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1krzmlysr | maths | vector-algebra | algebra-and-modulus-of-vectors | Let a, b and c be distinct positive numbers. If the vectors $$a\widehat i + a\widehat j + c\widehat k,\widehat i+\widehat k$$ and $$c\widehat i + c\widehat j + b\widehat k$$ are co-planar, then c is equal to : | [{"identifier": "A", "content": "$${2 \\over {{1 \\over a} + {1 \\over b}}}$$"}, {"identifier": "B", "content": "$${{a + b} \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over a} + {1 \\over b}$$"}, {"identifier": "D", "content": "$$\\sqrt {ab} $$"}] | ["D"] | null | Because vectors are coplanar<br><br>Hence, $$\left| {\matrix{
a & a & c \cr
1 & 0 & 1 \cr
c & c & b \cr
} } \right| = 0$$<br><br>$$ \Rightarrow {c^2} = ab \Rightarrow c = \sqrt {ab} $$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1l54taa8j | maths | vector-algebra | algebra-and-modulus-of-vectors | Let A, B, C be three points whose position vectors respectively are
<p>$$\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$$</p>
<p>$$\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k,\,\alpha \in R$$</p>
<p>$$\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$$</p>
<p>If $$\alpha$$ is the smallest positive integer for which $$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are noncollinear, then the length of the median, in $$\Delta$$ABC, through A is :</p> | [{"identifier": "A", "content": "$${{\\sqrt {82} } \\over 2}$$"}, {"identifier": "B", "content": "$${{\\sqrt {62} } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {69} } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt {66} } \\over 2}$$"}] | ["A"] | null | $\overrightarrow{A B} \| \overrightarrow{A C}$ if
<br/><br/>
$\frac{1}{2}=\frac{\alpha-4}{-6}=\frac{1}{2}$
<br/><br/>
$\Rightarrow \alpha=1$
<br/><br/>
$\vec{a}, \vec{b}, \vec{c}$ are non-collinear for $\alpha=2$ (smallest positive integer)
<br/><br/>
Mid point of $B C=M\left(\frac{5}{2}, 0, \frac{9}{2}\right)$
<br/><br/>
$A M=\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$ | mcq | jee-main-2022-online-29th-june-evening-shift |
1l6m59jzo | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>Let the vectors $$\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$$ and $$\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$$ be such that for $$\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0$$. Then, the set of all values of $$t$$ is :</p> | [{"identifier": "A", "content": "a non-empty finite set"}, {"identifier": "B", "content": "equal to $$\\mathbf{N}$$"}, {"identifier": "C", "content": "equal to $$\\mathbf{R}-\\{0\\}$$"}, {"identifier": "D", "content": "equal to $$\\mathbf{R}$$"}] | ["C"] | null | <p>Clearly $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ are non-coplanar</p>
<p>$$\left| {\matrix{
{1 + t} & {1 - t} & 1 \cr
{1 - t} & {1 + t} & 2 \cr
t & { - t} & 1 \cr
} } \right| \ne 0$$</p>
<p>$$ \Rightarrow (1 + t)(1 + t + 2t) - (1 - t)(1 - t - 2t) + 1({t^2} - t - t - {t^2}) \ne 0$$</p>
<p>$$ \Rightarrow (3{t^2} + 4t + 1) - (1 - t)(1 - 3t) - 2t \ne 0$$</p>
<p>$$ \Rightarrow (3{t^2} + 4t + 1) - (3{t^2} - 4t + 1) - 2t \ne 0$$</p>
<p>$$ \Rightarrow t \ne 0$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1ldybks1h | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that <br/><br/>$${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$$. Then $${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{5}{2}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Let the position vector of $P, Q, R$ be $\overrightarrow{0}, \overrightarrow{a}, \overrightarrow{b}$
<br><br>$\Rightarrow$ Position vector of $A=\frac{2 \overrightarrow{a}+\overrightarrow{b}}{3}, $
<br><br>Position vector of $B=\frac{2 \overrightarrow{b}}{3}$ and
<br><br>Position vector of $C=\frac{\overrightarrow{a}}{3}$
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lfhv3u5p/8ecc7bc1-f988-47d9-9271-a3d8261cd426/d568c1d0-c7af-11ed-b61f-698c390a560e/file-1lfhv3u5q.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lfhv3u5p/8ecc7bc1-f988-47d9-9271-a3d8261cd426/d568c1d0-c7af-11ed-b61f-698c390a560e/file-1lfhv3u5q.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Morning Shift Mathematics - Vector Algebra Question 61 English Explanation">
<br>$$
\begin{aligned}
& \therefore \overrightarrow{A B}=\frac{2 \vec{b}}{3}-\left(\frac{2 \vec{a}+\vec{b}}{3}\right)=\frac{\vec{b}}{3}-\frac{2 \vec{a}}{3}=\frac{\vec{b}-2 \vec{a}}{3} \\\\
& \overrightarrow{C A}=\frac{2 \vec{a}+\vec{b}}{3}-\frac{\vec{a}}{3}=\frac{\vec{a}+\vec{b}}{3} \\\\
& \text { Area of } \triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\frac{1}{2}|\vec{a} \times \vec{b}| \\\\
& \text { Area of } \triangle A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{A B}| \\
& =\frac{1}{2}\left|\left(\frac{\vec{a}+\vec{b}}{3}\right) \times\left(\frac{\vec{b}-2 \vec{a}}{3}\right)\right|=\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right| \\\\
& \therefore \frac{\text { Area }(\triangle P Q R)}{\text { Area }(\triangle A B C)}=\frac{\frac{1}{2}|\vec{a} \times \vec{b}|}{\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right|}=3 \text { sq. units }
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnwmpwc | maths | vector-algebra | algebra-and-modulus-of-vectors | Let $\mathrm{ABCD}$ be a quadrilateral. If $\mathrm{E}$ and $\mathrm{F}$ are the mid points of the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ respectively and $(\overrightarrow{A B}-\overrightarrow{B C})+(\overrightarrow{A D}-\overrightarrow{D C})=k \overrightarrow{F E}$, then $k$ is equal to : | [{"identifier": "A", "content": "-2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "-4"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$, respectively.</p>
<p>Then the position vector of $E$ is:</p>
<p>$$
\vec{E} = \frac{\vec{a} + \vec{c}}{2}
$$</p>
<p>And the position vector of $F$ is:</p>
<p>$$
\vec{F} = \frac{\vec{b} + \vec{d}}{2}
$$</p>
<p>Now, we are given the equation:</p>
<p>$$
(\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k\overrightarrow{FE}
$$</p>
<p>We can rewrite this equation using the position vectors:</p>
<p>$$
(\vec{b} - \vec{a} - (\vec{c} - \vec{b})) + (\vec{d} - \vec{a} - (\vec{c} - \vec{d})) = k(\vec{E} - \vec{F})
$$</p>
<p>Simplifying the equation, we get:</p>
<p>$$
(2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}(2\vec{E} - 2\vec{F})
$$</p>
<p>Now substitute $\vec{E}$ and $\vec{F}$ expressions we found earlier:</p>
<p>$$
(2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}\left(2\left(\frac{\vec{a} + \vec{c}}{2}\right) - 2\left(\frac{\vec{b} + \vec{d}}{2}\right)\right)
$$</p>
<p>Simplifying the equation:</p>
<p>$$
(2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = -\frac{k}{2}(\vec{b} + \vec{d} - \vec{a} - \vec{c})
$$</p>
<p>Since both sides of the equation are equal:</p>
<p>$$
k = -4
$$</p>
| mcq | jee-main-2023-online-15th-april-morning-shift |
1lguu8s7l | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>For any vector $$\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$$, with $$10\left|a_{i}\right|<1, i=1,2,3$$, consider the following statements :</p>
<p>(A): $$\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\} \leq|\vec{a}|$$</p>
<p>(B) : $$|\vec{a}| \leq 3 \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$$</p> | [{"identifier": "A", "content": "Only (B) is true"}, {"identifier": "B", "content": "Only (A) is true"}, {"identifier": "C", "content": "Neither (A) nor (B) is true"}, {"identifier": "D", "content": "Both (A) and (B) are true"}] | ["D"] | null | We have,
<br/><br/>$$
\begin{aligned}
& 10\left|a_i\right|<1, i=1,2,3 \\\\
& \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\
& |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\
& \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \text {. }
\end{aligned}
$$
<br/><br/>Hence, (A) is true.
<br/><br/>$$
\begin{array}{rlrl}
& |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2} \leq \sqrt{a_1^2+a_1^2+a_1^2} \\\\
& =\sqrt{3}\left|a_1\right| \\\\
\therefore & |\vec{a}|=\sqrt{3}\left|a_1\right|<3\left|a_1\right| \\\\
\therefore & |\vec{a}|<3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\}
\end{array}
$$
<br/><br/>Hence, (B) is also true. | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvq2r23 | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>If the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ are respectively the circumcenter and the orthocentre of a $$\triangle \mathrm{ABC}$$, then $$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$$ is
equal to :</p> | [{"identifier": "A", "content": "$$\\overrightarrow {QP} $$"}, {"identifier": "B", "content": "$$\\overrightarrow {PQ} $$"}, {"identifier": "C", "content": "$$2\\overrightarrow {PQ} $$"}, {"identifier": "D", "content": "$$2\\overrightarrow {QP} $$"}] | ["B"] | null | 1. **Circumcenter $ P $**:
<br/><br/>The circumcenter of a triangle is equidistant from the vertices of the triangle. It is the center of the circumcircle, the circle that passes through all three vertices of the triangle.
<br/><br/>2. **Orthocenter $ Q $**:
<br/><br/>The orthocenter of a triangle is the point of intersection of its three altitudes. An altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side (or its extension).
<br/><br/>3. **Centroid $ G $**:
<br/><br/>The centroid of a triangle is the point of intersection of its medians. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The centroid always divides each median in a 2:1 ratio, with the larger segment being closer to the vertex.
<br/><br/>With the above definitions, it's known that the centroid divides the line segment joining the circumcenter and orthocenter in the ratio 2 : 1, meaning :
<br/><br/>$ \overrightarrow{PG} = \frac{2}{3} \overrightarrow{PQ} $
<br/><br/>$ \overrightarrow{PQ} = 3\overrightarrow{PG} $
<p>The position vector of the centroid $G$ in terms of the vertices $A, B,$ and $C$ is :
<br/><br/>$ \overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} $</p>
<p>Because $P$ is the circumcenter and is at the origin in this problem:
<br/><br/>$ \overrightarrow{PA} = \overrightarrow{A} $
<br/><br/>$ \overrightarrow{PB} = \overrightarrow{B} $
<br/><br/>$ \overrightarrow{PC} = \overrightarrow{C} $</p>
<p>Substituting these into the equation for $G$ :
<br/><br/>$ \overrightarrow{PG} = \frac{\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}}{3} $</p>
<p>Now, using the relationship between $PQ$ and $PG$ established earlier :
<br/><br/>$ \overrightarrow{PQ} = 3\overrightarrow{PG} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} $</p> | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxt48t5 | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $$\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $$, and $$\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \overrightarrow v $$, then $$\alpha ,{\beta ^2}$$ are the roots of the equation :</p> | [{"identifier": "A", "content": "$${x^2} + x - 2 = 0$$"}, {"identifier": "B", "content": "$$3{x^2} + 2x - 1 = 0$$"}, {"identifier": "C", "content": "$$3{x^2} - 2x - 1 = 0$$"}, {"identifier": "D", "content": "$${x^2} - x - 2 = 0$$"}] | ["D"] | null | An arc $P Q$ of a circle subtends a right angle at its centre $O$. The mid-point of an $\operatorname{arc} P Q$ is $R$. So, $P R=R Q$
<br><br>Also given that, $$\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $$, and $$\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \overrightarrow v $$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbavvev/3506a77b-c00f-43d4-873d-aea0b917a5f0/c874fd70-6275-11ee-af05-bf9054b926c1/file-6y3zli1lnbavvew.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnbavvev/3506a77b-c00f-43d4-873d-aea0b917a5f0/c874fd70-6275-11ee-af05-bf9054b926c1/file-6y3zli1lnbavvew.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Vector Algebra Question 46 English Explanation">
<br><br>Let $$\overrightarrow {OP} = \overrightarrow u$$
<br><br>$$
\begin{aligned}
& \overrightarrow{OQ}=\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v} \\\\
& \overrightarrow{O R}=\overrightarrow{v}
\end{aligned}
$$
<br><br>Clearly, $\overrightarrow{q}=\hat{\mathbf{j}}$
<br><br>$$
\overrightarrow{u}=\hat{\mathbf{i}}, \overrightarrow{v}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}
$$
<br><br>Now, given, $\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v}$
<br><br>$$
\begin{array}{ll}
\Rightarrow & \hat{\mathbf{j}}=\alpha(\hat{\mathbf{i}})+\beta\left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}\right) \\\\
\Rightarrow & \hat{\mathbf{j}}=\mathbf{i}\left(\alpha+\frac{\beta}{\sqrt{2}}\right)+\frac{\beta}{\sqrt{2}} \hat{\mathbf{j}}
\end{array}
$$
<br><br>On comparing both sides, we get
<br><br>$$
\begin{array}{rlrl}
\alpha+\frac{\beta}{\sqrt{2}} =0 \text { and } \frac{\beta}{\sqrt{2}}=1 \\\\
\Rightarrow \alpha =-\frac{\beta}{\sqrt{2}} \quad \therefore \beta =\sqrt{2} \\\\
\Rightarrow \alpha =-1, \beta^2=2
\end{array}
$$
<br><br>Now, $\alpha$ and $\beta^2$ are roots of the quadratic equation is $x^2-x-2=0$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgzzzo1k | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>If the points with position vectors $$\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}, \frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$$ are collinear, then $$(19 \alpha-6 \beta)^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "49"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "25"}] | ["C"] | null | Given : Points with position vectors
<br/><br/>$$
\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}
$$
<br/><br/>and $\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$ are collinear.
