
<p>and $$\overrightarrow a \,.\,\left( {3\widehat i - {1 \over 2}\widehat j + 2\widehat k} \right) = 0 \Rightarrow 3{a_1} + {{{a_2}} \over 2} + 2{a_3} = 0$$ ..... (i)</p>
<p>and $$\overrightarrow a \times (2\widehat i + \widehat k) = 2\widehat i - 13\widehat j - 4\widehat k$$</p>
<p>$$ \Rightarrow {a_2}\widehat i + (2{a_3} - {a_1})\widehat j - 2{a_2}\widehat k = 2\widehat i - 13\widehat j - 4\widehat k$$</p>
<p>$$\therefore$$ $${a_2} = 2$$ ..... (ii)</p>
<p>and $${a_1} - 2{a_3} = 13$$ ..... (iii)</p>
<p>From eq. (i) and (iii) : $${a_1} = 3$$ and $${a_3} = - 5$$</p>
<p>$$\therefore$$ $$\overrightarrow a = 3\widehat i + 2\widehat j - 5\widehat k$$</p>
<p>$$\therefore$$ projection of $$\overrightarrow a $$ on $$2\widehat i + 2\widehat j + \widehat k = {{6 + 4 - 5} \over 3} = {5 \over 3}$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l567mmea | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>If $$\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$$, $$\overrightarrow b = 3\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$$ are coplanar vectors and $$\overrightarrow a \,.\,\overrightarrow c = 5$$, $$\overrightarrow b \bot \overrightarrow c $$, then $$122({c_1} + {c_2} + {c_3})$$ is equal to ___________.</p> | [] | null | 150 | <p>$$2{C_1} + {C_2} + 3{C_3} = 5$$ ...... (i)</p>
<p>$$3{C_1} + 3{C_2} + {C_3} = 0$$ ...... (ii)</p>
<p>$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{
2 & 1 & 3 \cr
3 & 3 & 1 \cr
{{C_1}} & {{C_2}} & {{C_3}} \cr
} } \right|$$</p>
<p>$$ = 2(3{C_3} - {C_2}) - 1(3{C_3} - {C_1}) + 3(3{C_2} - 3{C_1})$$</p>
<p>$$ = 3{C_3} + 7{C_2} - 8{C_1}$$</p>
<p>$$ \Rightarrow 8{C_1} - 7{C_2} - 3{C_3} = 0$$ ...... (iii)</p>
<p>$${C_1} = {{10} \over {122}},{C_2} = {{ - 85} \over {122}},{C_3} = {{225} \over {122}}$$</p>
<p>So $$122({C_1} + {C_2} + {C_3}) = 150$$</p> | integer | jee-main-2022-online-28th-june-morning-shift |
1l56rew3i | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be the vectors along the diagonals of a parallelogram having area $$2\sqrt 2 $$. Let the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ be acute, $$|\overrightarrow a | = 1$$, and $$|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$$. If $$\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow b $$, then an angle between $$\overrightarrow b $$ and $$\overrightarrow c $$ is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$$-$$ $${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${{5\\pi } \\over 6}$$"}, {"identifier": "D", "content": "$${{3\\pi } \\over 4}$$"}] | ["D"] | null | <p>$$\because$$ $$\overrightarrow a $$ and $$\overrightarrow b $$ be the vectors along the diagonals of a parallelogram having area 2$$\sqrt2$$.</p>
<p>$$\therefore$$ $${1 \over 2}|\overrightarrow a \times \overrightarrow b | = 2\sqrt 2 $$</p>
<p>$$|\overrightarrow a ||\overrightarrow b |\sin \theta = 4\sqrt 2 $$</p>
<p>$$ \Rightarrow |\overrightarrow b |\sin \theta = 4\sqrt 2 $$ ..... (i)</p>
<p>and $$|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$$</p>
<p>$$|\overrightarrow a ||\overrightarrow b |\cos \theta = |\overrightarrow a ||\overrightarrow b |\sin \theta $$</p>
<p>$$ \Rightarrow \tan \theta = 1$$</p>
<p>$$\therefore$$ $$\theta = {\pi \over 4}$$</p>
<p>By (i) $$|\overrightarrow b | = 8$$</p>
<p>Now $$\overrightarrow c = 2\sqrt 2 (\overrightarrow a \times \overrightarrow b ) - 2\overrightarrow b $$</p>
<p>$$ \Rightarrow \overrightarrow c \,.\,\overrightarrow b = - 2|\overrightarrow b {|^2} = - 128$$ ...... (ii)</p>
<p>and $$\overrightarrow c \,.\,\overrightarrow c = 8|\overrightarrow a \times \overrightarrow b {|^2} + 4|\overrightarrow b {|^2}$$</p>
<p>$$ \Rightarrow |\overrightarrow c {|^2} = 8.32 + 4.64$$</p>
<p>$$ \Rightarrow |\overrightarrow c | = 16\sqrt 2 $$ ..... (iii)</p>
<p>From (ii) and (iii)</p>
<p>$$|\overrightarrow c ||\overrightarrow b |\cos \alpha = - 128$$</p>
<p>$$ \Rightarrow \cos \alpha = {{ - 1} \over {\sqrt 2 }}$$</p>
<p>$$\alpha = {{3\pi } \over 4}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l57okaze | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \widehat i + \widehat j - \widehat k$$ and $$\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$$. Then the number of vectors $$\overrightarrow b $$ such that $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$ and $$|\overrightarrow b | \in $$ {1, 2, ........, 10} is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["A"] | null | <p>$$\overrightarrow a = \widehat i + \widehat j - \widehat k$$</p>
<p>$$\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$$</p>
<p>Now, $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$</p>
<p>$$\overrightarrow c \,.\,(\overrightarrow b \times \overrightarrow c ) = \overrightarrow c \,.\,\overrightarrow a $$</p>
<p>$$\overrightarrow c \,.\,\overrightarrow a = 0$$</p>
<p>$$ \Rightarrow (\widehat i + \widehat j - \widehat k)(2\widehat i - 3\widehat j + 2\widehat k) = 0$$</p>
<p>$$ = 2 - 3 - 2 = 0$$</p>
<p>$$ \Rightarrow - 3 = 0$$ (Not possible)</p>
<p>$$\Rightarrow$$ No possible value of $$\overrightarrow b $$ is possible.</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1l59ldqt3 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow b = \widehat i + \widehat j + \lambda \widehat k$$, $$\lambda$$ $$\in$$ R. If $$\overrightarrow a $$ is a vector such that $$\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat k$$ and $$\overrightarrow a \,.\,\overrightarrow b + 21 = 0$$, then $$\left( {\overrightarrow b - \overrightarrow a } \right).\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right).\,\left( {\widehat i - \widehat k} \right)$$ is equal to _____________.</p> | [] | null | 14 | <p>Let $$\overrightarrow a = x\widehat i = y\widehat j + z\widehat k$$</p>
<p>So, $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
x & y & z \cr
1 & 1 & \lambda \cr
} } \right| = \widehat i(\lambda y - z) + \widehat j(z - \lambda x) + \widehat k(x - y)$$</p>
<p>$$ \Rightarrow \lambda y - z = 13,\,z - \lambda x = - 1,\,x - y = - 4$$</p>
<p>and $$x + y + \lambda z = - 21$$</p>
<p>$$\Rightarrow$$ Clearly, $$\lambda = 3$$, $$x = - 2$$, $$y = 2$$ and $$z = - 7$$</p>
<p>So, $$\overrightarrow b - \overrightarrow a = 3\widehat i - \widehat j + 10\widehat k$$</p>
<p>and $$\overrightarrow b + \overrightarrow a = - \widehat i + 3\widehat j - 4\widehat k$$</p>
<p>$$ \Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right)\,.\,\left( {\widehat i - \widehat k} \right) = 11 + 3 = 14$$</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l5ajnmz4 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\theta$$ be the angle between the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$, where $$|\overrightarrow a | = 4,$$ $$|\overrightarrow b | = 3$$ and $$\theta \in \left( {{\pi \over 4},{\pi \over 3}} \right)$$. Then $${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$$ is equal to __________.</p> | [] | null | 576 | <p>$${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$$</p>
<p>$$ \Rightarrow {\left| {\overrightarrow a \times \overrightarrow a + \overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$$</p>
<p>$$ \Rightarrow {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$$</p>
<p>$$ \Rightarrow 4{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$$</p>
<p>$$ \Rightarrow 4{\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2} = 4\,.\,16\,.\,9 = 576$$</p> | integer | jee-main-2022-online-25th-june-morning-shift |
1l5ban31s | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\widehat a$$ and $$\widehat b$$ be two unit vectors such that $$|(\widehat a + \widehat b) + 2(\widehat a \times \widehat b)| = 2$$. If $$\theta$$ $$\in$$ (0, $$\pi$$) is the angle between $$\widehat a$$ and $$\widehat b$$, then among the statements :</p>
<p>(S1) : $$2|\widehat a \times \widehat b| = |\widehat a - \widehat b|$$</p>
<p>(S2) : The projection of $$\widehat a$$ on ($$\widehat a$$ + $$\widehat b$$) is $${1 \over 2}$$</p> | [{"identifier": "A", "content": "Only (S1) is true."}, {"identifier": "B", "content": "Only (S2) is true."}, {"identifier": "C", "content": "Both (S1) and (S2) are true."}, {"identifier": "D", "content": "Both (S1) and (S2) are false."}] | ["C"] | null | <p>$$\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right| = 2,\,\theta \in (0,\,\pi )$$</p>
<p>$$ \Rightarrow {\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right|^2} = 4$$</p>
<p>$$ \Rightarrow {\left| {\widehat a} \right|^2} + {\left| {\widehat b} \right|^2} + 4{\left| {\widehat a \times \widehat b} \right|^2} + 2\widehat a\,.\,\widehat b = 4$$</p>
<p>$$\therefore$$ $$\cos \theta = \cos 2\theta $$</p>
<p>$$\therefore$$ $$\theta = {{2\pi } \over 3}$$</p>
<p>where $$\theta$$ is angle between $$\widehat a$$ and $$\widehat b$$.</p>
<p>$$\therefore$$ $$2\left| {\widehat a \times \widehat b} \right| = \sqrt 3 = \left| {\widehat a - \widehat b} \right|$$</p>
<p>(S1) is correct.</p>
<p>And projection of $$\widehat a$$ on $$(\widehat a + \widehat b) = \left| {{{\widehat a\,.\,(\widehat a + \widehat b)} \over {\left| {\widehat a + \widehat b} \right|}}} \right| = {1 \over 2}$$</p>
<p>(S2) is correct.</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c1or9b | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\widehat a$$, $$\widehat b$$ be unit vectors. If $$\overrightarrow c $$ be a vector such that the angle between $$\widehat a$$ and $$\overrightarrow c $$ is $${\pi \over {12}}$$, and $$\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right)$$, then $${\left| {6\overrightarrow c } \right|^2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$6\\left( {3 - \\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$3 + \\sqrt 3 $$"}, {"identifier": "C", "content": "$$6\\left( {3 + \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$6\\left( {\\sqrt 3 + 1} \\right)$$"}] | ["C"] | null | $\because \quad \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\
&\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a})
\end{aligned}
$$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdot \vec{c}=4|\vec{c}|^{2}|\vec{a}|^{2} \sin ^{2} \frac{\pi}{12} \\\\
&\Rightarrow 1+|\vec{c}|^{2}-2|c|^{2}=4|\vec{c}|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2} \\\\
&\Rightarrow 1=|\vec{c}|^{2}(3-\sqrt{3}) \\\\
&\Rightarrow 36|\vec{c}|^{2}=\frac{36}{3-\sqrt{3}}=6(3+\sqrt{3})
\end{aligned}
$$ | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6dwjgp3 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\mathrm{ABC}$$ be a triangle such that $$\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}$$ and $$\vec{b} \cdot \vec{c}=12$$. Consider the statements :
</p>
<p>$$(\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1)$$
</p>
<p>$$(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$$
</p>
<p>Then</p> | [{"identifier": "A", "content": "both (S1) and (S2) are true"}, {"identifier": "B", "content": "only (S1) is true"}, {"identifier": "C", "content": "only (S2) is true"}, {"identifier": "D", "content": "both (S1) and (S2) are false"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97sapyl/779dd912-c32e-40a3-8f62-d0f26f1743e1/b6acf9d0-4b5b-11ed-bfde-e1cb3fafe700/file-1l97sapym.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97sapyl/779dd912-c32e-40a3-8f62-d0f26f1743e1/b6acf9d0-4b5b-11ed-bfde-e1cb3fafe700/file-1l97sapym.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Vector Algebra Question 97 English Explanation"><br><br>
$$
\because \vec{a}+\vec{b}+\vec{c}=0
$$
<br><br>
then $\bar{a}+\vec{c}=-\vec{b}$
<br><br>
then $(\vec{a}+\vec{c}) \times \vec{b}=-\vec{b} \times \bar{b}$
<br><br>
$\therefore \quad \vec{a} \times \vec{b}+\vec{c} \times \vec{b}=\overline{0}\quad\dots(i)$
<br><br>
For $(S 1):|\vec{a} \times \vec{b}+\vec{c} \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1)$
<br><br>
$$
\begin{aligned}
&|(\vec{a}+\vec{c}) \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1) \\\\
&|\vec{c}|=6-12 \sqrt{2} \text { (not possible) }
\end{aligned}
$$
<br><br>
Hence (S1) is not correct
<br><br>
For (S2) : from (i) $\vec{b}+\vec{c}=-\vec{a}$
<br><br>
$\Rightarrow \vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}=-\vec{a} \cdot \vec{b}$
<br><br>
$\Rightarrow 12+12=-6 \sqrt{2} \cdot 2 \sqrt{3} \cos (\pi-\angle A C B)$
<br><br>
$\therefore \cos (\angle A C B)=\sqrt{\frac{2}{3}}$
<br><br>
$$
\begin{aligned}
&\therefore \angle A C B=\cos ^{-1} \sqrt{\frac{2}{3}} \\\\
&\therefore S(2) \text { is correct. }
\end{aligned}
$$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6f2wm5r | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$ and let $$\vec{b}$$ be a vector such that $$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$$ and $$\vec{a} \cdot \vec{b}=3$$. Then the projection of $$\vec{b}$$ on the vector $$\vec{a}-\vec{b}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{2}{\\sqrt{21}}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{\\frac{3}{7}}$$"}, {"identifier": "C", "content": "$$\n\\frac{2}{3} \\sqrt{\\frac{7}{3}}\n$$"}, {"identifier": "D", "content": "$$\\frac{2}{3}$$"}] | ["A"] | null | <p>$$\overrightarrow a = \widehat i - \widehat j + 2\widehat k$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = 2\widehat i - \widehat k$$</p>
<p>$$\overrightarrow a \,.\,\overrightarrow b = 3$$</p>
<p>$$|\overrightarrow a \times \overrightarrow b {|^2} + |\overrightarrow a \,.\,\overrightarrow b {|^2} = |\overrightarrow a {|^2}\,.\,|\overrightarrow b {|^2}$$</p>
<p>$$ \Rightarrow 5 + 9 = 6|\overrightarrow b {|^2}$$</p>
<p>$$ \Rightarrow |b {|^2} = {7 \over 3}$$</p>
<p>$$|\overrightarrow a - \overrightarrow b | = \sqrt {|\overrightarrow a {|^2} + |\overrightarrow b {|^2} - 2\overrightarrow a \,.\,\overrightarrow b } = \sqrt {{7 \over 3}} $$</p>
<p>projection of $$\overrightarrow b $$ on $$\overrightarrow a - \overrightarrow b = {{\overrightarrow b \,.\,(\overrightarrow a - \overrightarrow b )} \over {|\overrightarrow a - \overrightarrow b |}}$$</p>
<p>$$ = {{\overrightarrow b \,.\,\overrightarrow a - |\overrightarrow b {|^2}} \over {|\overrightarrow a - \overrightarrow b |}} = {{3 - {7 \over 3}} \over {\sqrt {{7 \over 3}} }}$$</p>
<p>$$ = {2 \over {\sqrt {21} }}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6giq99g | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}, \alpha>0$$. If the projection of $$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$$ on the vector $$-\hat{i}+2 \hat{j}-2 \hat{k}$$ is 30, then $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{15}{2}$$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "$$\\frac{13}{2}$$"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p>Given : $$\overrightarrow a = (\alpha ,1, - 1)$$ and $$\overrightarrow b = (2,1, - \alpha )$$</p>
<p>$$\overrightarrow c = \overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\alpha & 1 & { - 1} \cr
2 & 1 & { - \alpha } \cr
} } \right|$$</p>
<p>$$ = ( - \alpha + 1)\widehat i + ({\alpha ^2} - 2)\widehat j + (\alpha - 2)\widehat k$$</p>
<p>Projection of $$\overrightarrow c $$ on $$\overrightarrow d = - \widehat i + 2\widehat j - 2\widehat k$$</p>
<p>$$ = \left| {\overrightarrow c \,.\,{{\overrightarrow d } \over {|d|}}} \right| = 30$$ {Given}</p>
<p>$$ \Rightarrow \, = \left| {{{\alpha - 1 - 4 + 2{\alpha ^2} - 2\alpha + 4} \over {\sqrt {1 + 4 + 4} }}} \right| = 30$$</p>
<p>On solving $$\alpha = {{ - 13} \over 2}$$ (Rejected as $$\alpha > 0$$)</p>
<p>and $$\alpha = 7$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6jc2mfl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$$ and $$\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$$ be two vectors, such that $$\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$$. Then the projection of $$\vec{b}-2 \vec{a}$$ on $$\vec{b}+\vec{a}$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\frac{39}{5}$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "$$\\frac{46}{5}$$"}] | ["D"] | null | <p>$$\overrightarrow a = \alpha \widehat i + \widehat j + \beta \widehat k$$, $$\overrightarrow b = 3\widehat i - 5\widehat j + 4\widehat k$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = - \widehat i + 9\widehat j + 12\widehat k$$</p>
<p>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\alpha & 1 & \beta \cr
3 & { - 5} & 4 \cr
} } \right| = - \widehat i + 9\widehat j + 12\widehat k$$</p>
<p>$$4 + 5\beta = - 1 \Rightarrow \beta = - 1$$</p>
<p>$$ - 5\alpha - 3 = 12 \Rightarrow \alpha = - 3$$</p>
<p>$$\overrightarrow b - 2\overrightarrow a = 3\widehat i - 5\widehat j + 4\widehat k - 2\left( { - 3\widehat i + \widehat j - \widehat k} \right)$$</p>
<p>$$\overrightarrow b - 2\overrightarrow a = 9\widehat i - 7\widehat j + 6\widehat k$$</p>
<p>$$\overrightarrow b + \overrightarrow a = \left( {3\widehat i - 5\widehat j + 4\widehat k} \right) + \left( { - 3\widehat i + \widehat j - \widehat k} \right)$$</p>
<p>$$\overrightarrow b + \overrightarrow a = - 4\widehat j + 3\widehat k$$</p>
<p>Projection of $$\overrightarrow b - 2\overrightarrow a $$ on $$\overrightarrow b + \overrightarrow a $$ is $$ = {{\left( {\overrightarrow b - 2\overrightarrow a } \right)\,.\,\left( {\overrightarrow b + \overrightarrow a } \right)} \over {\left| {\overrightarrow b + \overrightarrow a } \right|}}$$</p>
<p>$$ = {{28 + 18} \over 5} = {{46} \over 5}$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6km5v7z | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ be three non-coplanar vectors such that $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ = 4$$\overrightarrow c $$, $$\overrightarrow b $$ $$\times$$ $$\overrightarrow c $$ = 9$$\overrightarrow a $$ and $$\overrightarrow c $$ $$\times$$ $$\overrightarrow a $$ = $$\alpha$$$$\overrightarrow b $$, $$\alpha$$ > 0. If $$\left| {\overrightarrow a } \right| + \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right| = {1 \over {36}}$$, then $$\alpha$$ is equal to __________.</p> | [] | null | 36 | <p>Given,</p>