<br/><br/>So, $\frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11+8}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{2(\alpha-6)}{3}=\frac{-1}{11-\beta}=\frac{2}{19} \\\\
& \Rightarrow \frac{2}{3}(\alpha-6)=\frac{2}{19}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 19 \alpha-114=3 \Rightarrow 19 \alpha=117 \\\\
& \Rightarrow \alpha=\frac{117}{19}
\end{aligned}
$$
<br/><br/>And, $\frac{-1}{11-\beta}=\frac{2}{19}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow-19=22-2 \beta \\\\
& \Rightarrow 2 \beta=41 \\\\
& \Rightarrow \beta=\frac{41}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore(19 \alpha-6 \beta)^2=\left(19 \times \frac{117}{19}-\frac{6 \times 41}{2}\right)^2 \\\\
& =(117-123)^2=36
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
jaoe38c1lsco116r | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>The position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle are $$2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k}$$ and $$-\hat{i}+\hat{j}+3 \hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector $$\mathrm{AD}$$ of $$\angle \mathrm{BAC}$$ where $$\mathrm{D}$$ is on the line segment $$\mathrm{BC}$$, then $$2 l^2$$ equals :</p> | [{"identifier": "A", "content": "45"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "42"}, {"identifier": "D", "content": "49"}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{AB}=5 \\
& \mathrm{AC}=5
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1v5c9e/0c2c210a-ab17-46c0-813d-6d826d2adf6f/7c1dc320-d40e-11ee-b9d5-0585032231f0/file-1lt1v5c9f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1v5c9e/0c2c210a-ab17-46c0-813d-6d826d2adf6f/7c1dc320-d40e-11ee-b9d5-0585032231f0/file-1lt1v5c9f.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 31 English Explanation"></p>
<p>$$\therefore \mathrm{D}$$ is midpoint of $$\mathrm{BC}$$</p>
<p>$$\begin{aligned}
& \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) \\
& \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} \\
& l=\sqrt{\frac{45}{2}} \\
& \therefore 2 l^2=45
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lseyalpj | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that $$\vec{b}$$ and $$\vec{c}$$ are non-collinear. If $$\vec{a}+5 \vec{b}$$ is collinear with $$\vec{c}, \vec{b}+6 \vec{c}$$ is collinear with $$\vec{a}$$ and $$\vec{a}+\alpha \vec{b}+\beta \vec{c}=\overrightarrow{0}$$, then $$\alpha+\beta$$ is equal to</p> | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "$$-$$30"}, {"identifier": "C", "content": "$$-$$25"}, {"identifier": "D", "content": "35"}] | ["D"] | null | <p>$$\begin{aligned}
& \vec{a}+5 \vec{b}=\lambda \vec{c} \\
& \vec{b}+6 \vec{c}=\mu \vec{a}
\end{aligned}$$</p>
<p>Eliminating $$\vec{a}$$</p>
<p>$$\begin{aligned}
& \lambda \overrightarrow{\mathrm{c}}-5 \overrightarrow{\mathrm{b}}=\frac{6}{\mu} \overrightarrow{\mathrm{c}}+\frac{1}{\mu} \overrightarrow{\mathrm{b}} \\
& \therefore \mu=\frac{-1}{5}, \lambda=-30 \\
& \alpha=5, \beta=30
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
sQ5ibg4HiGWlSCmT | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a \,\,,\,\,\overrightarrow b \,\,,\,\,\overrightarrow c $$ are vectors such that $$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = 4$$ then $$\left[ {\overrightarrow a \, \times \overrightarrow b \,\,\overrightarrow b \times \,\overrightarrow c \,\,\overrightarrow c \, \times \overrightarrow a } \right] = $$ | [{"identifier": "A", "content": "$$16$$ "}, {"identifier": "B", "content": "$$64$$ "}, {"identifier": "C", "content": "$$4$$ "}, {"identifier": "D", "content": "$$8$$ "}] | ["A"] | null | We have, $$\left[ {\overrightarrow a \times \overrightarrow b \,\,\overrightarrow b \times \overrightarrow c \,\,\overrightarrow c \times \overrightarrow a } \right]$$
<br><br>$$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\,\,\left\{ {\left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow c \times \overrightarrow a } \right)} \right\}$$
<br><br>$$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\,\,\left\{ {\left( {\overrightarrow m \,.\,\overrightarrow a } \right)\overrightarrow c - \left( {\overrightarrow m \,.\,\overrightarrow c } \right)\overrightarrow a } \right\}$$
<br><br>( where $$\overrightarrow m = \overrightarrow b \times \overrightarrow c $$ )
<br><br>$$ = \left\{ {\left( {\overrightarrow a \times \overrightarrow b } \right).\overrightarrow c } \right\}.\,\left\{ {\overrightarrow a \,.\,\left( {\overrightarrow b \times \overrightarrow c } \right)} \right\}$$
<br><br>$$ = {\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]^2} = {4^2} = 16.$$ | mcq | aieee-2002 |
D4Bcevo00Vrh7Gww | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow u \,,\overrightarrow v $$ and $$\overrightarrow w $$ are three non-coplanar vectors, then $$\,\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u - \overrightarrow v } \right) \times \left( {\overrightarrow v - \overrightarrow w} \right)$$ equals : | [{"identifier": "A", "content": "$$3\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ "}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ "}, {"identifier": "D", "content": "$$\\overrightarrow u .\\overrightarrow w \\times \\overrightarrow v $$ "}] | ["C"] | null | $$\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u \times \overrightarrow v - \overrightarrow u \times \overrightarrow w - \overrightarrow v \times \overrightarrow v + \overrightarrow v \times \overrightarrow w } \right)$$
<br><br>$$ = \left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u \times \overrightarrow v - \overrightarrow u \times \overrightarrow w + \overrightarrow v \times \overrightarrow w } \right) = \overrightarrow u .\left( {\overrightarrow u \times \overrightarrow v } \right)$$
<br><br>$$ - \overrightarrow u .\left( {\overrightarrow u \times \overrightarrow w } \right) + \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right) + \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow v } \right) - \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow w } \right)$$
<br><br>$$ + \overrightarrow v .\left( {\overrightarrow v \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow v } \right) + \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow w } \right)$$
<br><br>$$ = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right) - \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow v } \right)$$
<br><br>$$ = \left[ {\left. {\overrightarrow u \overrightarrow v \overrightarrow w } \right)} \right] + \left[ {\left. {\overrightarrow v \overrightarrow w \overrightarrow u } \right)} \right] - \left[ {\overrightarrow w \overrightarrow u \overrightarrow v } \right]$$
<br><br>$$ = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right)$$ | mcq | aieee-2003 |
jZuo6aOIGIMwAj1z | maths | vector-algebra | scalar-and-vector-triple-product | If $${\overrightarrow a ,\overrightarrow b ,\overrightarrow c }$$ are non-coplanar vectors and $$\lambda $$ is a real number, then the vectors $${\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ,\,\,\lambda \overrightarrow b + 4\overrightarrow c }$$ and $$\left( {2\lambda - 1} \right)\overrightarrow c $$ are non coplanar for : | [{"identifier": "A", "content": "no value of $$\\lambda $$ "}, {"identifier": "B", "content": "all except one value of $$\\lambda $$ "}, {"identifier": "C", "content": "all except two values of $$\\lambda $$ "}, {"identifier": "D", "content": "all values of $$\\lambda $$ "}] | ["C"] | null | Vectors $$\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ,\lambda \overrightarrow b + 4\overrightarrow c ,\,\,\,$$ <br><br>and $$\left( {2\lambda - 1} \right)\overrightarrow c $$ are
<br><br>coplanar if $$\left| {\matrix{
1 & 2 & 3 \cr
0 & \lambda & 4 \cr
0 & 0 & {2\lambda - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \lambda \left( {2\lambda - 1} \right) = 0 \Rightarrow \lambda = 0$$ or $${1 \over 2}$$
<br><br>$$\therefore$$ Forces are noncoplanar for all $$\lambda ,$$
<br><br>except $$\lambda = 0,{1 \over 2}$$ | mcq | aieee-2004 |
BRq9yBuutoyT3OZ3 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a \,\, = \,\,\widehat i - \widehat k,\,\,\,\,\,\overrightarrow b \,\,\, = \,\,\,x\widehat i + \widehat j\,\,\, + \,\,\,\left( {1 - x} \right)\widehat k$$ and $$\overrightarrow c \,\, = \,\,y\widehat i + x\widehat j + \left( {1 + x - y} \right)\widehat k.$$ Then $$\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right]$$ depends on : | [{"identifier": "A", "content": "only $$y$$ "}, {"identifier": "B", "content": "only $$x$$ "}, {"identifier": "C", "content": "both $$x$$ and $$y$$ "}, {"identifier": "D", "content": "neither $$x$$ nor $$y$$ "}] | ["D"] | null | $$\overrightarrow a = \widehat j - \widehat k,\overrightarrow b = x\widehat i + \overrightarrow j + \left( {1 - x} \right)\widehat k$$
<br><br>and $$\overrightarrow c = y\widehat i + x\widehat j + \left( {1 + x - y} \right)\widehat k$$
<br><br>$$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = \overrightarrow a .\overrightarrow b \times \overrightarrow c = \left| {\matrix{
1 & 0 & { - 1} \cr
x & 1 & {1 - x} \cr
y & x & {1 + x - y} \cr
} } \right|$$
<br><br>$$ = 1\left[ {1 + x - y - x + {x^2}} \right] - \left[ { - {x^2} - y} \right]$$
<br><br>$$ = 1 - y + {x^2} - {x^2} + y = 1$$
<br><br>Hence $$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$$ is independent of $$x$$ and $$y$$ both. | mcq | aieee-2005 |
uM4lvmwUCURnQ18y | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ are non coplanar vectors and $$\lambda $$ is a real number then <br/><br/>$$\left[ {\lambda \left( {\overrightarrow a + \overrightarrow b } \right)\,\,\,\,\,\,\,\,{\lambda ^2}\overrightarrow b \,\,\,\,\,\,\,\,\lambda \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\,\,\,\,\,\,\overrightarrow b + \overrightarrow c \,\,\,\,\,\,\,\,\overrightarrow b } \right]$$ for : | [{"identifier": "A", "content": "exactly one value of $$\\lambda $$ "}, {"identifier": "B", "content": "no value of $$\\lambda $$"}, {"identifier": "C", "content": "exactly three values of $$\\lambda $$"}, {"identifier": "D", "content": "exactly two values of $$\\lambda $$"}] | ["B"] | null | $$\left[ {\lambda \left( {\overrightarrow a + \overrightarrow b } \right){\lambda ^2}\overrightarrow b \,\,\,\lambda \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow b } \right]$$
<br><br>$$ \Rightarrow {\lambda ^4}\left[ {\overrightarrow a + \overrightarrow b \,\,\overrightarrow b \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow b } \right]$$
<br><br>$$ \Rightarrow {\lambda ^4}\left\{ {\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] + \left[ {\overrightarrow b \,\overrightarrow b \,\overrightarrow c } \right]} \right\} = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow b } \right] + \left[ {\overrightarrow a \,\overrightarrow c \,\overrightarrow b } \right]$$
<br><br>$$ \Rightarrow {\lambda ^4}\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = - \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$$
<br><br>$$ \Rightarrow {\lambda ^4} = - 1$$
<br><br>$$ \Rightarrow \lambda \,\,$$ has no real values. | mcq | aieee-2005 |
zsaWIZzGBykVkw8A | maths | vector-algebra | scalar-and-vector-triple-product | If $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)$$ where $${\overrightarrow a ,\overrightarrow b }$$ and $${\overrightarrow c }$$ are any three vectors such that $$\overrightarrow a .\overrightarrow b \ne 0,\,\,\overrightarrow b .\overrightarrow c \ne 0$$ then $${\overrightarrow a }$$ and $${\overrightarrow c }$$ are : | [{"identifier": "A", "content": "inclined at an angle of $${\\pi \\over 3}$$ between them "}, {"identifier": "B", "content": "inclined at an angle of $${\\pi \\over 6}$$ between them "}, {"identifier": "C", "content": "perpendicular "}, {"identifier": "D", "content": "parallel "}] | ["D"] | null | $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\overrightarrow a .\overrightarrow b \ne 0,\,\,\overrightarrow b .\overrightarrow c \ne 0$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right).\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\overrightarrow a .\overrightarrow c } \right).\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c $$
<br><br>$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c = \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a $$
<br><br>$$ \Rightarrow \overrightarrow a ||\overrightarrow c .$$ | mcq | aieee-2006 |
lWNMlgYnWH2QteNt | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat i - \widehat j + 2\widehat k$$ and $$\overrightarrow c = x\widehat i + \left( {x - 2} \right)\widehat j - \widehat k\,\,.$$ If the vectors $$\overrightarrow c $$ lies in the plane of $$\overrightarrow a $$ and $$\overrightarrow b $$, then $$x$$ equals : | [{"identifier": "A", "content": "$$-4$$ "}, {"identifier": "B", "content": "$$-2$$"}, {"identifier": "C", "content": "$$0$$ "}, {"identifier": "D", "content": "$$1.$$"}] | ["B"] | null | Given $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat i - \widehat j + 2\widehat k$$
<br><br>and $$\overrightarrow c = x\widehat i + \left( {x - 2} \right)\widehat j - \widehat k$$
<br><br>If $$\overrightarrow c $$ lies in the plane of $$\overrightarrow a $$ and $$\overrightarrow b ,$$
<br><br>then $$\left[ {\overrightarrow a \,\overrightarrow {b\,} \overrightarrow c } \right] = 0$$
<br><br>i.e.$$\left| {\matrix{
1 & 1 & 1 \cr
1 & { - 1} & 2 \cr
x & {\left( {x - 2} \right)} & { - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow 1\left[ {1 - 2\left( {x - 2} \right)} \right] - 1\left[ { - 1 - 2x} \right]$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 1\left[ {x - 2 + x} \right] = 0$$
<br><br>$$ \Rightarrow 1 - 2x + 4 + 1 + 2x + 2x - 2 = 0$$
<br><br>$$ \Rightarrow 2x = - 4\,\,\,\,\,\,\,\,\, \Rightarrow x = - 2$$ | mcq | aieee-2007 |
QgpDQ7Vm5ZN1ORQI | maths | vector-algebra | scalar-and-vector-triple-product | The vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$$ lies in the plane of the vectors
<br/>$$\overrightarrow b = \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat j + \widehat k$$ and bisects the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$.Then which one of the following gives possible values of $$\alpha $$ and $$\beta $$ ? | [{"identifier": "A", "content": "$$\\alpha = 2,\\,\\,\\beta = 2$$ "}, {"identifier": "B", "content": "$$\\alpha = 1,\\,\\,\\beta = 2$$"}, {"identifier": "C", "content": "$$\\alpha = 2,\\,\\,\\beta = 1$$"}, {"identifier": "D", "content": "$$\\alpha = 1,\\,\\,\\beta = 1$$"}] | ["D"] | null | As $$\overrightarrow a $$ lies in the plane of $$\overrightarrow b $$ and $$\overrightarrow c $$
<br><br>$$\therefore$$ $$\overrightarrow a = \overrightarrow b + \lambda \overrightarrow c $$
<br><br>$$ \Rightarrow \alpha \widehat i + 2\widehat j + \beta \widehat k = \widehat i + \widehat j + \lambda \left( {\widehat j + \widehat k} \right)$$
<br><br>$$ \Rightarrow \alpha = 1,2 = 1 + \lambda ,\,\beta = \lambda \Rightarrow \alpha = 1,\beta = 1$$ | mcq | aieee-2008 |