<p>$$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c $$ ..... (i)</p>
<p>$$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a $$ ..... (ii)</p>
<p>$$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b $$ .... (iii)</p>
<p>Taking dot products with $$\overrightarrow c ,\overrightarrow a ,\overrightarrow b $$ we get</p>
<p>$$\overrightarrow a \,.\,\overrightarrow b = \overrightarrow b \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow a = 0$$</p>
<p>Hence,</p>
<p>(i) $$ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b | = 4\,.\,|\overrightarrow c |$$ ..... (iv)</p>
<p>(ii) $$ \Rightarrow |\overrightarrow b |\,.\,|\overrightarrow c | = 9\,.\,|\overrightarrow a |$$ ..... (v)</p>
<p>(iii) $$ \Rightarrow |\overrightarrow c |\,.\,|\overrightarrow a | = \alpha \,.\,|\overrightarrow b |$$ .... (vi)</p>
<p>Multiplying (iv), (v) and (vi)</p>
<p>$$ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b |\,.\,|\overrightarrow c | = 36\alpha $$ ..... (vii)</p>
<p>Dividing (vii) by (iv) $$ \Rightarrow |\overrightarrow c {|^2} = 9\alpha \Rightarrow |\overrightarrow c | = 3\sqrt \alpha $$ ..... (viii)</p>
<p>Dividing (vii) by (v) $$ \Rightarrow |\overrightarrow a {|^2} = 4\alpha \Rightarrow |\overrightarrow a | = 2\sqrt \alpha $$</p>
<p>Dividing (viii) by (vi) $$ \Rightarrow |\overrightarrow b {|^2} = 36 \Rightarrow |\overrightarrow b | = 6$$</p>
<p>Now, as given, $$3\sqrt \alpha + 2\sqrt \alpha + 6 = {1 \over {36}} \Rightarrow \sqrt \alpha = {{ - 43} \over {36}}$$</p> | integer | jee-main-2022-online-27th-july-evening-shift |
1l6m5gnnw | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let a vector $$\vec{a}$$ has magnitude 9. Let a vector $$\vec{b}$$ be such that for every $$(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}$$, the vector $$(x \vec{a}+y \vec{b})$$ is perpendicular to the vector $$(6 y \vec{a}-18 x \vec{b})$$. Then the value of $$|\vec{a} \times \vec{b}|$$ is equal to :</p> | [{"identifier": "A", "content": "$$9 \\sqrt{3}$$"}, {"identifier": "B", "content": "$$27 \\sqrt{3}$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "81"}] | ["B"] | null | <p>$$\left( {x\overrightarrow a + y\overrightarrow b } \right).\left( {6y\overrightarrow a - 18x\overrightarrow b } \right) = 0$$</p>
<p>$$ \Rightarrow \left( {6xy|\overrightarrow a {|^2} - 18xy|\overrightarrow b {|^2}} \right) + \left( {6{y^2} - 18{x^2}} \right)\overrightarrow a .\overrightarrow b = 0$$</p>
<p>As given equation is identity</p>
<p>Coefficient of $${x^2} = $$ coefficient of $${y^2} = $$ coefficient of $$xy = 0$$</p>
<p>$$ \Rightarrow |\overrightarrow a {|^2} = 3|\overrightarrow b {|^2} \Rightarrow |\overrightarrow b | = 3\sqrt 3 $$</p>
<p>and $$\overrightarrow a .\overrightarrow b = 0$$</p>
<p>$$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a ||\overrightarrow b |\sin \theta $$</p>
<p>$$ = 9.\,3\sqrt 3 .1 = 27\sqrt 3 $$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6p2fvoj | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\hat{a}$$ and $$\hat{b}$$ be two unit vectors such that the angle between them is $$\frac{\pi}{4}$$. If $$\theta$$ is the angle between the vectors $$(\hat{a}+\hat{b})$$ and $$(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$$, then the value of $$164 \,\cos ^{2} \theta$$ is equal to :</p> | [{"identifier": "A", "content": "$$90+27 \\sqrt{2}$$"}, {"identifier": "B", "content": "$$45+18 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$90+3 \\sqrt{2}$$"}, {"identifier": "D", "content": "$$54+90 \\sqrt{2}$$"}] | ["A"] | null | <p>$$\widehat a\,.\,\widehat b = {1 \over {\sqrt 2 }}$$ and $$|\widehat a \times \widehat b| = {1 \over {\sqrt 2 }}$$</p>
<p>$${{\left( {\widehat a + \widehat b} \right)\,.\,\left( {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right)} \over {\left| {\widehat a + \widehat b} \right|\left| {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right|}} = \cos \theta $$</p>
<p>$$ \Rightarrow \cos \theta = {{1 + 3\widehat a\widehat b + 2} \over {|\widehat a + \widehat b||\widehat a + 2\widehat b + 2(\widehat a \times \widehat b)|}}$$</p>
<p>$$|\widehat a + \widehat b{|^2} = 2 + \sqrt 2 $$</p>
<p>$$|\widehat a + 2\widehat b + 2(\widehat a \times \widehat b){|^2} = 1 + 4 + 4|\widehat a \times \widehat b{|^2} + 4\widehat a\widehat b$$</p>
<p>$$ = 5 + 4\,.\,{1 \over 2} + {4 \over {\sqrt 2 }} = 7 + 2\sqrt 2 $$</p>
<p>So, $${\cos ^2}\theta = {{{{\left( {3 + {3 \over {\sqrt 2 }}} \right)}^2}} \over {(2 + \sqrt 2 )(7 + 2\sqrt 2 )}} = {{9\sqrt 2 (5\sqrt 2 + 3)} \over {164}}$$</p>
<p>$$ \Rightarrow 164{\cos ^2}\theta = 90 + 27\sqrt 2 $$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1l6rf9wxz | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}, \vec{b}, \vec{c}$$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and
$$(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168$$, then $$|\vec{a}|+|\vec{b}|+|\vec{c}|$$ is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "18"}] | ["C"] | null | $|\vec{a}||\vec{b}||\vec{c}|=14$
<br/><br/>
$$
\begin{aligned}
& \vec{a} \wedge \vec{b}=\vec{b} \wedge \vec{c}=\vec{c} \wedge \vec{a}=\theta=\frac{2 \pi}{3} \\\\
& \vec{a} \cdot \vec{b}=-\frac{1}{2}|\vec{a}||\vec{b}| \\\\
& \vec{b} \cdot \vec{c}=-\frac{1}{2}|\vec{b}||\vec{c}| \\\\
& \vec{c} \cdot \vec{a}=-\frac{1}{2}|\vec{c}||\vec{a}|
\end{aligned}
$$
<br/><br/>
Now,
<br/><br/>
$$
\begin{aligned}
& \begin{aligned}
&(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b}) \\\\
&=168 \quad\quad...(i)\\\\
&(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})-(\vec{a} \cdot \vec{c})|\vec{b}|^{2} \\\\
&= \frac{1}{4}|\vec{b}|^{2}|\vec{a}||\vec{c}|+\frac{1}{2}|\vec{a}||\vec{b}|^{2}|\vec{c}| \\\\
&= \frac{3}{4}|\vec{a}||\vec{b}|^{2}|\vec{c}| \quad\quad...(ii)
\end{aligned}
\end{aligned}
$$
<br/><br/>
Similarly $(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})=\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|^{2} \quad\quad...(iii)$
<br/><br/>
$(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=\frac{3}{4}|\vec{a}|^{2}|\vec{b}||\vec{c}| \quad\quad...(iv)$
<br/><br/>
Substitute (ii), (iii), (iv) in (i)
<br/><br/>
$\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168$
<br/><br/>
$\frac{3}{4} \times 14[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168$
<br/><br/>
$|\vec{a}|+|\vec{b}|+|\vec{c}|=16$
| mcq | jee-main-2022-online-29th-july-evening-shift |
1l6rg17yk | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3$$ and $$|\vec{a} \times \vec{b}|^{2}=75$$. Then $$|\vec{a}|^{2}$$ is equal to __________.</p> | [] | null | 14 | $\because|\vec{a}+\dot{b}|^{2}=|\vec{a}|^{2}+2|b|^{2}$
<br/><br/>or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
<br/><br/>$\therefore|\vec{b}|^{2}=6$
<br/><br/>Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
<br/><br/>$$
75=|\vec{a}|^{2} \cdot 6-9
$$
<br/><br/>$\therefore|\vec{a}|^{2}=14$ | integer | jee-main-2022-online-29th-july-evening-shift |
ldo7cuo5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be three vectors. If $\vec{r}$ is a vector such
that, $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a}=0$, then $25|\vec{r}|^{2}$ is equal to : | [{"identifier": "A", "content": "336"}, {"identifier": "B", "content": "449"}, {"identifier": "C", "content": "339"}, {"identifier": "D", "content": "560"}] | ["C"] | null | $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
<br/><br/>$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
<br/><br/>$\overrightarrow{\mathrm{c}}=\hat{5 \mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
<br/><br/>$(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \vec{b}=0, \quad \vec{r} \cdot \vec{a}=0$
<br/><br/>$\Rightarrow \vec{r}-\vec{c}=\lambda \vec{b}$
<br/><br/>Also, $(\vec{c}+\lambda \vec{b}) \cdot \vec{a}=0$
<br/><br/>$\Rightarrow \vec{a} \cdot \vec{c}+\lambda(\vec{a} \cdot \vec{b})=0$
<br/><br/>$\therefore \lambda=\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}=\frac{-8}{5}$
<br/><br/>$\overrightarrow{\mathrm{r}}=\frac{5(5 \hat{\mathrm{i}}-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})-8(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{5}$
<br/><br/>$$ \Rightarrow $$ $\overrightarrow{\mathrm{r}}=\frac{17 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{5}$
<br/><br/>$$ \Rightarrow $$ $|\overrightarrow{\mathrm{r}}|^{2}=\frac{1}{25}(289+50)$
<br/><br/>$$ \Rightarrow $$ $25|\overrightarrow{\mathrm{r}}|^{2}=339$ | mcq | jee-main-2023-online-31st-january-evening-shift |
ldoaq5ul | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that
<br/><br/>$|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$.