hZS5SQ5907JAkQoO | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w $$ are non-coplanar vectors and $$p,q$$ are real numbers, then the equality $$\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow w } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow w \,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow w \,\,q\overrightarrow v \,\,q\overrightarrow u } \right] = 0$$ holds for : | [{"identifier": "A", "content": "exactly two values of $$(p,q)$$"}, {"identifier": "B", "content": "more than two but not all values of $$(p,q)$$ "}, {"identifier": "C", "content": "all values of $$(p,q)$$ "}, {"identifier": "D", "content": "exactly one value of $$(p,q)$$"}] | ["D"] | null | $$\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow \omega } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow \omega \,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow \omega \,\,q\overrightarrow v \,\,q\overrightarrow u } \right] = 0$$
<br><br>$$ \Rightarrow \left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow \omega } \right] = 0$$
<br><br>$$ \Rightarrow 3{p^2} - pq + 2{q^2} = 0\,\,$$ $$\,\,\,\,\left( \, \right.$$ As $$\,\,\,\,\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow \omega } \right] \ne 0$$ $$\left. {} \right)$$
<br><br>$$ \Rightarrow 2{p^2} + {p^2} - pq + {{{q^2}} \over 4} + {{7{q^2}} \over 4} = 0$$
<br><br>$$ \Rightarrow 2{p^2} + {\left( {p - {q \over 2}} \right)^2} + {7 \over 4}{q^2} = 0$$
<br><br>$$ \Rightarrow p = 0,q = 0,p = {q \over 2}$$ $$ \Rightarrow p = 0,q = 0$$
<br><br>$$\therefore$$ Exactly one value of $$\left( {p,q} \right)$$ | mcq | aieee-2009 |
FVPtJqT2k8NUVozx | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat j - \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j - \widehat k.$$ Then the vector $$\overrightarrow b $$ satisfying $$\overrightarrow a \times \overrightarrow b + \overrightarrow c = \overrightarrow 0 $$ and $$\overrightarrow a .\overrightarrow b = 3$$ : | [{"identifier": "A", "content": "$$2\\widehat i - \\widehat j + 2\\widehat k$$ "}, {"identifier": "B", "content": "$$\\widehat i - \\widehat j - 2\\widehat k$$"}, {"identifier": "C", "content": "$$\\widehat i + \\widehat j - 2\\widehat k$$"}, {"identifier": "D", "content": "$$-\\widehat i +\\widehat j - 2\\widehat k$$"}] | ["D"] | null | $$\overrightarrow c = \overrightarrow b \times \overrightarrow a $$
<br><br>$$ \Rightarrow \overrightarrow b .\overrightarrow c = \overrightarrow b .\left( {\overrightarrow b \times \overrightarrow a } \right) \Rightarrow \overrightarrow b .\overrightarrow c = 0$$
<br><br>$$ \Rightarrow \left( {{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k} \right).\left( {\widehat i - \widehat j - \widehat k} \right) = 0,$$
<br><br>where $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$$
<br><br>$${b_1} - {b_2} - {b_3} = 0\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$
<br><br>and $$\,\,\,\,\overrightarrow a .\overrightarrow b = 3$$
<br><br>$$ \Rightarrow \left( {\widehat j - \widehat k} \right).\left( {{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k} \right) = 3$$
<br><br>$$ \Rightarrow {b_2} - {b_3} = 3$$
<br><br>From equation $$(i)$$
<br><br>$${b_1} = {b_2} + {b_3} = \left( {3 + {b_3}} \right) + {b_3} = 3 + 2{b_3}$$
<br><br>$$\overrightarrow b = \left( {3 + 2{b_3}} \right)\widehat i + \left( {3 + {b_3}} \right)\widehat j + {b_3}\widehat k$$
<br><br>From the option given, it is clear that $${b_3}$$ equal to either $$2$$ or $$-2.$$
<br><br>$${b_3} = 2$$
<br><br>then $$\overrightarrow b = 7\widehat i + 5\widehat j + 2\widehat k$$ which is not possible
<br><br>If $${b_3} = - 2,$$ then $$\overrightarrow b = - \widehat i + \widehat j - 2\widehat k$$ | mcq | aieee-2010 |
tPY4NnwJZCoCQIE5 | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a = {1 \over {\sqrt {10} }}\left( {3\widehat i + \widehat k} \right)$$ and $$\overrightarrow b = {1 \over 7}\left( {2\widehat i + 3\widehat j - 6\widehat k} \right),$$ then the value
<br/><br>of $$\left( {2\overrightarrow a - \overrightarrow b } \right)\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right)} \right]$$ is :</br> | [{"identifier": "A", "content": "$$-3$$ "}, {"identifier": "B", "content": "$$5$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$-5$$ "}] | ["D"] | null | We have $$\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow a .\overrightarrow a = 1,\,\,\overrightarrow b .\overrightarrow b = 1$$
<br><br>$$\left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right)} \right]$$
<br><br>$$ = \left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\left\{ {\overrightarrow a .\left( {\overrightarrow a + 2\overrightarrow b } \right)} \right\}\overrightarrow b - \left\{ {\overrightarrow b .\left( {\overrightarrow a + 2\overrightarrow b } \right)\overrightarrow a } \right\}} \right]$$
<br><br>$$ = \left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow a .\overrightarrow a + 2\overrightarrow a .\overrightarrow b } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b + 2\overrightarrow b .\overrightarrow b } \right)\overrightarrow a } \right]$$
<br><br>$$ = \left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\overrightarrow b - 2\overrightarrow a } \right]$$
<br><br>$$ = 4\overrightarrow a .\overrightarrow b - \overrightarrow b .\overrightarrow b - 4\overrightarrow a .\overrightarrow a $$
<br><br>$$ = - 5$$ | mcq | aieee-2011 |
fnnbNmLCR5jL4JWe | maths | vector-algebra | scalar-and-vector-triple-product | If $$\left[ {\overrightarrow a \times \overrightarrow b \,\,\,\,\overrightarrow b \times \overrightarrow c \,\,\,\,\overrightarrow c \times \overrightarrow a } \right] = \lambda {\left[ {\overrightarrow a\,\,\,\,\,\,\,\, \overrightarrow b \,\,\,\,\,\,\,\,\overrightarrow c } \right]^2}$$ then $$\lambda $$ is equal to : | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$3$$"}] | ["B"] | null | $$L.H.S$$ $$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow c \times \overrightarrow a } \right)} \right]$$
<br><br>$$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left( {\overrightarrow b \times \overrightarrow c .\overrightarrow a } \right)} \right]\overrightarrow c - \left( {\overrightarrow b \times \overrightarrow c .\overrightarrow c } \right)\left. {\overrightarrow a } \right]$$
<br><br>$$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left[ {\overrightarrow b \,\overrightarrow c \,\overrightarrow a } \right]\overrightarrow c } \right]$$ $$\,\,\,\,\,\,\left[ \, \right.$$As $$\overrightarrow b \times \overrightarrow c .\overrightarrow c = 0$$ $$\left. \, \right]$$
<br><br>$$ = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right].\left( {\overrightarrow a \times \overrightarrow b .\overrightarrow c } \right) = {\left[ {\overrightarrow a \,\,\,\,\,\overrightarrow b \,\,\,\,\,\overrightarrow c } \right]^2}$$
<br><br>$$\left[ {\overrightarrow a \times \overrightarrow b \,\,\,\overrightarrow b \times \overrightarrow c \,\,\,\overrightarrow c \times \overrightarrow a } \right] = {\left[ {\overrightarrow a \,\,\,\,\,\overrightarrow b \,\,\,\,\,\overrightarrow c } \right]^2}$$
<br><br>So $$\,\,\,\,\,\lambda = 1$$ | mcq | jee-main-2014-offline |
GHozmbvUXpiFybyD | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be three non-zero vectors such that no two of them are collinear and <br/><br/>$$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a .$$ If $$\theta $$ is the angle between vectors $$\overrightarrow b $$ and $${\overrightarrow c }$$ , then a value of sin $$\theta $$ is : | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${{ - 2\\sqrt 3 } \\over 3}$$ "}, {"identifier": "C", "content": "$${{ 2\\sqrt 2 } \\over 3}$$"}, {"identifier": "D", "content": "$${{ - \\sqrt 2 } \\over 3}$$ "}] | ["C"] | null | $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>$$ \Rightarrow - \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>$$ \Rightarrow - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a + \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>$$ \Rightarrow - \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\cos \theta \overrightarrow a + \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>$$\therefore$$ $$\,\,\,\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$ are non collinear, the above equation is possible only when
<br><br>$$ - \cos \theta = {1 \over 3}$$ and $$\overrightarrow c .\overrightarrow a = 0$$
<br><br>$$ \Rightarrow \cos \theta = - {1 \over 3}$$
<br><br>$$ \Rightarrow \sin \theta = {{2\sqrt 2 } \over 3};\theta \in \,{\rm I}{\rm I}$$ quad | mcq | jee-main-2015-offline |
6uIl4ZYYXZz8Y3pd | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors such that $$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = {{\sqrt 3 } \over 2}\left( {\overrightarrow b + \overrightarrow c } \right).$$ If $${\overrightarrow b }$$ is not parallel to $${\overrightarrow c },$$ then the angle between $${\overrightarrow a }$$ and $${\overrightarrow b }$$ is: | [{"identifier": "A", "content": "$${{2\\pi } \\over 3}$$ "}, {"identifier": "B", "content": "$${{5\\pi } \\over 6}$$"}, {"identifier": "C", "content": "$${{3\\pi } \\over 4}$$"}, {"identifier": "D", "content": "$${{\\pi } \\over 2}$$"}] | ["B"] | null | $$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = {{\sqrt 3 } \over 2}\left( {\overrightarrow b + \overrightarrow c } \right)$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c = {{\sqrt 3 } \over 2}\overrightarrow b + {{\sqrt 3 } \over 2}\overrightarrow c $$
<br><br>On comparing both sides
<br><br>$$\overrightarrow a .\overrightarrow b = - {{\sqrt 3 } \over 2} \Rightarrow \cos \theta = - {{\sqrt 3 } \over 2}$$
<br><br>$$\left[ \, \right.$$ As $$\overrightarrow a $$ and $$\overrightarrow b $$ are unit vectors $$\left. \, \right]$$
<br><br>where $$\theta $$ is the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$
<br><br>$$\theta = {{5\pi } \over 6}$$ | mcq | jee-main-2016-offline |
EtDY30nGqeWkJ7RGPLaJz | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors, out of which vectors $$\overrightarrow b $$ and $$\overrightarrow c $$ are non-parallel. If $$\alpha $$ and $$\beta $$ are the angles which vector $$\overrightarrow a $$ makes with vectors $$\overrightarrow b $$ and $$\overrightarrow c $$ respectively and $$\overrightarrow a $$ $$ \times $$ ($$\overrightarrow b $$ $$ \times $$ $$\overrightarrow c $$) = $${1 \over 2}\overrightarrow b $$, then $$\left| {\alpha - \beta } \right|$$ is equal to : | [{"identifier": "A", "content": "90<sup>o</sup>"}, {"identifier": "B", "content": "30<sup>o</sup>"}, {"identifier": "C", "content": "45<sup>o</sup>"}, {"identifier": "D", "content": "60<sup>o</sup>"}] | ["B"] | null | $$\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c = {1 \over 2}\overrightarrow b $$
<br><br>$$ \because $$ $$\overrightarrow b \,\,$$ & $$\overrightarrow c \,\,$$ are linearly independent
<br><br>$$ \therefore $$ $$\overrightarrow a \,$$.$$\overrightarrow c \,$$ = $${1 \over 2}$$ & $$\overrightarrow a .\overrightarrow b $$ = 0
<br><br>(All given vectors are unit vectors)
<br><br>$$ \therefore $$ $$\overrightarrow a $$^$$\overrightarrow c $$ = 60<sup>o</sup> & $$\overrightarrow a $$^$$\overrightarrow b $$ = 90<sup>o</sup>
<br><br>$$ \therefore $$ $$\left| {\alpha - \beta } \right| = {30^o}$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
K0mbLDiKg5frblM18B3rsa0w2w9jxaz0egd | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\alpha $$ $$ \in $$ R and the three vectors <br/><br/>$$\overrightarrow a = \alpha \widehat i + \widehat j + 3\widehat k$$, $$\overrightarrow b = 2\widehat i + \widehat j - \alpha \widehat k$$ <br/><br/>and $$\overrightarrow c = \alpha \widehat i - 2\widehat j + 3\widehat k$$. <br/><br/>Then the set
S = {$$\alpha $$ :
$$\overrightarrow a $$ ,
$$\overrightarrow b $$ and
$$\overrightarrow c $$ are coplanar} : | [{"identifier": "A", "content": "contains exactly two numbers only one of which is positive"}, {"identifier": "B", "content": "is singleton"}, {"identifier": "C", "content": "contains exactly two positive numbers"}, {"identifier": "D", "content": "is empty"}] | ["D"] | null | Since these vectors are coplanar then,<br><br>
$$\left| {\matrix{
\alpha & 1 & 3 \cr
2 & 1 & { - \alpha } \cr
\alpha & { - 2} & 3 \cr
} } \right| = 0$$<br><br>
Now, $$\alpha (3 - 2\alpha ) - 1\left( {6 + {\alpha ^2}} \right) + 3\left( { - 4 - \alpha } \right) = 0$$<br><br>
$$ - 3{\alpha ^2} - 18 = 0$$ $$ \Rightarrow $$ $${\alpha ^2}$$ = -6
| mcq | jee-main-2019-online-12th-april-evening-slot |
0v0x2Efs8pBEclG5EP3rsa0w2w9jx6glnbj | maths | vector-algebra | scalar-and-vector-triple-product | If the volume of parallelopiped formed by the vectors $$\widehat i + \lambda \widehat j + \widehat k$$, $$\widehat j + \lambda \widehat k$$ and $$\lambda \widehat i + \widehat k$$ is minimum, then $$\lambda $$ is
equal to : | [{"identifier": "A", "content": "$$ - {1 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${\\sqrt 3 }$$"}, {"identifier": "C", "content": "$$-{\\sqrt 3 }$$"}, {"identifier": "D", "content": "$$ {1 \\over {\\sqrt 3 }}$$"}] | ["D"] | null | $$V = \left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{
1 & \lambda & 1 \cr
0 & 1 & \lambda \cr
\lambda & 0 & 1 \cr
} } \right|$$<br><br>
$$ \Rightarrow 1 - \lambda \left( { - {\lambda ^2}} \right) + 1.\left( {0 - \lambda } \right) = {\lambda ^3} - \lambda + 1$$<br><br>
Whose minimum value occur at $$\lambda $$ = $${1 \over {\sqrt 3 }}$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
XR9wDoMoZhk3UqVQLWL6Q | maths | vector-algebra | scalar-and-vector-triple-product | The sum of the distinct real values of $$\mu $$, for which the vectors, $$\mu \widehat i + \widehat j + \widehat k,$$ $$\widehat i + \mu \widehat j + \widehat k,$$ $$\widehat i + \widehat j + \mu \widehat k$$ are co-planar, is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $$\left| {\matrix{
\mu & 1 & 1 \cr
1 & \mu & 1 \cr
1 & 1 & \mu \cr
} } \right| = 0$$
<br><br>$$\mu \left( {{\mu ^2} - 1} \right) - 1\left( {\mu - 1} \right) + 1\left( {1 - \mu } \right) = 0$$
<br><br>$${\mu ^3} - \mu - \mu + 1 + 1\mu = 0$$
<br><br>$${\mu ^3} - 3\mu + 2 = 0$$
<br><br>$${\mu ^3} - 1 - 3\left( {\mu - 1} \right) = 0$$
<br><br>$$u = 1,\,\,{\mu ^2} + \mu - 2 = 0$$
<br><br>$$\mu = 1,\,\,\mu = - 2$$
<br><br>sum of distinct solutions $$=$$ $$-$$ 1 | mcq | jee-main-2019-online-12th-january-morning-slot |
NaM2UpfKHhYSEwnbzsWn0 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$ = $$\widehat i - \widehat j$$, $$\overrightarrow b $$ = $$\widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c $$
<br/><br/>be a vector such that $$\overrightarrow a $$ × $$\overrightarrow c $$ + $$\overrightarrow b $$ = $$\overrightarrow 0 $$ <br/><br/>and $$\overrightarrow a $$ . $$\overrightarrow c $$ = 4, then |$$\overrightarrow c $$|<sup>2</sup> is equal to : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "$$19 \\over 2$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "$$17 \\over 2$$"}] | ["B"] | null | Given that,
<br><br>$$\overrightarrow a \times \overrightarrow c + \overrightarrow b = \overrightarrow 0 $$
<br><br>$$ \Rightarrow $$ $$\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow a \times \overrightarrow b = \overrightarrow 0 $$
<br><br>$$ \Rightarrow $$ $$\left( {\overrightarrow a \cdot \overrightarrow c } \right)\overrightarrow a - \left( {\overrightarrow a \cdot \overrightarrow a } \right)\overrightarrow c + \overrightarrow a \times \overrightarrow b = \overrightarrow 0 $$