<br/><br/>If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^{2}$ is equal to __________. | [] | null | 3 | $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$
<br/><br/>$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$
<br/><br/>$$
\begin{aligned}
& \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\
& |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\
& 31=31 \lambda^{2} \\\\
& \lambda=\pm 1 \\\\
& \vec{a}=\pm(2 \vec{b}+3 \vec{c}) \\\\
& \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}} \\\\
& |\vec{b} \times \vec{c}|^{2}=\frac{1}{4} \cdot 4-\left(1-\frac{1}{2}\right)^{2} \\\\
& =\frac{3}{4} \\\\
& \therefore \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{\sqrt{3}}{2 \cdot \frac{1}{4}-\frac{3}{2}}=\frac{\sqrt{3}}{-1} \\\\
& \left(\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}\right)^{2}=3
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-evening-shift |
1ldooe6vv | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>$$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$$ and $$D(4,5,0),|\lambda| \leq 5$$ are the vertices of a quadrilateral $$A B C D$$. If its area is 18 square units, then $$5-6 \lambda$$ is equal to __________.</p> | [] | null | 11 | $$
\begin{aligned}
& \mathrm{A}(2,6,2) \quad \mathrm{B}(-4,0, \lambda), \mathrm{C}(2,3,-1) \mathrm{D}(4,5,0) \\\\
& \text { Area }=\frac{1}{2}|\overrightarrow{B D} \times \overrightarrow{A C}|=18 \\\\
& \overrightarrow{A C} \times \overrightarrow{B D}=\left|\begin{array}{ccc}
\hat{i} & j & k \\\\
0 & -3 & -3 \\\\
8 & 5 & -\lambda
\end{array}\right|
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24) \\\\
& \overrightarrow{A C} \times \overrightarrow{B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k} \\\\
& =\sqrt{(3 \lambda+15)^2+(24)^2+(24)^2}=36 \\\\
& =\lambda^2+10 \lambda+9=0 \\\\
& =\lambda=-1,-9 \\\\
& |\lambda| \leq 5 \Rightarrow \lambda=-1 \\\\
& 5-6 \lambda=5-6(-1)=11
\end{aligned}
$$ | integer | jee-main-2023-online-1st-february-morning-shift |
1ldptp6vk | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$$ and $$|\vec{a} \times \vec{b}|=\sqrt{48}$$. Then $$(\vec{a} \cdot \vec{b})^{2}$$ is equal to ___________.</p> | [] | null | 36 | $|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\
& \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\
& \Rightarrow (\vec{a} \cdot \vec{b})^{2}=84-48 \\\\
&=36
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-morning-shift |
ldqx819b | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\vec{a}$ and $\vec{b}$ be two vectors, Let $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$, then the value of $\vec{b} \cdot \vec{c}$ is : | [{"identifier": "A", "content": "$-48$"}, {"identifier": "B", "content": "$-60$"}, {"identifier": "C", "content": "$-84$"}, {"identifier": "D", "content": "$-24$"}] | ["A"] | null | <p>$$\overrightarrow b .\overrightarrow c = \overrightarrow b .(2\overline a \times \overrightarrow b ) - 3\overline b .\overrightarrow b $$</p>
<p>$$ = 0 - 3|\overline b {|^2} = - 48$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldr7kwam | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let a unit vector $$\widehat{O P}$$ make angles $$\alpha, \beta, \gamma$$ with the positive directions of the co-ordinate axes $$\mathrm{OX}$$, $$\mathrm{OY}, \mathrm{OZ}$$ respectively, where $$\beta \in\left(0, \frac{\pi}{2}\right)$$. If $$\widehat{\mathrm{OP}}$$ is perpendicular to the plane through points $$(1,2,3),(2,3,4)$$ and $$(1,5,7)$$, then which one of the following is true?</p> | [{"identifier": "A", "content": "$$\\alpha \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$$ and $$\\gamma \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$$"}, {"identifier": "B", "content": "$$\\alpha \\in\\left(0, \\frac{\\pi}{2}\\right)$$ and $$\\gamma \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$$"}, {"identifier": "C", "content": "$$\\alpha \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$$ and $$\\gamma \\in\\left(0, \\frac{\\pi}{2}\\right)$$"}, {"identifier": "D", "content": "$$\\alpha \\in\\left(0, \\frac{\\pi}{2}\\right)$$ and $$\\gamma \\in\\left(0, \\frac{\\pi}{2}\\right)$$"}] | ["A"] | null | <p>Let $$A \equiv (1,2,3),B \equiv (2,3,4),C \equiv (1,5,7)$$</p>
<p>$$\overrightarrow n = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{
i & j & k \cr
1 & 1 & 1 \cr
0 & 3 & 4 \cr
} } \right|$$</p>
<p>$$ = \widehat i - 4\widehat j + 3\widehat k$$</p>
<p>$$\widehat {OP} = {{ \pm (\widehat i - 4\widehat j + 3\widehat k)} \over {\sqrt {26} }}$$</p>
<p>Since $$\cos \beta > 0$$, take $$-$$ sign</p>
<p>$$\widehat {OP} = {{\widehat i - 4\widehat j + 3\widehat k} \over {\sqrt {26} }}$$</p>
<p>$$ \Rightarrow \cos \alpha < 0,\cos \gamma < 0$$</p>
<p>$$\alpha ,\gamma \in \left( {{\pi \over 2},\pi } \right)$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldseppqy | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>If $$\overrightarrow a = \widehat i + 2\widehat k,\overrightarrow b = \widehat i + \widehat j + \widehat k,\overrightarrow c = 7\widehat i - 3\widehat j + 4\widehat k,\overrightarrow r \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0 $$ and $$\overrightarrow r \,.\,\overrightarrow a = 0$$. Then $$\overrightarrow r \,.\,\overrightarrow c $$ is equal to :</p> | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "34"}, {"identifier": "D", "content": "32"}] | ["C"] | null | <p>$$(\overrightarrow r - \overrightarrow c ) \times \overrightarrow b = 0$$</p>
<p>$$\overrightarrow r = \lambda \overrightarrow b + \overrightarrow c $$</p>
<p>$$ \Rightarrow \lambda \overrightarrow b \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a = 0$$</p>
<p>$$ \Rightarrow \lambda (3) + (7 + 8) = 0$$</p>
<p>$$ \Rightarrow \lambda = - 5$$</p>
<p>$$\overrightarrow r = 5\overrightarrow b + \overrightarrow c $$</p>
<p>$$ = - 5\widehat i - 5\widehat j - 5\widehat k + (7\widehat i + 3\widehat j + 4\widehat k)$$</p>
<p>$$ = 2\widehat i - 8\widehat j - \widehat k$$</p>
<p>$$\therefore$$ $$\overrightarrow r \,.\,\overrightarrow c = 17 + 24 - 4 = 34$$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsfbgwo | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = 4\widehat i + 3\widehat j$$ and $$\overrightarrow b = 3\widehat i - 4\widehat j + 5\widehat k$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow c .\left( {\overrightarrow a \times \overrightarrow b } \right) + 25 = 0,\overrightarrow c \,.(\widehat i + \widehat j + \widehat k) = 4$$, and projection of $$\overrightarrow c $$ on $$\overrightarrow a $$ is 1, then the projection of $$\overrightarrow c $$ on $$\overrightarrow b $$ equals :</p> | [{"identifier": "A", "content": "$$\\frac{3}{\\sqrt2}$$"}, {"identifier": "B", "content": "$$\\frac{1}{\\sqrt2}$$"}, {"identifier": "C", "content": "$$\\frac{1}{5}$$"}, {"identifier": "D", "content": "$$\\frac{5}{\\sqrt2}$$"}] | ["D"] | null | <p>$$[\matrix{
{\overrightarrow c } & {\overrightarrow a } & {\overrightarrow b } \cr
} ] = - 25$$</p>
<p>Let $$\overrightarrow c = l\widehat i + n\widehat j + n\widehat k$$</p>
<p>$$\left| {\matrix{
l & m & n \cr
4 & 3 & 0 \cr
3 & { - 4} & 5 \cr
} } \right| = - 25$$</p>
<p>$$ \Rightarrow 3l - 4m - 5n = - 5$$ ..... (i)</p>
<p>$$\overrightarrow c \,.\,(\widehat i + \widehat j + \widehat k) = 4$$</p>
<p>$$ \Rightarrow l + m + n = 4$$ ..... (ii)</p>
<p>$${{\overrightarrow c \,.\,\overrightarrow a } \over {|\overrightarrow a |}} = 1 \Rightarrow \overrightarrow c \,.\,\overrightarrow a = 5$$</p>
<p>$$ \Rightarrow 4l + 3m = 5$$ ...... (iii)</p>
<p>Using (i), (ii) and (iii)</p>
<p>$$l = 2,m = - 1,n = 3$$</p>
<p>Now, $${{\overrightarrow c \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = {{25} \over {5\sqrt 2 }} = {5 \over {\sqrt 2 }}$$</p>
<p>$$\therefore$$ Option (2) is correct.</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldwxj5lf | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = \widehat i + 2\widehat j + \lambda \widehat k,\overrightarrow b = 3\widehat i - 5\widehat j - \lambda \widehat k,\overrightarrow a \,.\,\overrightarrow c = 7,2\overrightarrow b \,.\,\overrightarrow c + 43 = 0,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow c $$. Then $$\left| {\overrightarrow a \,.\,\overrightarrow b } \right|$$ is equal to :</p> | [] | null | 8 | $$
\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7 \\\\
& \vec{a} \times \vec{c}-\vec{b} \times \vec{c}=\overrightarrow{0} \\\\
& (\vec{a}-\vec{b}) \times \vec{c}=0 \Rightarrow(\vec{a}-\vec{b}) \text { is paralleled to } \vec{c} \\\\
& \vec{a}-\vec{b}=\mu \vec{c}, \text { where } \mu \text { is a scalar } \\\\
& -2 \hat{i}+7 \hat{\mathrm{j}}+2 \lambda \hat{k}=\mu \cdot \overrightarrow{\mathrm{c}}
\end{aligned}
$$<br/><br/>
Now $\vec{a} \cdot \overrightarrow{\mathbf{c}}=7$ gives $2 \lambda^2+12=7 \mu$<br/><br/>
And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^2+82=43 \mu$<br/><br/> $\mu=2$ and $\lambda^2=1$<br/><br/>
$$
|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=8
$$ | integer | jee-main-2023-online-24th-january-evening-shift |
1lgow35tq | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$|\vec{a}|=2,|\vec{b}|=3$$ and the angle between the vectors $$\vec{a}$$ and $$\vec{b}$$ be $$\frac{\pi}{4}$$. Then $$|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "441"}, {"identifier": "B", "content": "482"}, {"identifier": "C", "content": "841"}, {"identifier": "D", "content": "882"}] | ["D"] | null | $$
\begin{aligned}
& |\vec{a}|=2 \\\\
& |\vec{b}|=3 \\\\
& \vec{a} \cdot \vec{b}=\frac{\pi}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\
& = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\
& = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2 \\\\
& = |-7 \vec{a} \times \vec{b}|^2 \\\\
& = \left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2
\end{aligned}
$$
<br/><br/>= $$
49 \times 4 \times 9 \times \frac{1}{2}=882
$$ | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgoxv6q0 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let for a triangle $$\mathrm{ABC}$$,</p>
<p>$$\overrightarrow{\mathrm{AB}}=-2 \hat{i}+\hat{j}+3 \hat{k}$$</p>
<p>$$\overrightarrow{\mathrm{CB}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$$</p>
<p>$$\overrightarrow{\mathrm{CA}}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$$</p>
<p>If $$\delta > 0$$ and the area of the triangle $$\mathrm{ABC}$$ is $$5 \sqrt{6}$$, then $$\overrightarrow{C B} \cdot \overrightarrow{C A}$$ is equal to</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "120"}, {"identifier": "D", "content": "108"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh1phdby/e74db975-74e1-4987-80c4-b8ae4380dc69/bb54dde0-e665-11ed-9cfb-9bcc868d407f/file-1lh1phdbz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh1phdby/e74db975-74e1-4987-80c4-b8ae4380dc69/bb54dde0-e665-11ed-9cfb-9bcc868d407f/file-1lh1phdbz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 13th April Evening Shift Mathematics - Vector Algebra Question 58 English Explanation">
<br><br>$$
\begin{aligned}
& C A+A B=C B \\\\
& \Rightarrow +2 i+4 j+(\delta+3) k=\alpha i+\beta j+\gamma k \\\\
& \Rightarrow \alpha=+2, \beta=4, \gamma=\delta+3
\end{aligned}
$$
<br><br>$$
\begin{aligned}
\text { Area } & =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|=\left| {{1 \over 2}\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{ - 2} & 1 & 3 \cr
{ + 2} & 4 & \gamma \cr
} } \right|} \right|=5 \sqrt{6} \\\\
& =(\gamma-12)^2+(6+2 \gamma)^2+100=(10 \sqrt{6})^2 \\\\
& \Rightarrow 5 \gamma^2=320
\end{aligned}
$$
<br><br>$$
\begin{aligned}
\Rightarrow\gamma^2 =64 \\\\
\Rightarrow\gamma =8 \\\\
\Rightarrow\delta =5 \\\\
\overrightarrow{C B} \cdot \overrightarrow{C A} & =(2 i+4 j+8 k)(4 i+3 j+5 k) \\\\
& =8+12+40 \\\\
& =60
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpxqwcg | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$ and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$. If a vector $$\vec{d}$$ satisfies $$\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$$ and $$\vec{d} \cdot \vec{a}=24$$, then $$|\vec{d}|^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "313"}, {"identifier": "B", "content": "413"}, {"identifier": "C", "content": "423"}, {"identifier": "D", "content": "323"}] | ["B"] | null | Given that $$\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$$, we can rewrite this as:
<br/><br/>$$(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$$
<br/><br/>This implies that the vector $$\vec{d} - \vec{c}$$ is a scalar multiple of $$\vec{b}$$:
<br/><br/>$$\vec{d} = \vec{c} + \lambda \vec{b}$$
<br/><br/>Also, we are given that $$\vec{d} \cdot \vec{a} = 24$$:
<br/><br/>$$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24$$
<br/><br/>Now, we can find the value of $$\lambda$$ :
<br/><br/>$$\lambda = \frac{24 - \vec{a} \cdot \vec{c}}{\vec{b} \cdot \vec{a}} = \frac{24 - 6}{9} = 2$$
<br/><br/>Therefore, we have :
<br/><br/>$$\vec{d} = \vec{c} + 2(\vec{b}) = 8\hat{i} - 5\hat{j} + 18\hat{k}$$
<br/><br/>Now, we can find the squared magnitude of $$\vec{d}$$ :
<br/><br/>$$|\vec{d}|^2 = 64 + 25 + 324 = 413$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgq10ehr | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$$. If $$\vec{b}$$ is a vector such that $$\vec{a}=\vec{b} \times \vec{c}$$ and $$|\vec{b}|^{2}=50$$, then $$|72-| \vec{b}+\left.\vec{c}\right|^{2} \mid$$ is equal to __________.</p> | [] | null | 66 | <p>Given that $$\vec{a} = \vec{b} \times \vec{c}$$, we can find the magnitudes of $$\vec{a}$$ and $$\vec{c}$$:</p>
<p>$$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$$
<br/><br/>$$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$$</p>
<p>We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :</p>
<p>$$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$$</p>
<p>Plugging in the known values :</p>
<p>$$\sqrt{11} = \sqrt{50}\sqrt{22}\sin\theta$$</p>
<p>Solving for the sine of the angle between the vectors :</p>
<p>$$\sin\theta = \frac{1}{10}$$</p>
<p>Now we can find $$|\vec{b} + \vec{c}|^2$$ using the formula :</p>
<p>$$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$$</p>
<p>We have the dot product $$\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$$, and we can use the relationship between sine and cosine: $$\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{99}}{10}$$.</p>
<p>Substitute the values into the formula :</p>
<p>$$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10} = 72 + 66$$</p>
<p>Finally, we need to find the absolute value of the difference :</p>
<p>$$|72 - |\vec{b} + \vec{c}|^2| = |72 - (72 + 66)| = 66$$</p>
| integer | jee-main-2023-online-13th-april-morning-shift |
1lgsw25ll | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$$ and $$\vec{b}=\hat{i}+\hat{j}-\hat{k}$$. If $$\vec{c}$$ is a vector such that $$\vec{a} \cdot \vec{c}=11,
\vec{b} \cdot(\vec{a} \times \vec{c})=27$$ and $$\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$$, then $$|\vec{a} \times \vec{c}|^{2}$$ is equal to _________.</p> | [] | null | 285 | Given,
<br/><br/>$$
\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\
& \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\
& \vec{a} \cdot \vec{c}=11 \\\\
& \vec{b} \cdot(\vec{a} \times \vec{c})=27 \\\\
& \vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}| \\\\
& (\vec{b} \times \vec{a}) \cdot \vec{c}=27
\end{aligned}
$$
<br/><br/>$$
\text { Let } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}
$$
<br/><br/>As $$\vec{a} \cdot \vec{c}=11$$
<br/><br/>$$ \therefore $$ $$
c_1+2 c_2+3 c_3=11
$$ ......(i)
<br/><br/>Also, $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$
<br/><br/>$$
\begin{aligned}
& \therefore c_1+c_2-c_3=-\sqrt{3} \sqrt{3} \\\\
& \Rightarrow c_1+c_2-c_3=-3 ......(ii)
\end{aligned}
$$
<br/><br/>Also, $\vec{b} \cdot(\vec{a} \times \vec{c})=27$
<br/><br/>$$ \therefore $$ $$
5 c_1-4 c_2+c_3=27
$$ ...........(iii)
<br/><br/>From (i), (ii) & (iii)
<br/><br/>$$
\vec{c}=3 \hat{i}-2 \hat{j}+4 \hat{k}
$$
<br/><br/>$$
\begin{aligned}
& |\vec{a} \times \vec{c}|^2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
3 & -2 & +4
\end{array}\right|^2 \\\\
& =|14 \hat{i}+5 \hat{j}-8 \hat{k}|^2 \\\\
& =14^2+5^2+8^2=285
\end{aligned}
$$ | integer | jee-main-2023-online-11th-april-evening-shift |
1lguvx5sl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}$$ be a non-zero vector parallel to the line of intersection of the two planes described by $$\hat{i}+\hat{j}, \hat{i}+\hat{k}$$ and $$\hat{i}-\hat{j}, \hat{j}-\hat{k}$$. If $$\theta$$ is the angle between the vector $$\vec{a}$$ and the vector $$\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}$$ and $$\vec{a} \cdot \vec{b}=6$$, then the ordered pair $$(\theta,|\vec{a} \times \vec{b}|)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\left(\\frac{\\pi}{3}, 3 \\sqrt{6}\\right)$$"}, {"identifier": "B", "content": "$$\\left(\\frac{\\pi}{3}, 6\\right)$$"}, {"identifier": "C", "content": "$$\\left(\\frac{\\pi}{4}, 3 \\sqrt{6}\\right)$$"}, {"identifier": "D", "content": "$$\\left(\\frac{\\pi}{4}, 6\\right)$$"}] | ["D"] | null | We have, $$\vec{a}$$ is non-zero vector parallel to the line of intersection of the two planes described by $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$.