<br><br>given that
<br><br>$$\overrightarrow a \cdot \overrightarrow c = 4$$
<br><br>and $$\overrightarrow a \cdot \overrightarrow a = {\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt 2 } \right)^2} = 2$$
<br><br>$$ \Rightarrow $$ $$4\overrightarrow a - 2\overrightarrow c + \overrightarrow a \times \overrightarrow b = 0$$
<br><br>Now $$\overrightarrow a \times \overrightarrow b $$
<br><br>$$ = \left| {\matrix{
i & j & k \cr
1 & { - 1} & 0 \cr
1 & 1 & 1 \cr
} } \right|$$
<br><br>$$ = - \widehat i - \widehat j + 2\widehat k$$
<br><br>$$ \therefore $$ $$2\overrightarrow c = 4\left( {\widehat i - \widehat j} \right) + \left( { - \widehat i - \widehat j + \widehat k} \right)$$
<br><br>$$ = 4\widehat i - 4\widehat j - \widehat i - \widehat j + \widehat k$$
<br><br>$$ = 3\widehat i - 5\widehat j + \widehat k$$
<br><br>$$ \therefore $$ $$\overrightarrow c = {3 \over 2}\widehat i - {5 \over 2}\widehat j + \widehat k$$
<br><br>$$ \therefore $$ $$\left| {\overrightarrow c } \right| = \sqrt {{9 \over 4} + {{25} \over 4} + 1} $$
<br><br>$$ = \sqrt {{{38} \over 4}} $$
<br><br>$$ = \sqrt {{{19} \over 2}} $$
<br><br>$$ \therefore $$ $${\left| {\overrightarrow c } \right|^2} = {{19} \over 2}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
l8CGSsy1iXuyJWw16U7k9k2k5gpqnac | maths | vector-algebra | scalar-and-vector-triple-product | Let the volume of a parallelopiped whose
coterminous edges are given by
<br/><br>$$\overrightarrow u = \widehat i + \widehat j + \lambda \widehat k$$, $$\overrightarrow v = \widehat i + \widehat j + 3\widehat k$$ and
<br/><br>$$\overrightarrow w = 2\widehat i + \widehat j + \widehat k$$ be 1 cu. unit. If $$\theta $$ be the angle between the
edges $$\overrightarrow u $$ and $$\overrightarrow w $$ , then cos$$\theta $$ can be :</br></br> | [{"identifier": "A", "content": "$${7 \\over {6\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${7 \\over {6\\sqrt 6 }}$$"}, {"identifier": "C", "content": "$${5 \\over 7}$$"}, {"identifier": "D", "content": "$${5 \\over {3\\sqrt 3 }}$$"}] | ["A"] | null | Volume of parallelopiped = 1
<br><br>$$\left| {\left[ {\matrix{
{\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr
} } \right]} \right|$$ = 1
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
1 & 1 & \lambda \cr
1 & 1 & 3 \cr
2 & 1 & 1 \cr
} } \right|$$ = $$ \pm $$1
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 2, 4
<br><br>$$\overrightarrow u = \widehat i + \widehat j + 2 \widehat k$$ or
<br>$$\overrightarrow u = \widehat i + \widehat j + 4 \widehat k$$
<br><br>$$ \therefore $$ cos $$\theta $$ = $${{{\overrightarrow u .\overrightarrow w } \over {\left| {\overrightarrow u } \right|\left| {\overrightarrow w } \right|}}}$$
<br><br>= $${{2 + 1 + 4} \over {\sqrt {18} \sqrt 6 }}$$ or $${{2 + 1 + 2} \over {\sqrt 6 \sqrt 6 }}$$
<br><br>= $${7 \over {6\sqrt 3 }}$$ or $${5 \over 6}$$ | mcq | jee-main-2020-online-8th-january-morning-slot |
uH7b90Y6mDcfmjaFSn7k9k2k5iu6y6w | maths | vector-algebra | scalar-and-vector-triple-product | If the vectors, $$\overrightarrow p = \left( {a + 1} \right)\widehat i + a\widehat j + a\widehat k$$,
<br/><br>
$$\overrightarrow q = a\widehat i + \left( {a + 1} \right)\widehat j + a\widehat k$$ and
<br/><br>
$$\overrightarrow r = a\widehat i + a\widehat j + \left( {a + 1} \right)\widehat k\left( {a \in R} \right)$$
<br/><br/>are coplanar
and $$3{\left( {\overrightarrow p .\overrightarrow q } \right)^2} - \lambda \left| {\overrightarrow r \times \overrightarrow q } \right|^2 = 0$$, then the value of $$\lambda $$ is ______.</br></br> | [] | null | 1 | $$ \because $$ $$\overrightarrow p$$, $$\overrightarrow q$$, $$\overrightarrow r$$ are coplanar
<br><br>$$ \therefore $$ $$\left[ {\matrix{
{\overrightarrow p } & {\overrightarrow q } & {\overrightarrow r } \cr
} } \right]$$ = 0
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{a + 1} & a & a \cr
a & {a + 1} & a \cr
a & a & {a + 1} \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ (a + 1) + a + a = 0
<br><br>$$ \Rightarrow $$ a = $$ - {1 \over 3}$$
<br><br>$$\overrightarrow p .\overrightarrow q $$ = $${1 \over 9}\left( { - 2 - 2 + 1} \right)$$ = $$ - {1 \over 3}$$
<br><br>$$\overrightarrow r \times \overrightarrow q $$ = $${1 \over 9}\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{ - 1} & 2 & { - 1} \cr
{ - 1} & { - 1} & 2 \cr
} } \right|$$
<br><br>= $${1 \over 9}\left( {3\widehat i + 3\widehat j + 3\widehat k} \right)$$
<br><br>= $${{\widehat i + \widehat j + \widehat k} \over 3}$$
<br><br>$$ \Rightarrow $$ $${\left| {\overrightarrow r \times \overrightarrow q } \right|^2} = {1 \over 3}$$
<br><br>Also $$3{\left( {\overrightarrow p .\overrightarrow q } \right)^2} - \lambda \left| {\overrightarrow r \times \overrightarrow q } \right|^2 = 0$$
<br><br>$$ \Rightarrow $$ $$3\left( {{1 \over 9}} \right) - \lambda \left( {{1 \over 3}} \right)$$ = 0
<br><br> $$ \Rightarrow $$ $$\lambda $$ = 1 | integer | jee-main-2020-online-9th-january-morning-slot |
9QNY5X8lrChVbAhY95jgy2xukf7gthwr | maths | vector-algebra | scalar-and-vector-triple-product | Let x<sub>0</sub> be the point of Local maxima of $$f(x) = \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$$, where <br/>$$\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k$$, $$\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k$$, $$\overrightarrow c = 7\widehat i - 2\widehat j + x\widehat k$$. Then the value of <br/>$$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a $$ at x = x<sub>0</sub> is : | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "-30"}, {"identifier": "C", "content": "-4"}, {"identifier": "D", "content": "-22"}] | ["D"] | null | $$f(x) = \overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )$$<br><br>$$ = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$$<br><br>$$ = \left| {\matrix{
x & { - 2} & 3 \cr
{ - 2} & x & { - 1} \cr
7 & { - 2} & x \cr
} } \right|$$<br><br>$$ = x({x^2} + 2) + 2( - 2a + 7) + 3(4 - 7x)$$<br><br>$$ \Rightarrow f(x) = {x^3} - 27x + 26$$<br><br>$$f'(x) = 3{x^2} - 27$$<br><br>For maxima or minima $$f'(x) = 0$$<br><br>$$ \therefore $$ $$3{x^2} - 27 = 0$$<br><br>$$ \Rightarrow x = \pm \,3$$<br><br>Now, $$f''(x) = 6x$$<br><br>f''(x) at x = $$-$$3 is 6($$-$$3) = $$-$$18 < 0<br><br>$$ \therefore $$ At x = $$-$$3 f(x) is maximum.<br><br>So, x<sub>0</sub> = $$-$$3<br><br>$$ \therefore $$ $$\overrightarrow a = - 3\widehat i - 2\widehat j + 3\widehat k$$<br><br>$$\overrightarrow b = - 2\widehat i - 3\widehat j - \widehat k$$<br><br>$$\overrightarrow c = 7\widehat i - 2\widehat j - 3\widehat k$$<br><br>Now, $$\overrightarrow a \,.\,\overrightarrow b \, + \,\overrightarrow b \,.\,\overrightarrow c \, + \,\overrightarrow c \,.\,\overrightarrow a $$<br><br>$$ = (6 + 6 - 3) + ( - 14 + 6 + 3) + ( - 21 + 4 - 9)$$<br><br>$$ = 9 - 5 - 26$$<br><br>$$ = - 22$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
WGSrrJfn4qBtQkI2kKjgy2xukfg6dkb0 | maths | vector-algebra | scalar-and-vector-triple-product | If the volume of a parallelopiped, whose<br/> coterminus edges are given by the <br/>vectors $$\overrightarrow a = \widehat i + \widehat j + n\widehat k$$, <br/>$$\overrightarrow b = 2\widehat i + 4\widehat j - n\widehat k$$ and <br/>$$\overrightarrow c = \widehat i + n\widehat j + 3\widehat k$$ ($$n \ge 0$$), is 158 cu. units, then : | [{"identifier": "A", "content": "n = 7"}, {"identifier": "B", "content": "$$\\overrightarrow b .\\overrightarrow c = 10$$"}, {"identifier": "C", "content": "$$\\overrightarrow a .\\overrightarrow c = 17$$"}, {"identifier": "D", "content": "n = 9"}] | ["B"] | null | We know, Volume(V) = $$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]$$
<br><br>$$ \Rightarrow $$ 158 = $$\left| {\matrix{
1 & 1 & n \cr
2 & 4 & { - n} \cr
1 & n & 3 \cr
} } \right|$$
<br><br>$$ \Rightarrow $$ (12 + n<sup>2</sup>) – (6 + n) + n(2n–4)=158
<br><br>$$ \Rightarrow $$ 3n<sup>2</sup>
–5n + 6 –158 = 0
<br><br>$$ \Rightarrow $$ 3n<sup>2</sup>
– 5n – 152 = 0
<br><br>$$ \Rightarrow $$ 3n<sup>2</sup>
– 24n + 19n – 152 = 0
<br><br>$$ \Rightarrow $$ (3n + 19) (n–8) = 0
<br><br>$$ \Rightarrow $$ n = 8, $$ - {{19} \over 3}$$ (rejected)
<br><br>$$ \therefore $$ $$\overrightarrow a = \widehat i + \widehat j + 8\widehat k$$, <br><br>$$\overrightarrow b = 2\widehat i + 4\widehat j - 8\widehat k$$ and <br><br>$$\overrightarrow c = \widehat i + 8\widehat j + 3\widehat k$$
<br><br>Now $${\overrightarrow a .\overrightarrow c }$$ = 1 + 8 + 24 = 33
<br><br>$${\overrightarrow b .\overrightarrow c }$$ = 2 + 32 - 24 = 10 | mcq | jee-main-2020-online-5th-september-morning-slot |
MgYtxj9nMPDMzcRPl01klrie7jw | maths | vector-algebra | scalar-and-vector-triple-product | Let three vectors $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be such that $$\overrightarrow c $$ is coplanar <br/>with $$\overrightarrow a $$ and $$\overrightarrow b $$,
$$\overrightarrow a .\overrightarrow c $$ = 7 and
$$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$, where <br/>$$\overrightarrow a = - \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow b = 2\widehat i + \widehat k$$ , then the <br/>value of $$2{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$$ is _____. | [] | null | 75 | $$\overrightarrow c = \lambda (\overrightarrow b \times (\overrightarrow a \times \overrightarrow b ))$$<br><br>$$ = \lambda ((\overrightarrow b \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow b \,.\,\overrightarrow a )\overrightarrow b )$$<br><br>$$ = \lambda (5( - \widehat i + \widehat j + \widehat k) + 2\widehat i + \widehat k)$$<br><br>$$ = \lambda ( - 3\widehat i + 5\widehat j + 6\widehat k)$$<br><br>$$\overrightarrow c \,.\,\overrightarrow a = 7 $$
<br/><br/>$$\Rightarrow 3\lambda + 5\lambda + 6\lambda = 7$$<br><br>$$ \Rightarrow $$ $$\lambda = {1 \over 2}$$<br><br>$$ \therefore $$ $$2{\left| {\left( {{{ - 3} \over 2} - 1 + 2} \right)\widehat i + \left( {{5 \over 2} + 1} \right)\widehat j + (3 + 1 + 1)\widehat k} \right|^2}$$<br><br>$$ = 2\left( {{1 \over 4} + {{49} \over 4} + 25} \right) = 25 + 50 = 75$$ | integer | jee-main-2021-online-24th-february-morning-slot |
obXnX2oiZZRcp8NgLj1klugp3ss | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a $$ and $$\overrightarrow b $$ are perpendicular, then <br/>$$\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2}|\\overrightarrow a {|^4}\\overrightarrow b $$"}, {"identifier": "B", "content": "$$\\overrightarrow 0 $$"}, {"identifier": "C", "content": "$$\\overrightarrow a \\times \\overrightarrow b $$"}, {"identifier": "D", "content": "$$|\\overrightarrow a {|^4}\\overrightarrow b $$"}] | ["D"] | null | $$\overrightarrow a \,.\,\overrightarrow b = 0$$<br><br>$$\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ) = (\overrightarrow a \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow a \,.\,\overrightarrow a )\overrightarrow b = - |\overrightarrow a {|^2}\overrightarrow b $$<br><br>Now, $$\overrightarrow a \times (\overrightarrow a \times ( - |\overrightarrow a {|^2}\overrightarrow b ))$$<br><br>$$ = - |\overrightarrow a {|^2}(\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ))$$<br><br>$$ = - |\overrightarrow a {|^2}( - |\overrightarrow a {|^2}\overrightarrow b ) = |\overrightarrow a {|^4}\overrightarrow b $$ | mcq | jee-main-2021-online-26th-february-morning-slot |
XOTDvrbroQnWGWAp5G1kmizgckj | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow c $$ be a vector perpendicular to the vectors, $$\overrightarrow a $$ = $$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$ and <br/>$$\overrightarrow b $$ = $$\widehat i$$ + 2$$\widehat j$$ + $$\widehat k$$. If $$\overrightarrow c \,.\,\left( {\widehat i + \widehat j + 3\widehat k} \right)$$ = 8 then the value of <br/>$$\overrightarrow c $$ . $$\left( {\overrightarrow a \times \overrightarrow b } \right)$$ is equal to __________. | [] | null | 28 | $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 1 & { - 1} \cr
1 & 2 & 1 \cr
} } \right| = (3, - 2,1)$$<br><br>$$\overrightarrow c \bot \overrightarrow a ,\overrightarrow c \bot \overrightarrow b \Rightarrow C||\overrightarrow a \times \overrightarrow b $$<br><br>$$\overrightarrow c = \lambda (\overrightarrow a \times \overrightarrow b )$$<br><br>$$ \Rightarrow \overrightarrow c = \lambda (3\widehat i - 2\widehat j + \widehat k)$$<br><br>Given, $$\overrightarrow c .(\widehat i + \widehat j + 3\widehat k) = 8$$<br><br>$$ \Rightarrow 3\lambda - 2\lambda + 3\lambda = 8$$<br><br>$$ \Rightarrow 4\lambda = 8 \Rightarrow \lambda = 2$$<br><br>$$ \therefore $$ $$\overrightarrow c = 6\widehat i - 4\widehat j + 2\widehat k$$<br><br>$$\overrightarrow c \,.\,(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow c \overrightarrow a \overrightarrow b ] = \left| {\matrix{
6 & { - 4} & 2 \cr
1 & 1 & { - 1} \cr
1 & 2 & 1 \cr
} } \right|$$<br><br>$$ \Rightarrow $$ 18 + 8 + 2 = 28 | integer | jee-main-2021-online-16th-march-evening-shift |
Z7BG0LplvDpHW9S7EN1kmjafjsy | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$ = 2$$\widehat i$$ $$-$$ 3$$\widehat j$$ + 4$$\widehat k$$ and $$\overrightarrow b $$ = 7$$\widehat i$$ + $$\widehat j$$ $$-$$ 6$$\widehat k$$.<br/><br/>If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow r $$ $$\times$$ $$\overrightarrow b $$, $$\overrightarrow r $$ . ($$\widehat i$$ + 2$$\widehat j$$ + $$\widehat k$$) = $$-$$3, then $$\overrightarrow r $$ . (2$$\widehat i$$ $$-$$ 3$$\widehat j$$ + $$\widehat k$$) is equal to : | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "12"}] | ["D"] | null | $$\overrightarrow a = (2, - 3,4)$$, $$\overrightarrow b = (7,1, - 6)$$<br><br>$$\overrightarrow r \times \overrightarrow a - \overrightarrow r \times \overrightarrow b = 0$$<br><br>$$\overrightarrow r \times (\overrightarrow a - \overrightarrow b ) = 0$$<br><br>$$\overrightarrow r = \lambda (\overrightarrow a - \overrightarrow b )$$<br><br>$$\overrightarrow r = \lambda ( - 5\widehat i - 4\widehat j + 10\widehat k)$$<br><br>$$\overrightarrow r \,.\,(2, - 3,1) = ?$$<br><br>Given $$\overrightarrow r \,.\,(1,2,1) = - 3$$<br><br>$$\lambda ( - 5 - 8 + 10) = - 3 \Rightarrow \lambda = 1$$<br><br>$$ \therefore $$ $$( - 5, - 4,10)\,.\,(2, - 3,1)$$<br><br>= - 10 + 12 + 10 = 12 | mcq | jee-main-2021-online-17th-march-morning-shift |
ALfCwHSHBoNiDqWmYY1kmjckog0 | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a = \alpha \widehat i + \beta \widehat j + 3\widehat k$$,<br/><br/>$$\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k$$ and <br/><br/>$$\overrightarrow c = \widehat i - 2\widehat j - \widehat k$$<br/><br/>such that $$\overrightarrow a \,.\,\overrightarrow b = 1$$ and $$\overrightarrow b \,.\,\overrightarrow c = - 3$$, then $${1 \over 3}\left( {\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c } \right)$$ is equal to _____________. | [] | null | 2 | $$\overrightarrow a .\overrightarrow b = 1 \Rightarrow - \alpha \beta - \alpha \beta - 3 = 1$$<br><br>$$ \Rightarrow \alpha \beta = - 2$$ .... (i)<br><br>$$\overrightarrow b .\overrightarrow c = - 3 \Rightarrow - \beta + 2\alpha + 1 = - 3$$<br><br>$$2\alpha - \beta = - 4$$ ..... (ii)<br><br>Solving (i) & (ii) $$\alpha$$ = $$-$$1, $$\beta$$ = 2,<br><br>$${1 \over 3}((\overrightarrow a \, \times \overrightarrow b )\,.\,\overrightarrow c ) = {1 \over 3}\left| {\matrix{