<br/><br/>Let $\mathbf{n}_1$ and $\mathbf{n}_2$ are the normal vector to the plane $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$, respectively.
<br/><br/>$$
\begin{aligned}
& n_1=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\
& n_2=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & -1 & 0 \\
1 & 0 & -1
\end{array}\right|=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\
& \vec{a}=\lambda\left|n_1 \times n_2\right|
\end{aligned}
$$
<br/><br/>[ $\because \mathbf{a}$ is parallel to line of intersection of both planes]
<br/><br/>$$
\begin{aligned}
& =\lambda\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & -1 & -1 \\
1 & 1 & 1
\end{array}\right| \\\\
& =\lambda(-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})
\end{aligned}
$$
<br/><br/>$$\vec{a} \cdot \vec{b} =6 [Given]$$
<br/><br/>$$\lambda(0+4+2) =6 $$
<br/><br/>$$\lambda =1$$
<br/><br/>$$
\begin{aligned}
\therefore \vec{a} & =-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\
\cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{6}{\sqrt{4+4} \sqrt{4+4+1}} \\\\
& =\frac{1}{\sqrt{2}}
\end{aligned}
$$
<br/><br/>$$
\therefore \theta=\frac{\pi}{4}
$$
<br/><br/>$$
\begin{aligned}
\vec{a} \times \vec{b} & =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
0 & -2 & 2 \\
2 & -2 & 1
\end{array}\right| \\\\
& =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\
|\vec{a} \times \vec{b}| & =\sqrt{4+16+16}=6 \\\\
\text { Hence, }(\theta,|\vec{a} \times \vec{b}|) & =\left(\frac{\pi}{4}, 6\right)
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvq0gag | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$$ and $$\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{d}$$ be a vector which is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$\vec{c} \cdot \vec{d}=12$$. Then $$(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$$ is equal to :</p> | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "42"}, {"identifier": "C", "content": "44"}, {"identifier": "D", "content": "48"}] | ["C"] | null | If $\vec{d}$ is $\perp$ to both $\vec{a}$ and $\vec{b}$ then
<br/><br/>$$
\vec{d}=\lambda(\vec{a} \times \vec{b})=\lambda\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 7 & -1 \\
3 & 0 & 5
\end{array}\right|=(35 \hat{i}-13 \hat{j}-21 \hat{k}) \lambda
$$
<br/><br/>$$
\begin{aligned}
& \text { but } \vec{c} \cdot \vec{d}=12 \Rightarrow \lambda(35 \times 1+13 \times 1-21 \times 2)=12 \\\\
& \Rightarrow \lambda(6)=12 \Rightarrow \lambda=2 \\\\
& \vec{\lambda}=2(35 \hat{i}-13 \hat{j}-21 \hat{k})
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, }(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d}) \\\\
& =\left|\begin{array}{ccc}
-1 & 1 & -1 \\
1 & -1 & 2 \\
70 & -26 & -42
\end{array}\right| \\\\
& =-94+182-44=44
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxh78lc | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let O be the origin and the position vector of the point P be $$ - \widehat i - 2\widehat j + 3\widehat k$$. If the position vectors of the points A, B and C are $$ - 2\widehat i + \widehat j - 3\widehat k,2\widehat i + 4\widehat j - 2\widehat k$$ and $$ - 4\widehat i + 2\widehat j - \widehat k$$ respectively, then the projection of the vector $$\overrightarrow {OP} $$ on a vector perpendicular to the vectors $$\overrightarrow {AB} $$ and $$\overrightarrow {AC} $$ is :</p> | [{"identifier": "A", "content": "$$\\frac{7}{3}$$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$$\\frac{10}{3}$$"}, {"identifier": "D", "content": "$$\\frac{8}{3}$$"}] | ["B"] | null | Given, the position vector of point P is :
$ \overrightarrow{OP} = -\widehat{i} - 2\widehat{j} + 3\widehat{k} $
<br/><br/>Position vectors of points A, B, and C are :
<br/><br/>$ \overrightarrow{OA} = -2\widehat{i} + \widehat{j} - 3\widehat{k} $
<br/><br/>$ \overrightarrow{OB} = 2\widehat{i} + 4\widehat{j} - 2\widehat{k} $
<br/><br/>$ \overrightarrow{OC} = -4\widehat{i} + 2\widehat{j} - \widehat{k} $
<br/><br/>Now, vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ can be calculated as :
<br/><br/>$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} $
<br/><br/>$ = (2 + 2)\widehat{i} + (4 - 1)\widehat{j} - (-2 + 3)\widehat{k} $
<br/><br/>$ = 4\widehat{i} + 3\widehat{j} - \widehat{k} $
<br/><br/>$ \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} $
<br/><br/>$ = (-4 + 2)\widehat{i} + (2 - 1)\widehat{j} - (-1 + 3)\widehat{k} $
<br/><br/>$ = -2\widehat{i} + \widehat{j} - 2\widehat{k} $
<br/><br/>$$
\begin{aligned}
\text { Now, } \overrightarrow{A B} \times \overrightarrow{A C} & =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
4 & 3 & 1 \\
-2 & 1 & 2
\end{array}\right| \\\\
& =\hat{\mathbf{i}}(5)+\hat{\mathbf{j}}(-2-8)+\hat{\mathbf{k}}(4+6) \\\\
& =5 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \\\\
&\overrightarrow{O P} =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\end{aligned}
$$
<br/><br/>To find the projection of $ \overrightarrow{OP} $ onto $ \overrightarrow{AB} \times \overrightarrow{AC} $, we need to find the dot product between $ \overrightarrow{OP} $ and the normalized vector $ \overrightarrow{AB} \times \overrightarrow{AC} $.
<br/><br/>First, find the magnitude of $ \overrightarrow{AB} \times \overrightarrow{AC} $ :
<br/><br/>$ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{5^2 + (-10)^2 + 10^2} $
<br/><br/>$ = \sqrt{25 + 100 + 100} $
<br/><br/>$ = \sqrt{225} $
<br/><br/>$ = 15 $
<br/><br/>The projection of vector $ \overrightarrow{OP} $ onto a vector perpendicular to both $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ (which is $ \overrightarrow{AB} \times \overrightarrow{AC}$) is given by :
<br/><br/>$ \frac{\overrightarrow{OP} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})}{| \overrightarrow{AB} \times \overrightarrow{AC} |} $
<br/><br/>= $ \frac{(-\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (5\hat{i} - 10\hat{j} + 10\hat{k})}{\sqrt{25 + 100 + 100}} $
<br/><br/>$ = \frac{-5 + 20 + 30}{15} $
<br/><br/>$ = \frac{45}{15} = 3 $ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgylle5f | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>The area of the quadrilateral $$\mathrm{ABCD}$$ with vertices $$\mathrm{A}(2,1,1), \mathrm{B}(1,2,5), \mathrm{C}(-2,-3,5)$$ and $$\mathrm{D}(1,-6,-7)$$ is equal to :</p> | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "$$8 \\sqrt{38}$$"}, {"identifier": "C", "content": "54"}, {"identifier": "D", "content": "$$9 \\sqrt{38}$$"}] | ["B"] | null | $$
\begin{aligned}
& \text { Here } \overrightarrow{\mathrm{AC}}=(-2-2) \hat{i}+(-3-1) \hat{j}+(5-1) \hat{k} \\\\
& =-4 \hat{i}-4 \hat{j}+4 \hat{k} \\\\
& \overrightarrow{\mathrm{BD}}=(1-1) \hat{i}+(-6-2) \hat{j}+(-7-5) \hat{k} \\\\
& =-8 \hat{j}-12 \hat{k}
\end{aligned}
$$
<br/><br/>So, area of quadrilateral $=\frac{1}{2}| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} \mid$
<br/><br/>$$
\begin{aligned}
& =\frac{1}{2}\left\|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-4 & -4 & 4 \\
0 & -8 & -12
\end{array}\right\| \\\\
& =\frac{1}{2}|(48+32) \hat{i}-(48-0) \hat{j}+(32-0) \hat{k}| \\\\
& =\frac{1}{2}|80 \hat{i}-48 \hat{j}+32 \hat{k}| \\\\
& =\frac{1}{2} 16|15 \hat{i}-3 \hat{j}+2 \hat{k}| \\\\
& =8 \sqrt{25+9+4}=8 \sqrt{38} \text { sq units. }
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh00gxyv | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}, \vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$$ and $$\vec{c}$$ be vectors such that $$\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$$. If <br/><br/>$$\vec{a} \cdot \vec{c}=-12, \vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$$, then $$\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$$ is equal to _______________.</p> | [] | null | 11 | Let $\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$
<br/><br/>Now, $\vec{a} \cdot \vec{c}=-12$
<br/><br/>$$
\Rightarrow 6 c_1+9 c_2+12 c_3=-12
$$ ..............(i)
<br/><br/>Also, $\vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$
<br/><br/>$$
\Rightarrow c_1-2 c_2+c_3=5
$$ ................(ii)
<br/><br/>$$
\begin{aligned}
& \text { Now, } \vec{a} \times \vec{c}=\vec{a} \times \vec{b} \\\\
& \Rightarrow \vec{a} \times(\vec{c}-\vec{b})=0 \\\\
& \Rightarrow \vec{a} \text { is parallel to }(\vec{c}-\vec{b}) \\\\
& \Rightarrow \vec{a}=\lambda(\vec{c}-\vec{b}) \\\\
& \Rightarrow 6 \hat{i}+9 \hat{j}+12 \hat{k}=\lambda\left(c_1-\alpha\right) \hat{i}+\lambda\left(c_2-11\right) \hat{j}+\lambda\left(c_3+2\right) \hat{k}
\end{aligned}
$$
<br/><br/>On comparing, we get
<br/><br/>$$
c_1=\frac{6}{\lambda}+\alpha, c_2=\frac{9}{\lambda}+11, c_3=\frac{12}{\lambda}-2
$$
<br/><br/>Put there values in (ii), we get
<br/><br/>$$
\begin{aligned}
& \frac{6}{\lambda}+\alpha-\frac{18}{\lambda}-22+\frac{12}{\lambda}-2=5 \\\\
& \Rightarrow \alpha=29
\end{aligned}
$$
<br/><br/>From (i) and values of $\mathrm{c}_1, \mathrm{c}_2, \mathrm{c}_3$, and $\alpha$ we have
<br/><br/>$$
\begin{aligned}
& 6\left(\frac{6}{\lambda}+29\right)+9\left(\frac{9}{\lambda}+11\right)+12\left(\frac{12}{\lambda}-2\right)=-12 \\\\
& \Rightarrow \frac{261}{\lambda}=-261 \Rightarrow \lambda=-1
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { So, } c_1=23, c_2=2, c_3=-14 \\\\
& \therefore \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=(23 \hat{i}+2 \hat{j}+-14 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\\\
& =23+2-14=11
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-morning-shift |
1lh21kev1 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$$ and $$\vec{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$$.
If $$\vec{d}$$ is a vector perpendicular to both $$\vec{b}$$ and $$\vec{c}$$, and $$\vec{a} \cdot \vec{d}=18$$, then $$|\vec{a} \times \vec{d}|^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "680"}, {"identifier": "B", "content": "720"}, {"identifier": "C", "content": "760"}, {"identifier": "D", "content": "640"}] | ["B"] | null | <p>Given vectors :
<br/><br/>$ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} $
<br/><br/>$ \vec{b} = \hat{i} - 2\hat{j} - 2\hat{k} $
<br/><br/>$ \vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k} $</p>
<p>Since $ \vec{d} $ is perpendicular to both $ \vec{b} $ and $ \vec{c} $, its direction is given by their cross product :</p>
<p>$ \vec{d} = \lambda(\vec{b} \times \vec{c}) $</p>
<p>$$
\begin{aligned}
\vec{b} \times \vec{c} & =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & -2 & -2 \\
-1 & 4 & 3
\end{array}\right| \\\\
& =\hat{\mathbf{i}}(-6+8)-\hat{\mathbf{j}}(3-2)+\hat{\mathbf{k}}(4-2)=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\end{aligned}
$$</p>
<p>Thus,
$ \vec{b} \times \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $</p>
<p>Given this, $ \vec{d} $ can be expressed as :
<br/><br/>$ \vec{d} = \lambda(2\hat{i} - \hat{j} + 2\hat{k}) $</p>
<p>Using the given condition that $ \vec{a} \cdot \vec{d} = 18 $ :
<br/><br/>$ (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda(2\hat{i} - \hat{j} + 2\hat{k}) = 18 $</p>
<p>$ \Rightarrow 2\lambda(2) - 3\lambda(1) + 4\lambda(2) = 18 $
<br/><br/>$ \Rightarrow \lambda(4 + 8 - 3) = 18 $
<br/><br/>$ \Rightarrow 9\lambda = 18 $
<br/><br/>$ \Rightarrow \lambda = 2 $</p>
<p>So,
$ \vec{d} = 4\hat{i} - 2\hat{j} + 4\hat{k} $</p>
<p>Now, using the identity :
<br/><br/>$ |\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2 $</p>
<p>Given $ |\vec{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29} $
<br/><br/>and $ |\vec{d}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$</p>
<p>Using your corrected calculations :
<br/><br/>$ |\vec{a} \times \vec{d}|^2 = 29 \times 36 - 18^2 = 1044 - 324 = 720 $</p>
| mcq | jee-main-2023-online-6th-april-morning-shift |
lsan8kau | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\overrightarrow{\mathrm{a}}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=-\hat{i}-8 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}=4 \hat{i}+\mathrm{c}_2 \hat{j}+\mathrm{c}_3 \hat{k}$ be three vectors such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$. If the angle between the vector $\overrightarrow{\mathrm{c}}$ and the vector $3 \hat{i}+4 \hat{j}+\hat{k}$ is $\theta$, then the greatest integer less than or equal to $\tan ^2 \theta$ is _______________. | [] | null | 38 | $\begin{aligned} & \vec{a}=\hat{i}+\hat{j}+k \\\\ & \vec{b}=\hat{i}+8 \hat{j}+2 k \\\\ & \vec{c}=4 \hat{i}+c_2 \hat{j}+c_3 k \\\\ & \vec{b} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & (\vec{b}-\vec{c}) \times \vec{a}=0 \\\\ & \vec{b}-\vec{c}=\lambda \vec{\alpha} \\\\ & \vec{b}=\vec{c}+\lambda \vec{\alpha}\end{aligned}$
<br/><br/>$\begin{aligned} & -\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \mathrm{k}=\left(4 \hat{\mathrm{i}}+\mathrm{c}_2 \hat{\mathrm{j}}+\mathrm{c}_3 \mathrm{k}\right)+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}) \\\\ & \lambda+4=-1 \Rightarrow \lambda=-5 \\\\ & \lambda+\mathrm{c}_2=-8 \Rightarrow \mathrm{c}_2=-3 \\\\ & \lambda+\mathrm{c}_3=2 \Rightarrow \mathrm{c}_3=7 \\\\ & \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k}\end{aligned}$
<br/><br/>$\begin{aligned} & \cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}} \\\\ & \tan ^2 \theta=\frac{625 \times 3}{49} \\\\ & {\left[\tan ^2 \theta\right]=38}\end{aligned}$ | integer | jee-main-2024-online-1st-february-evening-shift |
lsaojnb9 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | Let $\overrightarrow{\mathrm{a}}=-5 \hat{i}+\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-4 \hat{k}$ and
<br/><br/>$\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to : | [{"identifier": "A", "content": "-12"}, {"identifier": "B", "content": "-10"}, {"identifier": "C", "content": "-13"}, {"identifier": "D", "content": "-15"}] | ["A"] | null | $\begin{aligned} & \vec{a}=-5 \cdot \hat{i}+\hat{j}-3 \hat{k}, \vec{b}=\hat{i}+2 \hat{j}-4 \hat{k} \\\\ & \vec{c}=(((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i} \\\\ & =(((\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}) \times \hat{i}) \times \hat{i} \\\\ & =((-5 \vec{b}-\vec{a}) \times \hat{i}) \times \hat{i}\end{aligned}$
<br/><br/>$\begin{aligned} & =((-11 \hat{j}+23 \hat{k}) \times \hat{i}) \times \hat{i} \\\\ & = (11 \hat{k}+23 \hat{j}) \times \hat{i} \\\\ & = (11 \hat{j}-23 \hat{k})\end{aligned}$
<br/><br/>$\begin{aligned} & \vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})=0+11-23 \\\\ & =-12\end{aligned}$ | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lscmwyjl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let the position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle be $$2 \hat{i}+2 \hat{j}+\hat{k}, \hat{i}+2 \hat{j}+2 \hat{k}$$ and $$2 \hat{i}+\hat{j}+2 \hat{k}$$ respectively. Let $$l_1, l_2$$ and $$l_3$$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $$\mathrm{AB}, \mathrm{BC}$$ and $$\mathrm{CA}$$ respectively, then $$l_1^2+l_2^2+l_3^2$$ equals:</p> | [{"identifier": "A", "content": "$$\\frac{1}{4}$$"}, {"identifier": "B", "content": "$$\\frac{1}{5}$$"}, {"identifier": "C", "content": "$$\\frac{1}{3}$$"}, {"identifier": "D", "content": "$$\\frac{1}{2}$$"}] | ["D"] | null | <p>$$\triangle \mathrm{ABC}$$ is equilateral</p>
<p>Orthocentre and centroid will be same</p>
<p>$$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1vv91f/efa66592-d248-4f32-8e31-2416eaa514c4/4caa5d30-d411-11ee-b9d5-0585032231f0/file-1lt1vv91g.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1vv91f/efa66592-d248-4f32-8e31-2416eaa514c4/4caa5d30-d411-11ee-b9d5-0585032231f0/file-1lt1vv91g.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Vector Algebra Question 30 English Explanation"></p>
<p>Mid-point of $$\mathrm{AB}$$ is $$\mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right)$$</p>
<p>$$\begin{aligned}
& \therefore \ell_1=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}} \\
& \ell_1=\sqrt{\frac{1}{6}}=\ell_2=\ell_3 \\
& \therefore \ell_1^2+\ell_2^2+\ell_3^2=\frac{1}{2}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4wwd2 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$$ and $$\vec{c}$$ be a vector such that $$(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$$ and $$(\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c}=-3$$. Then $$|\vec{c}|^2$$ is equal to ________.</p> | [] | null | 38 | <p>$$\begin{aligned}
& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
& (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{c}}=2(7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
& \left|\begin{array}{lrr}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
5 & 1 & 4 \\
x & y & \mathrm{z}
\end{array}\right|=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\
& \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\
& \mathrm{z}-4 \mathrm{y}=14,4 \mathrm{x}-5 \mathrm{z}=10,5 \mathrm{y}-\mathrm{x}=-20 \\
& (\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \overrightarrow{\mathrm{c}}=-3 \\
& (2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \overrightarrow{\mathrm{c}}=-3 \\
& 2 x+3 y-2 z=-3 \\
& \therefore x=5, y=-3, z=2 \\
& |\overrightarrow{\mathrm{c}}|^2=25+9+4=38
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse565zc | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=4 \hat{i}+\hat{j}+7 \hat{k}$$ and $$\vec{c}=\hat{i}-3 \hat{j}+4 \hat{k}$$ be three vectors. If a vectors $$\vec{p}$$ satisfies $$\vec{p} \times \vec{b}=\vec{c} \times \vec{b}$$ and $$\vec{p} \cdot \vec{a}=0$$, then $$\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$$ is equal to</p> | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "28"}] | ["B"] | null | <p>$$\begin{aligned}