{ - 1} & 2 & 3 \cr
{ - 2} & 1 & { - 1} \cr
1 & { - 2} & { - 1} \cr
} } \right| = 2$$ | integer | jee-main-2021-online-17th-march-morning-shift |
7hj16RzMi655FEX1h41kmkmjktz | maths | vector-algebra | scalar-and-vector-triple-product | Let O be the origin. Let $$\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k$$ and $$\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$$, x, y$$\in$$R, x > 0, be such that $$\left| {\overrightarrow {PQ} } \right| = \sqrt {20} $$ and the vector $$\overrightarrow {OP} $$ is perpendicular $$\overrightarrow {OQ} $$. If $$\overrightarrow {OR} $$ = $$3\widehat i + z\widehat j - 7\widehat k$$, z$$\in$$R, is coplanar with $$\overrightarrow {OP} $$ and $$\overrightarrow {OQ} $$, then the value of x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $$\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k\,$$
<br/><br/>$$\overrightarrow {OP} \bot \overrightarrow {OQ} $$<br><br>$$\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$$<br><br>$$\overrightarrow {PQ} = \left( { - 1 - x} \right)\widehat i + \left( {2 - y} \right)\widehat j + \left( {3x + 1} \right)\widehat k$$<br><br>$$\left| {\overrightarrow {PQ} } \right| = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} $$ <br><br>$$\sqrt {20} = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} $$<br><br>20 = 1 + x<sup>2</sup> + 2x + 4 + y<sup>2</sup> $$-$$ 4y + 9x<sup>2</sup> + 1 + 6x<br><br>20 = 10x<sup>2</sup> + y<sup>2</sup> + 8x + 6 $$-$$ 4y<br><br>20 = 10x<sup>2</sup> + 4x<sup>2</sup> + 8x + 6 $$-$$ 8x<br><br>14 = 14x<sup>2</sup> $$ \Rightarrow $$ x<sup>2</sup> = 1
<br><br>Also, $$\overrightarrow {OP} .\,\overrightarrow {OQ} = 0$$<br><br>$$ - x + 2y - 3x = 0$$<br><br>$$4x = 2y$$<br><br>y = 2x
<br><br>$$ \therefore $$ y<sup>2</sup> = 4x<sup>2</sup> $$ \Rightarrow $$ y<sup>2</sup> = 4<br><br>x = 1 as x > 0 and y = 2<br><br>$$ \therefore $$ $$\left| {\matrix{
x & y & { - 1} \cr
{ - 1} & 2 & {3x} \cr
3 & z & { - 7} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
1 & 2 & { - 1} \cr
{ - 1} & 2 & 3 \cr
3 & z & { - 7} \cr
} } \right|$$ = 0<br><br>$$ \Rightarrow $$ 1($$-$$14 $$-$$3z) $$-$$ 2(7 $$-$$ 9) $$-$$ 1($$-$$z $$-$$6) = 0<br><br>$$ \Rightarrow $$ $$-$$14 $$-$$3z + 4 + z + 6 = 0<br><br>$$ \Rightarrow $$ 2z = $$-$$4 $$ \Rightarrow $$ z = $$-$$2<br><br>$$ \therefore $$ x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 9 | mcq | jee-main-2021-online-17th-march-evening-shift |
1krpwmm6x | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow a .\,\overrightarrow c = \left| {\overrightarrow c } \right|,\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $$ and the angle between $$(\overrightarrow a \times \overrightarrow b )$$ and $$\overrightarrow c $$ is $${\pi \over 6}$$, then the value of $$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right|$$ is : | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["D"] | null | $$\left| {\overrightarrow a } \right| = 3 = a;\overrightarrow a \,.\,\overrightarrow c = c$$<br><br>Now, $$\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $$<br><br>$$ \Rightarrow {c^2} + {a^2} - 2\overrightarrow c \,.\,\overrightarrow a = 8$$<br><br>$$ \Rightarrow {c^2} + 9 - 2(c) = 8$$<br><br>$$ \Rightarrow {c^2} - 2c + 1 = 0 \Rightarrow c = 1 = \left| {\overrightarrow c } \right|$$<br><br>Also, $$\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$$<br><br>Given, $$(\overrightarrow a \times \overrightarrow b ) = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin {\pi \over 6}$$<br><br>$$ = (3)(1)(1/2)$$<br><br>$$ = 3/2$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krtcv0nx | maths | vector-algebra | scalar-and-vector-triple-product | Let a vector $${\overrightarrow a }$$ be coplanar with vectors $$\overrightarrow b = 2\widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j + \widehat k$$. If $${\overrightarrow a}$$ is perpendicular to $$\overrightarrow d = 3\widehat i + 2\widehat j + 6\widehat k$$, and $$\left| {\overrightarrow a } \right| = \sqrt {10} $$. Then a possible value of $$[\matrix{
{\overrightarrow a } & {\overrightarrow b } & {\overrightarrow c } \cr
} ] + [\matrix{
{\overrightarrow a } & {\overrightarrow b } & {\overrightarrow d } \cr
} ] + [\matrix{
{\overrightarrow a } & {\overrightarrow c } & {\overrightarrow d } \cr
} ]$$ is equal to : | [{"identifier": "A", "content": "$$-$$42"}, {"identifier": "B", "content": "$$-$$40"}, {"identifier": "C", "content": "$$-$$29"}, {"identifier": "D", "content": "$$-$$38"}] | ["A"] | null | $$\overrightarrow a = \lambda \overrightarrow b + \mu \overrightarrow c = \widehat i(2\lambda + \mu ) + \widehat j(\lambda - \mu ) + \widehat k(\lambda + \mu )$$<br><br>$$\overrightarrow a \,.\,\overrightarrow d = 0 = 3(2\lambda + \mu ) + 2(\lambda - \mu ) + 6(\lambda + \mu )$$<br><br>$$ \Rightarrow 14\lambda + 7\mu = 0 \Rightarrow \mu = - 2\lambda $$<br><br>$$ \Rightarrow \overrightarrow a = (0)\widehat i - 3\lambda \widehat j + ( - \lambda )\widehat k$$<br><br>$$ \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {10} \left| \lambda \right| = \sqrt {10} \Rightarrow \left| \lambda \right| = 1$$<br><br>$$\lambda = 1$$ or $$ - 1$$<br><br>$$[\overrightarrow a \overrightarrow b \overrightarrow c ]$$ = 0<br><br>$$[\overrightarrow a \overrightarrow b \overrightarrow c ] + [\overrightarrow a \overrightarrow b \overrightarrow d ] + [\overrightarrow a \overrightarrow c \overrightarrow d ] = [\overrightarrow a \overrightarrow b + \overrightarrow c \overrightarrow d ]$$<br><br>$$ = \left| {\matrix{
0 & { - 3\lambda } & \lambda \cr
3 & 0 & 2 \cr
3 & 2 & 6 \cr
} } \right|$$<br><br>$$ = 3\lambda (12) + \lambda (6) = 42\lambda = - 42$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krthnmq3 | maths | vector-algebra | scalar-and-vector-triple-product | Let three vectors $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$, $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$ and $$\left| {\overrightarrow a } \right| = 2$$. Then which one of the following is not true? | [{"identifier": "A", "content": "$$\\overrightarrow a \\times \\left( {(\\overrightarrow b + \\overrightarrow c ) \\times (\\overrightarrow b \\times \\overrightarrow c )} \\right) = \\overrightarrow 0 $$"}, {"identifier": "B", "content": "Projection of $$\\overrightarrow a $$ on $$(\\overrightarrow b \\times \\overrightarrow c )$$ is 2"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n {\\overrightarrow a } & {\\overrightarrow b } & {\\overrightarrow c } \\cr \n\n } } \\right] + \\left[ {\\matrix{\n {\\overrightarrow c } & {\\overrightarrow a } & {\\overrightarrow b } \\cr \n\n } } \\right] = 8$$"}, {"identifier": "D", "content": "$${\\left| {3\\overrightarrow a + \\overrightarrow b - 2\\overrightarrow c } \\right|^2} = 51$$"}] | ["D"] | null | (1) $$\overrightarrow a \times \left( {(\overrightarrow b + \overrightarrow c ) \times (\overrightarrow b \times \overrightarrow c )} \right)$$<br><br>$$ = \overrightarrow a ( - \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow b ) = - 2\left( {\overrightarrow a \times (\overrightarrow b \times \overrightarrow c )} \right)$$<br><br>$$ = - 2(\overrightarrow a \times \overrightarrow a ) = \overrightarrow 0 $$<br><br>(2) Projection of $$\overrightarrow a $$ on $$\overrightarrow b \times \overrightarrow c $$<br><br>$$ = {{\overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )} \over {\left| {\overrightarrow b \times \overrightarrow c } \right|}} = {{\overrightarrow a .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right| = 2$$<br><br>(3) $$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] + \left[ {\overrightarrow c \overrightarrow a \overrightarrow b } \right] = 2\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = 2\overrightarrow a .(\overrightarrow b \times \overrightarrow c )$$<br><br>$$ = 2\overrightarrow a .\overrightarrow a = 2{\left| {\overrightarrow a } \right|^2} = 8$$<br><br>(4) $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$<br><br>$$ \Rightarrow \overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ are mutually $$ \bot $$ vectors.<br><br>$$\therefore$$ $$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow c } \right| \Rightarrow \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| \Rightarrow \left| {\overrightarrow b } \right| = {{\left| {\overrightarrow c } \right|} \over 2}$$<br><br>Also, $$\left| {\overrightarrow b \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right| \Rightarrow \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| = 2 \Rightarrow \left| {\overrightarrow c } \right| = 2$$ & $$\left| {\overrightarrow b } \right| = 1$$<br><br>$${\left| {3\overrightarrow a + \overrightarrow b - 2\overrightarrow c } \right|^2} = (3\overrightarrow a + \overrightarrow b - 2\overrightarrow c ).(3\overrightarrow a + \overrightarrow b - 2\overrightarrow c )$$<br><br>$$ = 9{\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2}$$<br><br>$$ = (9 \times 4) + 1 + (4 \times 4)$$<br><br>$$ = 36 + 1 + 16 = 53$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krvuv970 | maths | vector-algebra | scalar-and-vector-triple-product | Let the vectors<br/><br/>$$(2 + a + b)\widehat i + (a + 2b + c)\widehat j - (b + c)\widehat k,(1 + b)\widehat i + 2b\widehat j - b\widehat k$$ and $$(2 + b)\widehat i + 2b\widehat j + (1 - b)\widehat k$$, $$a,b,c, \in R$$<br><br/> be co-planar. Then which of the following is true?</br> | [{"identifier": "A", "content": "2b = a + c"}, {"identifier": "B", "content": "3c = a + b"}, {"identifier": "C", "content": "a = b + 2c"}, {"identifier": "D", "content": "2a = b + c"}] | ["A"] | null | If the vectors are co-planar,<br><br>$$\left| {\matrix{
{a + b + 2} & {a + 2b + c} & { - b - c} \cr
{b + 1} & {2b} & { - b} \cr
{b + 2} & {2b} & {1 - b} \cr
} } \right| = 0$$<br><br>Now, $${R_3} \to {R_3} - {R_2},{R_1} \to {R_1} - {R_2}$$<br><br>So, $$\left| {\matrix{
{a + 1} & {a + c} & { - c} \cr
{b + 1} & {2b} & { - b} \cr
1 & 0 & 1 \cr
} } \right| = 0$$<br><br>$$ = (a + 1)2b - (a + c)(2b + 1) - c( - 2b)$$<br><br>$$ = 2ab + 2b - 2ab - a - 2bc - c + 2bc$$<br><br>$$ = 2b - a - c = 0$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1kryf0e1y | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three vectors such that $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ ($$\overrightarrow b $$ $$\times$$ $$\overrightarrow c $$). If magnitudes of the vectors $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ are $$\sqrt 2 $$, 1 and 2 respectively and the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$ is $$\theta \left( {0 < \theta < {\pi \over 2}} \right)$$, then the value of 1 + tan$$\theta$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt 3 + 1$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${{\\sqrt 3 + 1} \\over {\\sqrt 3 }}$$"}] | ["B"] | null | $$\overrightarrow a = \left( {\overrightarrow b .\,\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b \,.\,\overrightarrow b } \right)\overrightarrow c $$<br><br>$$ = 1.2\cos \theta \overrightarrow b - \overrightarrow c $$<br><br>$$ \Rightarrow \overrightarrow a = 2\cos \theta \overrightarrow b - \overrightarrow c $$<br><br>$${\left| {\overrightarrow a } \right|^2} = {(2\cos \theta )^2} + {2^2} - 2.2\cos \theta \overrightarrow b \,.\,\overrightarrow c $$<br><br>$$ \Rightarrow 2 = 4{\cos ^2}\theta + 4 - 4\cos \theta .2\cos \theta $$<br><br>$$ \Rightarrow - 2 = - 4{\cos ^2}\theta $$<br><br>$$ \Rightarrow {\cos ^2}\theta = {1 \over 2}$$<br><br>$$ \Rightarrow {\sec ^2}\theta = 2$$<br><br>$$ \Rightarrow {\tan ^2}\theta = 1$$<br><br>$$ \Rightarrow \theta = {\pi \over 4}$$<br><br>$$ \therefore $$ $$1 + \tan \theta = 2$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1kryfi5s7 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat i - \alpha \widehat j + \beta \widehat k$$, $$\overrightarrow b = 3\widehat i + \beta \widehat j - \alpha \widehat k$$ and $$\overrightarrow c = -\alpha \widehat i - 2\widehat j + \widehat k$$, where $$\alpha$$ and $$\beta$$ are integers. If $$\overrightarrow a \,.\,\overrightarrow b = - 1$$ and $$\overrightarrow b \,.\,\overrightarrow c = 10$$, then $$\left( {\overrightarrow a \, \times \overrightarrow b } \right).\,\overrightarrow c $$ is equal to ___________. | [] | null | 9 | $$\overrightarrow a = (1, - \alpha ,\beta )$$<br><br>$$\overrightarrow b = (3,\beta , - \alpha )$$<br><br>$$\overrightarrow c = ( - \alpha , - 2,1);\alpha ,\beta \in I$$<br><br>$$\overrightarrow a \,.\,\overrightarrow b = - 1 \Rightarrow 3 - \alpha \beta - \alpha \beta = - 1$$<br><br>$$ \Rightarrow \alpha \beta = 2$$
<br><br>Possible value of <br>$$\alpha $$ and $$\beta $$
<br><br>$$\matrix{
1 & 2 \cr
2 & 1 \cr
{ - 1} & { - 2} \cr
{ - 2} & { - 1} \cr
} $$<br><br>$$\overrightarrow b \,.\,\overrightarrow c = 10$$<br><br>$$ \Rightarrow - 3\alpha - 2\beta - \alpha = 10$$<br><br>$$ \Rightarrow 2\alpha + \beta + 5 = 0$$<br><br>$$\therefore$$ $$\alpha$$ = $$-$$2; $$\beta$$ = $$-$$1<br><br>$$[\overrightarrow a \,\overrightarrow b \,\overrightarrow c ] = \left| {\matrix{
1 & 2 & { - 1} \cr
3 & { - 1} & 2 \cr
2 & { - 2} & 1 \cr
} } \right|$$<br><br>$$ = 1( - 1 + 4) - 2(3 - 4) - 1( - 6 + 2)$$<br><br>$$ = 3 + 2 + 4 = 9$$ | integer | jee-main-2021-online-27th-july-evening-shift |
1ktbebrpm | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow b = \widehat j - \widehat k$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow a \times \overrightarrow c = \overrightarrow b $$ and $$\overrightarrow a .\overrightarrow c = 3$$, then $$\overrightarrow a .(\overrightarrow b \times \overrightarrow c )$$ is equal to : | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "$$-$$6"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["A"] | null | $$\left| {\overrightarrow a } \right| = \sqrt 3 $$; $$\overrightarrow a .\overrightarrow c = 3$$; $$\overrightarrow a \times \overrightarrow b = - 2\widehat i + \widehat j + \widehat k$$, $$\overrightarrow a \times \overrightarrow c = \overrightarrow b $$<br><br>Cross with $$\overrightarrow a $$,<br><br>$$\overrightarrow a \times (\overrightarrow a \times \overrightarrow c ) = \overrightarrow a \times \overrightarrow b $$<br><br>$$ \Rightarrow (\overrightarrow a .\overrightarrow c )\overrightarrow a - {a^2}\overrightarrow c = \overrightarrow a \times \overrightarrow b $$<br><br>$$ \Rightarrow 3\overrightarrow a - 3\overrightarrow c = - 2\widehat i + \widehat j + \widehat k$$<br><br>$$ \Rightarrow 3\widehat i + 3\widehat j + 3\widehat k - 3\overrightarrow c = - 2\widehat i + \widehat j + \widehat k$$<br><br>$$ \Rightarrow \overrightarrow c = {{5\widehat i} \over 3} + {{2\widehat j} \over 3} + {{2\widehat k} \over 3}$$<br><br>$$\therefore$$ $$\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a \times \overrightarrow b ).\overrightarrow c = {{ - 10} \over 3} + {2 \over 3} + {2 \over 3} = - 2$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktk3sgt1 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ three vectors mutually perpendicular to each other and have same magnitude. If a vector $${ \overrightarrow r } $$ satisfies.
<br/><br/>$$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightarrow b \times \{ (\overrightarrow r - \overrightarrow c ) \times \overrightarrow b \} + \overrightarrow c \times \{ (\overrightarrow r - \overrightarrow a ) \times \overrightarrow c \} = \overrightarrow 0 $$, then $$\overrightarrow r $$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 3}(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c )$$"}, {"identifier": "B", "content": "$${1 \\over 3}(2\\overrightarrow a + \\overrightarrow b - \\overrightarrow c )$$"}, {"identifier": "C", "content": "$${1 \\over 2}(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c )$$"}, {"identifier": "D", "content": "$${1 \\over 2}(\\overrightarrow a + \\overrightarrow b + 2\\overrightarrow c )$$"}] | ["C"] | null | Suppose $$\overrightarrow r = x\overrightarrow a + y\overrightarrow b + 2\overrightarrow c $$<br><br>and $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = k$$<br><br>$$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightarrow b \times \{ (\overrightarrow r - \overrightarrow c ) \times \overrightarrow b \} + \overrightarrow c \times \{ (\overrightarrow r - \overrightarrow a ) \times \overrightarrow c \} = \overrightarrow 0 $$<br><br>$$ \Rightarrow {k^2}(\overrightarrow r - \overrightarrow b ) - {k^2}x\overrightarrow a + {k^2}(\overrightarrow r - \overrightarrow c ) - {k^2}y\overrightarrow b + {k^2}(\overrightarrow r - \overrightarrow a ) - {k^2}z\overrightarrow c = \overrightarrow 0 $$<br><br>$$ \Rightarrow 3\overrightarrow r -(\overrightarrow a + \overrightarrow b + \overrightarrow c ) - \overrightarrow r = \overrightarrow 0 $$<br><br>$$ \Rightarrow \overrightarrow r = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1l58a2gs8 | maths | vector-algebra | scalar-and-vector-triple-product | <p>If $$\overrightarrow a \,.\,\overrightarrow b = 1,\,\overrightarrow b \,.\,\overrightarrow c = 2$$ and $$\overrightarrow c \,.\,\overrightarrow a = 3$$, then the value of $$\left[ {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right),\,\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right)} \right]$$ is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$ - 6\\overrightarrow a \\,.\\,\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)$$"}, {"identifier": "C", "content": "$$ - 12\\overrightarrow c \\,.\\,\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)$$"}, {"identifier": "D", "content": "$$ - 12\\overrightarrow b \\,.\\,\\left( {\\overrightarrow c \\times \\overrightarrow a } \\right)$$"}] | ["A"] | null | <p>$$\because$$ $$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = 3\overrightarrow b - \overrightarrow c = \overrightarrow u $$</p>
<p>$$\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right) = \overrightarrow c - 2\overrightarrow a = \overrightarrow v $$</p>
<p>$$\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right) = 3\overrightarrow b - 2\overrightarrow a = \overrightarrow w $$</p>
<p>$$\therefore$$ $$\overrightarrow u + \overrightarrow v = \overrightarrow w $$</p>
<p>So, vectors $$\overrightarrow u $$, $$\overrightarrow v $$ and $$\overrightarrow w $$ are coplanar, hence their Scalar triple product will be zero.</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l5w0bd5u | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let a vector $$\overrightarrow c $$ be coplanar with the vectors $$\overrightarrow a = - \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow b = 2\widehat i + \widehat j - \widehat k$$. If the vector $$\overrightarrow c $$ also satisfies the conditions $$\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$$ and $$\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$$, then the value of $$|\overrightarrow c {|^2}$$ is equal to :</p> | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "29"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "42"}] | ["C"] | null | <p>Given,</p>
<p>$$\overrightarrow a = - \widehat i + \widehat j + \widehat k$$</p>
<p>$$\overrightarrow b = 2\widehat i + \widehat j - \widehat k$$</p>
<p>and let $$\overrightarrow c = x\widehat i + y\widehat j + z\widehat k$$</p>
<p>Now, $$\overrightarrow a + \overrightarrow b = \widehat i + 2\widehat j$$</p>
<p>and $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{ - 1} & 1 & 1 \cr
2 & 1 & { - 1} \cr
} } \right| = - 2\widehat i + \widehat j - 3\widehat k$$</p>
<p>$$\overrightarrow c $$ is coplanar with $$\overrightarrow a $$ and $$\overrightarrow b $$</p>
<p>$$\therefore$$ $$\left[ {\overrightarrow c \,\overrightarrow a \,\overrightarrow b } \right] = 0$$</p>
<p>$$ \Rightarrow \left| {\matrix{
x & y & z \cr
{ - 1} & 1 & 1 \cr
2 & 1 & { - 1} \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow x( - 2) - y( - 1) + z( - 3) = 0$$</p>
<p>$$ \Rightarrow - 2x + y - 3z = 0$$ ..... (1)</p>
<p>Now, $$\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)$$</p>
<p>$$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 0 \cr
{ - 2} & 1 & { - 3} \cr
} } \right|$$</p>
<p>$$ = - 6\widehat i + 3\widehat j + 5\widehat k$$</p>
<p>Given, $$\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$$</p>
<p>$$ \Rightarrow \left( {x\widehat i + y\widehat j + z\widehat k} \right)\,.\,\left( { - 6\widehat i + 3\widehat j + 5\widehat k} \right) = - 42$$</p>
<p>$$ \Rightarrow - 6x + 3y + 5z = - 42$$ ...... (2)</p>
<p>Now, $$\overrightarrow a - \overrightarrow b = \left( { - \widehat i + \widehat j + \widehat k} \right) - \left( {2\widehat i + \widehat j - \widehat k} \right)$$</p>
<p>$$ = - 3\widehat i + 0\widehat j + 2\widehat k$$</p>
<p>$$\therefore$$ $$\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)$$</p>
<p>$$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
x & y & z \cr
{ - 3} & 0 & 2 \cr
} } \right|$$</p>
<p>$$ = 2y\widehat i - \widehat j(2x + 3z) + 3y\widehat k$$</p>
<p>Given,</p>
<p>$$\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$$</p>
<p>$$ \Rightarrow \left( {2y\widehat i - \widehat j(2x + 3z) + 3y\widehat k} \right)\,.\,\widehat k = 3$$</p>
<p>$$ \Rightarrow 3y = 3$$</p>
<p>$$ \Rightarrow y = 1$$</p>
<p>Putting value of $$y = 1$$ in equation (1) and (2) we get,</p>
<p>$$ - 2x + 1 - 3z = 0$$ ..... (3)</p>
<p>and $$ - 6x + 3 + 5z = - 42$$</p>
<p>$$ \Rightarrow - 6x + 5z = - 45$$ ..... (4)</p>
<p>Solving (3) and (4), we get</p>
<p>$$x = 5$$ and $$z = - 3$$</p>
<p>$$\therefore$$ $$\overrightarrow c = 5\widehat i + \widehat j - 3\widehat k$$</p>
<p>$$ \Rightarrow {\left| {\overrightarrow c } \right|^2} = {5^2} + {1^2} + {( - 3)^2} = 35$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6jc594i | maths | vector-algebra | scalar-and-vector-triple-product | <p>$$
\text { Let } \vec{a}=2 \hat{i}-\hat{j}+5 \hat{k} \text { and } \vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k} \text {. If }((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2} \text {, then }|\vec{b} \times 2 \hat{j}|
$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$\\sqrt{21}$$"}, {"identifier": "D", "content": "$$\\sqrt{17}$$"}] | ["B"] | null | <p>Given, $$\overrightarrow a = 2\widehat i - \widehat j + 5\widehat k$$ and $$\overrightarrow b = \alpha \widehat i + \beta \widehat j + 2\widehat k$$</p>
<p>Also, $$\left( {\left( {\overrightarrow a \times \overrightarrow b } \right) \times i} \right)\,.\,\widehat k = {{23} \over 2}$$</p>
<p>$$ \Rightarrow \left( {\left( {\overrightarrow a \,.\,\widehat i} \right)\overrightarrow b - \left( {\overrightarrow b \,.\,\widehat i} \right)\,.\,\overline a } \right)\,.\,\widehat k = {{23} \over 2}$$</p>
<p>$$ \Rightarrow \left( {2\,.\,\overrightarrow b - \alpha \,.\,\overrightarrow a } \right)\,.\,\widehat k = {{23} \over 2}$$</p>
<p>$$ \Rightarrow 2\,.\,2 - 5\alpha = {{23} \over 2} \Rightarrow \alpha = {{ - 3} \over 2}$$</p>
<p>Now, $$\left| {\overrightarrow b \times 2j} \right| = \left| {\left( {\alpha \widehat i + \beta \widehat j + 2\widehat k} \right) \times 2\widehat j} \right|$$</p>
<p>$$ = \left| {2\alpha \widehat k + 0 - 4\widehat i} \right|$$</p>
<p>$$ = \sqrt {4{\alpha ^2} + 16} $$</p>
<p>$$ = \sqrt {4{{\left( {{{ - 3} \over 2}} \right)}^2} + 16} = 5$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6p23cfm | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow{\mathrm{a}}=3 \hat{i}+\hat{j}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}+\hat{k}$$. Let $$\overrightarrow{\mathrm{c}}$$ be a vector satisfying $$\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{c}}$$. If $$\overrightarrow{\mathrm{b}}$$ and $$\overrightarrow{\mathrm{c}}$$ are non-parallel, then the value of $$\lambda$$ is :</p> | [{"identifier": "A", "content": "$$-$$5"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$-$$1"}] | ["A"] | null | <p>$$\overrightarrow a = 3\widehat i + \widehat j$$ & $$\overrightarrow b = \widehat i + 2\widehat j + \widehat k$$</p>
<p>$$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a \,.\,\overrightarrow c )\overrightarrow b - (\overrightarrow a \,.\,\overrightarrow b )\overrightarrow c = \overrightarrow b + \lambda \overrightarrow c $$</p>
<p>If $$\overrightarrow b $$ & $$\overrightarrow c $$ are non-parallel</p>
<p>then $$\overrightarrow a \,.\,\overrightarrow c = 1$$ & $$\overrightarrow a \,.\,\overrightarrow b = - \lambda $$</p>
<p>but $$\overrightarrow a \,.\,\overrightarrow b = 5 \Rightarrow \lambda = - 5$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldoob9wo | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}, \vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{u}$$ be a vector such that $$|\vec{u}|=\alpha>0$$. If the minimum value of the scalar triple product $$\left[ {\matrix{
{\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr
} } \right]$$ is $$-\alpha \sqrt{3401}$$, and $$|\vec{u} \cdot \hat{i}|^{2}=\frac{m}{n}$$ where $$m$$ and $$n$$ are coprime natural numbers, then $$m+n$$ is equal to ____________.</p> | [] | null | 3501 | $\vec{v} \times \vec{w}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1\end{array}\right|=\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k}$
<br/><br/>$$\left[ {\matrix{
{\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr
} } \right] = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right)$$
<br/><br/>$=|\vec{u}||\vec{v} \times \vec{w}| \times \cos \theta$
<br/><br/>$=\alpha \sqrt{34 \alpha^{2}+1} \cos \theta$
<br/><br/>$[\vec{u} \vec{v} \vec{w}]_{\min }=-\alpha \sqrt{3401}$
<br/><br/>$\alpha \sqrt{34 \alpha^{2}+1} \times(-1)=-\alpha \sqrt{3401}$
<br/><br/>(taking $\cos \theta=1$ )
<br/><br/>$\Rightarrow \alpha=10$
<br/><br/>$\vec{v} \times \vec{w}=\hat{i}-50 \hat{j}-30 \hat{k}$
<br/><br/>$\cos \theta=-1 \Rightarrow \vec{u}$ is antiparallel to $\vec{v} \times \vec{w}$
<br/><br/>$\vec{u}=-|\vec{u}| \cdot \frac{\vec{v} \times \vec{w}}{|\vec{v} \times \vec{w}|}=\frac{-10(\hat{i}-50 \hat{j}-30 \hat{k})}{\sqrt{3401}}$
<br/><br/>$|\vec{u} \cdot \hat{i}|^{2}=\left|\frac{-10}{\sqrt{3401}}\right|^{2}=\frac{100}{3401}=\frac{m}{n}$
<br/><br/>$m+n=3501$ | integer | jee-main-2023-online-1st-february-morning-shift |
ldqwqsa2 | maths | vector-algebra | scalar-and-vector-triple-product | Let $\lambda \in \mathbb{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}$.