& \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\
& (\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\
& \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{b}}
\end{aligned}$$</p>
<p>Now, $$\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0$$ (given)</p>
<p>$$\begin{aligned}
& \text { So, } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \\
& (3-3-8)+\lambda(12+1-14)=0 \\
& \lambda=-8 \\
& \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}-8 \overrightarrow{\mathrm{b}} \\
& \overrightarrow{\mathrm{p}}=-31 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-52 \hat{\mathrm{k}} \\
& \text { So, } \overrightarrow{\mathrm{p}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\
& =-31+11+52 \\
& =32
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lse59x61 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>The distance of the point $$Q(0,2,-2)$$ form the line passing through the point $$P(5,-4, 3)$$ and perpendicular to the lines $$\vec{r}=(-3 \hat{i}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+5 \hat{k}), \lambda \in \mathbb{R}$$ and $$\vec{r}=(\hat{i}-2 \hat{j}+\hat{k})+\mu(-\hat{i}+3 \hat{j}+2 \hat{k}), \mu \in \mathbb{R}$$ is :</p> | [{"identifier": "A", "content": "$$\\sqrt{74}$$\n"}, {"identifier": "B", "content": "$$\\sqrt{86}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{54}$$\n"}, {"identifier": "D", "content": "$$\\sqrt{20}$$"}] | ["A"] | null | <p>A vector in the direction of the required line can be obtained by cross product of</p>
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 5 \\
-1 & 3 & 2
\end{array}\right| \\\\
& =-9 \hat{i}-9 \hat{j}+9 \hat{k}
\end{aligned}$$</p>
<p>Required line</p>
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda^{\prime}(-9 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}) \\\\
& \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})
\end{aligned}$$</p>
<p>Now distance of $$(0,2,-2)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjezh5/8e16ffb2-7d5b-416d-bc65-2de0bc58b08c/a1b0b090-cc2d-11ee-b20d-39b621d226e3/file-6y3zli1lsnjezh6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsnjezh5/8e16ffb2-7d5b-416d-bc65-2de0bc58b08c/a1b0b090-cc2d-11ee-b20d-39b621d226e3/file-6y3zli1lsnjezh6.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Morning Shift Mathematics - Vector Algebra Question 27 English Explanation"></p>
<p>$$\begin{aligned}
& \text { P.V. of } \mathrm{P} \equiv(5+\lambda) \hat{i}+(\lambda-4) \hat{j}+(3-\lambda) \hat{k} \\\\
& \overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{i}+(\lambda-6) \hat{j}+(5-\lambda) \hat{k} \\\\
& \overrightarrow{\mathrm{AP}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\\\
& 5+\lambda+\lambda-6-5+\lambda=0 \\\\
& \lambda=2 \\\\
& |\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9} \\\\
& |\overrightarrow{\mathrm{AP}}|=\sqrt{74}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lse5sysr | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}|=1,|\vec{b}|=4$$, and $$\vec{a} \cdot \vec{b}=2$$. If $$\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$$ and the angle between $$\vec{b}$$ and $$\vec{c}$$ is $$\alpha$$, then $$192 \sin ^2 \alpha$$ is equal to ________.</p> | [] | null | 48 | <p>$$\begin{aligned}
& \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{b}}-3|\mathrm{b}|^2 \\
& |\mathrm{~b}||c| \cos \alpha=-3|\mathrm{~b}|^2 \\
& |\mathrm{c}| \cos \alpha=-12 \text {, as }|\mathrm{b}|=4 \\
& \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=2 \\
& \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \\
& |c|^2=|(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-3 \overrightarrow{\mathrm{b}}|^2 \\
& =64 \times \frac{3}{4}+144=192 \\
& |c|^2 \cos ^2 \alpha=144 \\
& 192 \cos ^2 \alpha=144 \\
& 192 \sin ^2 \alpha=48
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsfkro4t | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b} \text { and } \overrightarrow{O C}=\vec{b}$$, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $$\mathrm{{{area\,of\,the\,quadrilateral\,OA\,BC} \over {area\,of\,S}}}$$ is equal to _________.</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "10"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8xnpy/836457b4-b43d-438c-b5ce-783fa67cdc3d/c728ff60-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8xnpz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8xnpy/836457b4-b43d-438c-b5ce-783fa67cdc3d/c728ff60-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8xnpz.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Evening Shift Mathematics - Vector Algebra Question 23 English Explanation"></p>
<p>Area of parallelogram, $$S=|\vec{a} \times \vec{b}|$$</p>
<p>Area of quadrilateral $$=\operatorname{Area}(\triangle \mathrm{OAB})+\operatorname{Area}(\triangle \mathrm{OBC})$$</p>
<p>$$\begin{aligned}
& =\frac{1}{2}\{|\vec{a} \times(12 \vec{a}+4 \vec{b})|+|\vec{b} \times(12 \vec{a}+4 \vec{b})|\} \\
& =8|(\vec{a} \times \vec{b})|
\end{aligned}$$</p>
<p>$$\text { Ratio }=\frac{8|(\vec{a} \times \vec{b})|}{|(\vec{a} \times \vec{b})|}=8$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg4dytc | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}, \alpha, \beta \in \mathbb{R}$$. Let a vector $$\vec{b}$$ be such that the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\frac{\pi}{4}$$ and $$|\vec{b}|^2=6$$. If $$\vec{a} \cdot \vec{b}=3 \sqrt{2}$$, then the value of $$\left(\alpha^2+\beta^2\right)|\vec{a} \times \vec{b}|^2$$ is equal to</p> | [{"identifier": "A", "content": "85"}, {"identifier": "B", "content": "90"}, {"identifier": "C", "content": "75"}, {"identifier": "D", "content": "95"}] | ["B"] | null | <p>$$\begin{aligned}
& |\overrightarrow{\mathrm{b}}|^2=6 ;|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=3 \sqrt{2} \\
& |\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{b}}|^2 \cos ^2 \theta=18 \\
& |\overrightarrow{\mathrm{a}}|^2=6
\end{aligned}$$</p>
<p>Also $$1+\alpha^2+\beta^2=6$$</p>
<p>$$\alpha^2+\beta^2=5$$</p>
<p>to find</p>
<p>$$\begin{aligned}
& \left(\alpha^2+\beta^2\right)|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta \\
& =(5)(6)(6)\left(\frac{1}{2}\right) \\
& =90
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsg4jz4o | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{b}|=1$$ and $$|\vec{b} \times \vec{a}|=2$$. Then $$|(\vec{b} \times \vec{a})-\vec{b}|^2$$ is equal to</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["C"] | null | <p>To find the value of $$|(\vec{b} \times \vec{a})-\vec{b}|^2$$, we can use properties of vector operations and magnitudes. Given $$|\vec{b}| = 1$$ and $$|\vec{b} \times \vec{a}| = 2$$, let's break down the calculation step by step:</p>
<p>Firstly, we observe that the cross product of two vectors $$\vec{b} \times \vec{a}$$ is orthogonal (perpendicular) to both $$\vec{b}$$ and $$\vec{a}$$. This means that when we take the dot product of $$\vec{b} \times \vec{a}$$ with either $$\vec{b}$$ or $$\vec{a}$$, the result will be zero due to the orthogonal property. Specifically,</p>
<p>$$(\vec{b} \times \vec{a}) \cdot \vec{b}= 0 $$(because $$\vec{b} \times \vec{a}$$ is perpendicular to both $$\vec{b}$$ and $$\vec{a}$$)</p>
<p>Next, to find the magnitude squared of the vector $$(\vec{b} \times \vec{a}) - \vec{b}$$, we apply the formula for the magnitude squared of a vector subtraction, which can be expressed as:</p>
<p>$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |-\vec{b}|^2 + 2(\vec{b} \times \vec{a}) \cdot (-\vec{b})$$</p>
<p>Since $$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$$ and $$|-\vec{b}| = |\vec{b}|$$, the equation simplifies to:</p>
<p>$$= |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$$</p>
<p>Given that $$|\vec{b} \times \vec{a}| = 2$$ and $$|\vec{b}| = 1$$, substituting these values gives:</p>
<p>$$= 2^2 + 1^2 = 4 + 1 = 5$$</p>
<p>Therefore, the value of $$|(\vec{b} \times \vec{a})-\vec{b}|^2$$ is <strong>5</strong>.</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
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