<br/><br/>If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$,<br/><br/> then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to : | [{"identifier": "A", "content": "136"}, {"identifier": "B", "content": "140"}, {"identifier": "C", "content": "144"}, {"identifier": "D", "content": "132"}] | ["B"] | null | <p>$$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right) + \overrightarrow b \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right) \times \left( {\overrightarrow a - \overrightarrow b } \right)$$</p>
<p>$$ = \left( {\overrightarrow a \left( {\overrightarrow a .\,\overrightarrow b } \right) - \overrightarrow b \left( {\overrightarrow a .\,\overrightarrow a } \right) + \overrightarrow a \left( {\overrightarrow b .\,\overrightarrow b } \right) - \overrightarrow b \left( {\overrightarrow a .\,\overrightarrow b } \right)} \right) \times \left( {\overrightarrow a - \overrightarrow b } \right)$$</p>
<p>$$ = \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow a - \overrightarrow a \times \overrightarrow b } \right) - \left( {\overrightarrow a .\,\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right) + \left( {\overrightarrow b .\,\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow a - \overrightarrow a \times \overrightarrow b } \right) - \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right)$$</p>
<p>$$ = \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) - \left( {\overrightarrow a .\,\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) + \left( {\overrightarrow b .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right) - \left( {\overrightarrow a .\,\overrightarrow b } \right)\left( {\overrightarrow b \times \overrightarrow a } \right)$$</p>
<p>$$ = \left( {\overrightarrow b \times \overrightarrow a } \right)\left( {\overrightarrow b .\,\overrightarrow b - \overrightarrow a .\,\overrightarrow a } \right)$$</p>
<p>$$ = \left( {5\overrightarrow b \times \overrightarrow a } \right)\left( {5 + {\lambda ^2} - 13 - {\lambda ^2}} \right)$$</p>
<p>$$ = 8\left( {\overrightarrow a \times \overrightarrow b } \right)$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\lambda & 2 & { - 3} \cr
1 & { - \lambda } & 2 \cr
} } \right|$$</p>
<p>$$ = \widehat i(4 - 3\lambda ) - \widehat j(2\lambda + 3) + \widehat k( - {\lambda ^2} - 2)$$</p>
<p>$$ \Rightarrow \lambda = 1$$</p>
<p>$${\left| {\overrightarrow a \times \left( {\overrightarrow a - \overrightarrow b } \right) + \overrightarrow b \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right|^2}$$</p>
<p>$$ = {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} = 4\,.\,35 = 140$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldr70m58 | maths | vector-algebra | scalar-and-vector-triple-product | <p>If $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ are three non-zero vectors and $$\widehat n$$ is a unit vector perpendicular to $$\overrightarrow c $$ such that $$\overrightarrow a = \alpha \overrightarrow b - \widehat n,(\alpha \ne 0)$$ and $$\overrightarrow b \,.\overrightarrow c = 12$$, then $$\left| {\overrightarrow c \times (\overrightarrow a \times \overrightarrow b )} \right|$$ is equal to :</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "12"}] | ["D"] | null | <p>$$\widehat n = \alpha \overrightarrow b - \overrightarrow a $$</p>
<p>$$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = \left( {\overrightarrow c \,.\,\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c \,.\,\overrightarrow a } \right)\overrightarrow b $$</p>
<p>$$ = 12\overrightarrow a - \left( {\overrightarrow c \,.\,\left( {\alpha \overrightarrow b - \widehat n} \right)} \right)\overrightarrow b $$</p>
<p>$$ = 12\overrightarrow a - (12\alpha - 0)\overrightarrow b $$</p>
<p>$$ = 12\left( {\overrightarrow a - \alpha \overrightarrow b } \right)$$</p>
<p>$$\therefore$$ $$\left| {\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right| = 12$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldsx2tkq | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three non-zero non-coplanar vectors. Let the position vectors of four points $$A,B,C$$ and $$D$$ be $$\overrightarrow a - \overrightarrow b + \overrightarrow c ,\lambda \overrightarrow a - 3\overrightarrow b + 4\overrightarrow c , - \overrightarrow a + 2\overrightarrow b - 3\overrightarrow c $$ and $$2\overrightarrow a - 4\overrightarrow b + 6\overrightarrow c $$ respectively. If $$\overrightarrow {AB} ,\overrightarrow {AC} $$ and $$\overrightarrow {AD} $$ are coplanar, then $$\lambda$$ is equal to __________.</p> | [] | null | 2 | $\overline{A B}=(\lambda-1) \bar{a}-2 \bar{b}+3 \bar{c}$
<br/><br/>
$$
\overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c}
$$<br/><br/>$$
\overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c}
$$<br/><br/>$$
\left|\begin{array}{ccc}
\lambda-1 & -2 & 3 \\
-2 & 3 & -4 \\
1 & -3 & 5
\end{array}\right|=0
$$<br/><br/>$$
\Rightarrow(\lambda-1)(15-12)+2(-10+4)+3(6-3)=0
$$<br/><br/>$$
\Rightarrow(\lambda-1)=1 \Rightarrow \lambda=2
$$
| integer | jee-main-2023-online-29th-january-morning-shift |
1ldu5mzow | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow a = - \widehat i - \widehat j + \widehat k,\overrightarrow a \,.\,\overrightarrow b = 1$$ and $$\overrightarrow a \times \overrightarrow b = \widehat i - \widehat j$$. Then $$\overrightarrow a - 6\overrightarrow b $$ is equal to :</p> | [{"identifier": "A", "content": "$$3\\left( {\\widehat i + \\widehat j + \\widehat k} \\right)$$"}, {"identifier": "B", "content": "$$3\\left( {\\widehat i - \\widehat j - \\widehat k} \\right)$$"}, {"identifier": "C", "content": "$$3\\left( {\\widehat i + \\widehat j - \\widehat k} \\right)$$"}, {"identifier": "D", "content": "$$3\\left( {\\widehat i - \\widehat j + \\widehat k} \\right)$$"}] | ["A"] | null | $$
\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})
$$<br/><br/>
Taking cross product with $\vec{a}$<br/><br/>
$$
\begin{aligned}
& \Rightarrow \vec{a} \times(\vec{a} \times \vec{b})=\vec{a} \times(\hat{i}-\hat{j}) \\\\
& \Rightarrow (\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=\hat{i}+\hat{j}+2 \hat{k} \\\\
& \Rightarrow \vec{a}-3 \vec{b}=\hat{i}+\hat{j}+2 \hat{k} \\\\
& \Rightarrow 2 \vec{a}-6 \vec{b}=2 \hat{i}+2 \hat{j}+4 \hat{k} \\\\
& \Rightarrow \vec{a}-6 \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldu5p9lo | maths | vector-algebra | scalar-and-vector-triple-product | <p>If the four points, whose position vectors are $$3\widehat i - 4\widehat j + 2\widehat k,\widehat i + 2\widehat j - \widehat k, - 2\widehat i - \widehat j + 3\widehat k$$ and $$5\widehat i - 2\alpha \widehat j + 4\widehat k$$ are coplanar, then $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "$${{73} \\over {17}}$$"}, {"identifier": "B", "content": "$$ - {{73} \\over {17}}$$"}, {"identifier": "C", "content": "$$ - {{107} \\over {17}}$$"}, {"identifier": "D", "content": "$${{107} \\over {17}}$$"}] | ["A"] | null | Let $\mathrm{A}:(3,-4,2) \quad \mathrm{C}:(-2,-1,3)$<br/><br/>
$$
\text { B : }(1,2,-1) \quad \text { D: }(5,-2 \alpha, 4)
$$<br/><br/>
A, B, C, D are coplanar points, then<br/><br/>
$$
\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
1-3 & 2+4 & -1-2 \\
-2-3 & -1+4 & 3-2 \\
5-3 & -2 \alpha+4 & 4-2
\end{array}\right|=0 \\\\
& \Rightarrow \alpha=\frac{73}{17}
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldv26m8w | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three non zero vectors such that $$\overrightarrow b $$ . $$\overrightarrow c $$ = 0 and $$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = {{\overrightarrow b - \overrightarrow c } \over 2}$$. If $$\overrightarrow d $$ be a vector such that $$\overrightarrow b \,.\,\overrightarrow d = \overrightarrow a \,.\,\overrightarrow b $$, then $$(\overrightarrow a \times \overrightarrow b )\,.\,(\overrightarrow c \times \overrightarrow d )$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "$$-\\frac{1}{4}$$"}, {"identifier": "C", "content": "$$\\frac{1}{4}$$"}, {"identifier": "D", "content": "$$\\frac{3}{4}$$"}] | ["C"] | null | $\vec{b}(\vec{a} \cdot \vec{c})-\vec{c}(\vec{a} \cdot \vec{b})=\frac{\vec{b}-\vec{c}}{2}$ $\vec{a} \cdot \vec{c}=\frac{1}{2}, \quad \vec{a} \cdot \vec{b}=\frac{1}{2}$
<br/><br/>
$$
\begin{aligned}
(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d}) & =(\vec{b} \cdot \vec{d})(\vec{a} \cdot \vec{c})-(\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) \\\\
& =(\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{c}) \\\\
& =\frac{1}{4}
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldybqbds | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow u = \widehat i - \widehat j - 2\widehat k,\overrightarrow v = 2\widehat i + \widehat j - \widehat k,\overrightarrow v .\,\overrightarrow w = 2$$ and $$\overrightarrow v \times \overrightarrow w = \overrightarrow u + \lambda \overrightarrow v $$. Then $$\overrightarrow u .\,\overrightarrow w $$ is equal to :</p> | [{"identifier": "A", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | $$
\begin{aligned}
&\begin{aligned}
& \vec{v} \times \vec{w}=(\vec{u}+\lambda \vec{v})=\hat{i}-\hat{j}-2 \hat{k}+\lambda(2 \hat{i}+\hat{j}-\hat{k}) \\\\
& =(2 \lambda+1) \hat{i}+(\lambda-1) \hat{j}-(2+\lambda) \hat{k} \\
&
\end{aligned}\\
&\begin{aligned}
& \text { Now, } \vec{v} \cdot(\vec{v} \times \vec{w})=0 \\\\
& \Rightarrow(2 \hat{i}+\hat{j}-\hat{k}) \cdot((2 \lambda+1) \hat{i}+(\lambda-1) \hat{j}-(\lambda+2) \hat{k})=0 \\\\
& \Rightarrow2(2 \lambda+1)+\lambda-1+\lambda+2=0 \Rightarrow 6 \lambda+3=0 \\\\
& \Rightarrow \lambda=-\frac{1}{2} \\\\
& \vec{w} \cdot(\vec{v} \times \vec{w})=\vec{w} \cdot(\vec{u}+\lambda \vec{v})=\vec{u} \cdot \vec{w}+\lambda \vec{v} \cdot \vec{w}=0 \\\\
& \vec{u} \cdot \vec{w}=-\lambda \vec{v} \cdot \vec{w} \\\\
& \vec{u} \cdot \vec{w}=\frac{1}{2} \times 2=1
\end{aligned}
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnw7ha3 | maths | vector-algebra | scalar-and-vector-triple-product | Let $S$ be the set of all $(\lambda, \mu)$ for which the vectors $\lambda \hat{i}-\hat{j}+\hat{k}, \hat{i}+2 \hat{j}+\mu \hat{k}$ and $3 \hat{i}-4 \hat{j}+5 \hat{k}$, where $\lambda-\mu=5$, are coplanar, then $\sum\limits_{(\lambda, \mu) \in S} 80\left(\lambda^2+\mu^2\right)$ is equal to : | [{"identifier": "A", "content": "2370"}, {"identifier": "B", "content": "2130"}, {"identifier": "C", "content": "2210"}, {"identifier": "D", "content": "2290"}] | ["D"] | null | Step 1: Given condition for coplanarity
<br/><br/>For three vectors to be coplanar, their scalar triple product must be zero. We have the vectors A, B, and C, and we know the given relation between λ and μ:
<br/><br/>$$A = \lambda \hat{i} - \hat{j} + \hat{k}$$
<br/><br/>$$B = \hat{i} + 2 \hat{j} + \mu \hat{k}$$
<br/><br/>$$C = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$$
<br/><br/>Scalar triple product condition:
<br/><br/>$$[A, B, C] = A \cdot (B \times C) = 0$$
<br/><br/>Step 2: Calculate the cross product of B and C
<br/><br/>$$B \times C = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & \mu \\
3 & -4 & 5 \\
\end{vmatrix}$$
<br/><br/>$$B \times C = (10 + 4 \mu) \hat{i} - (5 - 3 \mu) \hat{j} - 10 \hat{k}$$
<br/><br/>Step 3: Calculate the scalar triple product and apply the coplanarity condition
<br/><br/>$$[A, B, C] = A \cdot (B \times C) = \lambda(10 + 4 \mu) - 1(5 - 3 \mu) + 1(-10) = 0$$
<br/><br/>Step 4: Substitute the given relation between λ and μ
Given that:
<br/><br/>$$\lambda - \mu = 5$$
<br/><br/>From the coplanarity condition, we have:
<br/><br/>$$4 \lambda \mu + 10 \lambda - 3 \mu = 5$$
<br/><br/>Now, substitute λ in terms of μ:
<br/><br/>$$4(5 + \mu) \mu + 10(5 + \mu) - 3 \mu = 5$$
<br/><br/>Step 5: Solve for μ and λ
<br/><br/>Simplify the equation and solve for μ:
<br/><br/>$$4 \mu^2 + 27 \mu + 45 = 0$$
<br/><br/>Factor the equation:
<br/><br/>$$(4 \mu + 15)(\mu + 3) = 0$$
<br/><br/>So, the two possible values for μ are:
<br/><br/>$$\mu = -3, \frac{-15}{4}$$
<br/><br/>For each value of μ, find the corresponding value of λ using the given relation:
<br/><br/>$$\lambda = \mu + 5$$
<br/><br/>So, we get the values for λ:
<br/><br/>$$\lambda = 2, \frac{5}{4}$$
<br/><br/>Step 6: Calculate the sum
<br/><br/>Now, we need to find the sum:
<br/><br/>$$\sum\limits_{(\lambda, \mu) \in S} 80\left(\lambda^2 + \mu^2\right)$$
<br/><br/>Substitute the values of λ and μ, and simplify:
<br/><br/>$$80\left[(2^2 + (-3)^2) + \left(\frac{5}{4}\right)^2 + \left(\frac{-15}{4}\right)^2\right] = 80\left[13 + \frac{25}{16} + \frac{225}{16}\right]$$
<br/><br/>$$= 80\left[13 + \frac{250}{16}\right] = 10 \times 229$$
<br/><br/>$$= 2290$$
<br/><br/>So, the correct answer is 2290 (Option D). | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgrellln | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$a, b, c$$ be three distinct real numbers, none equal to one. If the vectors $$a \hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+b \hat{j}+\hat{\mathrm{k}}$$ and $$\hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}$$ are coplanar, then $$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $$
\left|\begin{array}{lll}
a & 1 & 1 \\\\
1 & \mathrm{~b} & 1 \\\\
1 & 1 & \mathrm{c}
\end{array}\right|=0
$$
<br/><br/>$$
\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1
$$
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{lll}
a & 1-a & 1-a \\
1 & b-1 & 0 \\
1 & 0 & c-1
\end{array}\right|=0 \\\\
& a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0 \\\\
& a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \frac{\mathrm{a}}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\
& \Rightarrow-1+\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\
& \Rightarrow \frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1
\end{aligned}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgremt4q | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\lambda \in \mathbb{Z}, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{c}$$ be a vector such that $$(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$$ and $$\vec{b} \cdot \vec{c}=-20$$. Then $$|\vec{c} \times(\lambda \hat{i}+\hat{j}+\hat{k})|^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "53"}, {"identifier": "B", "content": "62"}, {"identifier": "C", "content": "49"}, {"identifier": "D", "content": "46"}] | ["D"] | null | The given vectors are :
<br/><br/>$$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$$
<br/><br/>$$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$$
<br/><br/>We are given that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0$ which implies $(\vec{a} + \vec{b}) \times \vec{c} = 0$. So, $\vec{c}$ is in the direction of $\vec{a} + \vec{b}$.
<br/><br/>Let's denote $\vec{c} = \alpha (\vec{a} + \vec{b})$.
<br/><br/>Substituting values for $\vec{a}$ and $\vec{b}$, we get :
<br/><br/>$$\vec{c} = \alpha((\lambda + 3) \hat{i} + \hat{k})$$
<br/><br/>From the conditions $\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$, we can form two equations :
<br/><br/>$$\alpha \lambda(\lambda + 3) - \alpha = -17 \quad \text{and} \quad \alpha(3\lambda + 9 + 2) = -20.$$
<br/><br/>By solving the above equations, we find $\alpha = -1$ and $\lambda = 3$.
<br/><br/>Substituting these values back into $\vec{c}$ gives $\vec{c} = -6 \hat{i} - \hat{k}$.
<br/><br/>Next, we need to find $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$.
<br/><br/>This simplifies to :
<br/><br/>$$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2$$
<br/><br/>Calculate the cross product $\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})$ which gives :
<br/><br/>$$
=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
-6 & 0 & -1 \\
3 & 1 & 1
\end{array}\right|=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}
$$
<br/><br/>Finally, the square of the magnitude of this vector is:
<br/><br/>$$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2 = (1)^2 + 3^2 + (-6)^2 = 1 + 9 + 36 = 46.$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgsujylg | maths | vector-algebra | scalar-and-vector-triple-product | <p>If four distinct points with position vectors $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$ are coplanar, then $$[\vec{a} \,\,\vec{b} \,\,\vec{c}]$$ is equal to :</p> | [{"identifier": "A", "content": "$$[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{a}]+[\\vec{a} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{d}]+[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{c}]$$"}, {"identifier": "B", "content": "$$[\\vec{b} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{d}]+[\\vec{d} \\,\\,\\,\\,\\,\\vec{a} \\,\\,\\,\\,\\,\\vec{c}]+[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{a}]$$"}, {"identifier": "C", "content": "$$[\\vec{a} \\,\\,\\,\\,\\,\\vec{d} \\,\\,\\,\\,\\,\\vec{b}]+[\\vec{d} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{a}]+[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{c}]$$"}, {"identifier": "D", "content": "$$[\\vec{d} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{a}]+[\\vec{b} \\,\\,\\,\\,\\,\\vec{d} \\,\\,\\,\\,\\,\\vec{a}]+[\\vec{c} \\,\\,\\,\\,\\,\\vec{d} \\,\\,\\,\\,\\,\\vec{b}]$$"}] | ["D"] | null | $$
\begin{aligned}
& {[\vec{b}-\vec{a} \,\,\,\,\,\vec{c}-\vec{a} \,\,\,\,\,\vec{d}-\vec{a}]=0} \\\\
& (\vec{b}-\vec{a}) \cdot[(\vec{c}-\vec{a}) \times(\vec{d}-\vec{a})]=0 \\\\
& (\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{d}-\vec{c} \times \vec{a}-\vec{a} \times \vec{d})=0 \\\\
& {[\vec{b}\,\,\,\,\, \vec{c} \,\,\,\,\,\vec{d}]-[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]-[\vec{b} \,\,\,\,\,\vec{a} \,\,\,\,\,\vec{d}]-[\vec{a} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]=0} \\\\
& \therefore [\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{c}]=[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{c}] \\\\
& \quad=[\vec{d} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]+[\vec{b} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{a}]+[\vec{c} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{b}]
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgyl6v9x | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the vectors $$\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}, \vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}$$ and $$\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}$$ be coplanar. If the vectors $$\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}, \vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}$$ and $$\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{k}$$ are also coplanar, then $$6(\mathrm{a}+\mathrm{b}+\mathrm{c})$$ is equal to :</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "4"}] | ["A"] | null | Since, $\vec{u}_1, \vec{u}_2, \vec{u}_3$ are coplanar.
<br/><br/>So, $\left[\begin{array}{lll}\vec{u}_1 & \vec{u}_2 & \vec{u}_3\end{array}\right]=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\left|\begin{array}{lll}
1 & 1 & a \\
1 & b & 1 \\
c & 1 & 1
\end{array}\right|=0 \\\\
& \Rightarrow 1(b-1)-1(1-c)+a(1-b c)=0 \\\\
& \Rightarrow b-1-1+c+a-a b c=0 \\\\
& \Rightarrow a+b+c-2=a b c ........... (i)
\end{aligned}
$$
<br/><br/>Also, $\left[\begin{array}{lll}\vec{v}_1 & \vec{v}_2 & \vec{v}_3\end{array}\right]=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
a+b & c & c \\
a & b+c & a \\
b & b & c+a
\end{array}\right|=0 \\
& \Rightarrow(a+b)\left[b c+b a+c^2+c a-a b\right]-c\left[a c+a^2-a b\right] \\\\
& \quad+c\left[a b-b^2-b c\right]=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow a b c+a c^2+a^2 c+b^2 c+b c^2+a b c-a c^2-a^2 c \\\\
& +a b c+a b c-b^2 c-b c^2=0 \\\\
& \Rightarrow 4 a b c=0 \Rightarrow a b c=0 \\\\
& \begin{array}{lr}
\text { So, } a+b+c-2=0 [from (i)]\\\\
\Rightarrow a+b+c=2
\end{array} \\\\
& \Rightarrow 6(a+b+c)=12
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh21ilhv | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the position vectors of the points A, B, C and D be
$$5 \hat{i}+5 \hat{j}+2 \lambda \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k},-2 \hat{i}+\lambda \hat{j}+4 \hat{k}$$ and $$-\hat{i}+5 \hat{j}+6 \hat{k}$$. Let the set $$S=\{\lambda \in \mathbb{R}$$ :
the points A, B, C and D are coplanar $$\}$$.
<br/><br/>Then $$\sum_\limits{\lambda \in S}(\lambda+2)^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{37}{2}$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "41"}] | ["D"] | null | Given, position vectors of the points $A, B, C$ and $D$ be
<br/><br/>$5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
<br/><br/>$$
\begin{aligned}
\overrightarrow{A B} & =(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}})=-4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+(3-2 \lambda) \hat{\mathbf{k}} \\\\
\overrightarrow{A C} & =(-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\
& =-7 \hat{\mathbf{i}}+(\lambda-5) \hat{\mathbf{j}}+(4-2 \lambda) \hat{\mathbf{k}}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\text { and } \overrightarrow{A D} & =(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})-(5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}) \\\\
& =-6 \hat{\mathbf{i}}+(6-2 \lambda) \hat{\mathbf{k}}
\end{aligned}
$$
<br/><br/>Since, points $A, B, C$ and $D$ are coplanar
<br/><br/>$$
\begin{aligned}
& \therefore [ { \overrightarrow{AB}~ \overrightarrow{AC}~ \overrightarrow{AD}] }=0 \\\\
& \Rightarrow \left|\begin{array}{ccc}
-4 & -3 & (3-2 \lambda) \\
-7 & (\lambda-5) & (4-2 \lambda) \\
-6 & 0 & 6-2 \lambda
\end{array}\right|=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow -6\left\{(-12+6 \lambda)-\left(13 \lambda-15-2 \lambda^2\right)\right\} \\\\
& +(6-2 \lambda)\{-4 \lambda+20-21\} =0 \\\\
& \Rightarrow -6\left(2 \lambda^2-7 \lambda+3\right)+(6-2 \lambda)(-4 \lambda-1) =0 \\\\
& \Rightarrow -12 \lambda^2+42 \lambda-18+8 \lambda^2-22 \lambda-6 =0 \\\\
& \Rightarrow -4 \lambda^2+20 \lambda-24 =0 \\\\
& \Rightarrow \lambda^2-5 \lambda+6 =0 \\\\
& \Rightarrow (\lambda-2)(\lambda-3) =0 \\\\
& \Rightarrow \lambda =2,3
\end{aligned}
$$
<br/><br/>$$
\therefore \sum\limits_{\lambda \varepsilon S}(\lambda+2)^2=(2+2)^2+(3+2)^2=16+25=41
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2y5tet | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the vectors $$\vec{a}, \vec{b}, \vec{c}$$ represent three coterminous edges of a parallelopiped of volume V. Then the volume of the parallelopiped, whose coterminous edges are represented by $$\vec{a}, \vec{b}+\vec{c}$$ and $$\vec{a}+2 \vec{b}+3 \vec{c}$$ is equal to :</p> | [{"identifier": "A", "content": "3 V"}, {"identifier": "B", "content": "2 V"}, {"identifier": "C", "content": "6 V"}, {"identifier": "D", "content": "V"}] | ["D"] | null | Given that the volume $V$ of the parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is represented by the scalar triple product $[\vec{a},\vec{b},\vec{c}]$, which is the determinant of the 3 x 3 matrix with vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as its rows (or columns).
<br/><br/>When the vectors representing the coterminous edges of the parallelepiped are $\vec{a}$, $\vec{b} + \vec{c}$, and $\vec{a} + 2\vec{b} + 3\vec{c}$, the volume $V$ of the parallelepiped is represented by :
<br/><br/>$\begin{aligned} & V=[\vec{a}, \vec{b}+\vec{c}, \vec{a}+2 \vec{b}+3 \vec{c}] \\\\ & =\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3\end{array}\right|\left[\begin{array}{ll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] \\\\ & =1(3-2)[\vec{a}, \vec{b}, \vec{c}] \\\\ & =\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=V \\\\ & \end{aligned}$ | mcq | jee-main-2023-online-6th-april-evening-shift |
1lh2yfbhy | maths | vector-algebra | scalar-and-vector-triple-product | <p>The sum of all values of $$\alpha$$, for which the points whose position vectors are $$\hat{i}-2 \hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{j}+4 \hat{k},(\alpha+1) \hat{i}+2 \hat{k}$$ and $$9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$$ are coplanar, is equal to :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "2"}] | ["D"] | null | Let $\overrightarrow{O A}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
<br/><br/>$$
\begin{aligned}
& \overrightarrow{O B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\
& \overrightarrow{O C}=(a+1) \hat{\mathbf{i}}+2 \hat{\mathbf{k}}
\end{aligned}
$$
<br/><br/>and $ \overrightarrow{O D}=9 \hat{\mathbf{i}}+(a-8) \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
<br/><br/>$$
\begin{array}{ll}
&\therefore \overrightarrow{A B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\
& \overrightarrow{A C}=a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\
&\text { and } \overrightarrow{A D}=8 \hat{\mathbf{i}}+(a-6) \hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\end{array}
$$
<br/><br/>Since, given point are coplanar
<br/><br/>$$
\begin{aligned}
& \therefore \quad[\overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D}]=0 \\\\
& \Rightarrow\left|\begin{array}{ccr}
1 & -1 & 1 \\
a & 2 & -1 \\
8 & a-6 & 3
\end{array}\right|=0 \\\\
& \Rightarrow 1(6+a-6)+1(3 a+8)+1\left(a^2-6 a-16\right)=0 \\\\
& \Rightarrow a+3 a+8+a^2-6 a-16=0 \Rightarrow a^2-2 a-8=0 \\\\
& \Rightarrow a^2-4 a+2 a-8=0 \Rightarrow a(a-4)+2(a-4)=0 \\\\
& \Rightarrow(a-4)(a+2)=0 \Rightarrow a=4,-2 \\\\
& \therefore \text { Sum of all values of } a=4-2=2
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-evening-shift |
lsbl7k96 | maths | vector-algebra | scalar-and-vector-triple-product | Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}, $
<br/>$\overrightarrow{\mathrm{b}}=3(\hat{i}-\hat{j}+\hat{k})$.
<br/>Let $\overrightarrow{\mathrm{c}}$ be the vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$ and $\vec{a} \cdot \vec{c}=3$.
<br/> Then $\vec{a} \cdot((\vec{c} \times \vec{b})-\vec{b}-\vec{c})$ is equal to : | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "20"}] | ["C"] | null | <p>$$\begin{aligned}
& \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\
& \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { given } \vec{a} \times \vec{c}=\vec{b} \\
& \Rightarrow(\vec{a} \times \vec{c}) \cdot \vec{b}=\vec{b} \cdot \vec{b}=|\vec{b}|^2=27 \\
& \Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=[\vec{a} \quad \vec{c} \quad \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27 \quad \text{.... (ii)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Now } \vec{a} \cdot \vec{b}=3-6+3=0 \quad \text{.... (iii)}\\
& \vec{a} \cdot \vec{c}=3 \quad \text{.... (iv) (given)}
\end{aligned}$$</p>
<p>$$\begin{gathered}
\text { By (i), (ii), (iii) & (iv) } \\
27-0-3=24
\end{gathered}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
lv2er3tg | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$$ and $$\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$$.
If $$\vec{d}$$ is the unit vector in the direction of $$\vec{b}+\vec{c}$$ such that $$\vec{a} \cdot \vec{d}=1$$, then $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ is equal to</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>$$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}, \vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in R$$<?p>
<p>$$\text { also, } \vec{b}+\vec{c}=(x+2) \hat{i}+6 \hat{j}-2 \hat{k}$$</p>
<p>$$\vec{d} \text { is the unit vector in the direction of } \vec{b}+\vec{c}$$</p>
<p>$$\begin{aligned}
& |\vec{b}+\vec{c}|=\sqrt{(x+2)^2+6^2+2^2} \\
& =\sqrt{40+(x+2)^2} \\
& \vec{d}=\frac{x+2}{\sqrt{40+(x+2)^2}} \hat{i}+\frac{6}{\sqrt{40+(x+2)^2}} \hat{j} -\frac{2}{\sqrt{40+(x+2)^2}} \hat{k} \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \vec{a} \cdot \vec{d}=1 \\
& \frac{x+2+6-2}{\sqrt{40+(x+2)^2}}=1 \\
& x+6=\sqrt{40+(x+2)^2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
&(x+6)^2=40+(x+2)^2 \\
& x^2+36+ 12 x=40+x^2+4+4 x \\
& 8 x=8 \\
& \Rightarrow x=1 \\
&(\vec{a} \times \vec{b}) \cdot \vec{c}=[\vec{a} \vec{b} \vec{c}] \\
&=\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 4 & -5 \\
1 & 2 & 3
\end{array}\right] \\
&=1(12+10)-1(6+5)+1(4-4) \\
&=22-11=11
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
ntSo8ltNI3kPao4H | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\left| {\overrightarrow a } \right| = 5,\left| {\overrightarrow b } \right| = 4,\left| {\overrightarrow c } \right| = 3$$ thus what will be the value of $$\left| {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right|,$$ given that $$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$ : | [{"identifier": "A", "content": "$$25$$"}, {"identifier": "B", "content": "$$50$$ "}, {"identifier": "C", "content": "$$-25$$"}, {"identifier": "D", "content": "$$-50$$"}] | ["A"] | null | We have, $$\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow 0 $$
<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><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\, + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$$
<br><br>$$ \Rightarrow 25 + 16 + 9 + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = - 25.$$
<br><br>$$\therefore$$ $$\left| {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right| = 25.$$ | mcq | aieee-2002 |
LrCwfitL3Reaie1v | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | $$\overrightarrow a \,,\overrightarrow b \,,\overrightarrow c $$ are $$3$$ vectors, such that <br/><br/>$$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$ , $$\left| {\overrightarrow a } \right| = 1\,\,\,\left| {\overrightarrow b } \right| = 2,\,\,\,\left| {\overrightarrow c } \right| = 3,$$,
<br/><br/> then $${\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a }$$ is equal to : | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$-7$$ "}, {"identifier": "D", "content": "$$7$$"}] | ["C"] | null | $$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = 0$$
<br><br>$${\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) = 0$$
<br><br>$$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a = {{ - 1 - 4 - 9} \over 2}$$
<br><br>$$ = - 7$$ | mcq | aieee-2003 |
WBKKBa1dT8vFmppP | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A particle acted on by constant forces $$4\widehat i + \widehat j - 3\widehat k$$ and $$3\widehat i + \widehat j - \widehat k$$ is displaced from the point $$\widehat i + 2\widehat j + 3\widehat k$$ to the point $$\,5\widehat i + 4\widehat j + \widehat k.$$ The total work done by the forces is : | [{"identifier": "A", "content": "$$50$$ units "}, {"identifier": "B", "content": "$$20$$ units "}, {"identifier": "C", "content": "$$30$$ units "}, {"identifier": "D", "content": "$$40$$ units "}] | ["D"] | null | The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle. The displacement vector can be found by subtracting the initial position from the final position:
<br/><br/>
$$\mathbf{displacement} = \mathbf{final\ position} - \mathbf{initial\ position} = (5\widehat i + 4\widehat j + \widehat k) - (\widehat i + 2\widehat j + 3\widehat k) = 4\widehat i + 2\widehat j - 2\widehat k$$
<br/><br/>
The total work done by the two forces is equal to the sum of the work done by each force. The work done by each force can be calculated as the dot product of the force and the displacement:
<br/><br/>
$$\mathbf{work\ done\ by\ force\ 1} = (4\widehat i + \widehat j - 3\widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 4 \cdot 4 + 1 \cdot 2 - 3 \cdot -2 = 16 + 2 + 6 = 24$$
<br/><br/>
$$\mathbf{work\ done\ by\ force\ 2} = (3\widehat i + \widehat j - \widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 3 \cdot 4 + 1 \cdot 2 - 1 \cdot -2 = 12 + 2 + 2 = 16$$
<br/><br/>
The total work done by the forces is the sum of the work done by each force:
<br/><br/>
$$\mathbf{total\ work\ done} = 24 + 16 = 40$$
<br/><br/>
Therefore, the total work done by the forces is 40 J (joules) or 40 units. | mcq | aieee-2004 |
3XEW0uiZJ9xlVtmR | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w $$ be such that $$\left| {\overrightarrow u } \right| = 1,\,\,\,\left| {\overrightarrow v } \right|2,\,\,\,\left| {\overrightarrow w } \right|3.$$ If the projection $${\overrightarrow v }$$ along $${\overrightarrow u }$$ is equal to that of $${\overrightarrow w }$$ along $${\overrightarrow u }$$ and $${\overrightarrow v },$$ $${\overrightarrow w }$$ are perpendicular to each other then $$\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right|$$ equals : | [{"identifier": "A", "content": "$$14$$ "}, {"identifier": "B", "content": "$${\\sqrt {7} }$$"}, {"identifier": "C", "content": "$${\\sqrt {14} }$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Projection of $$\overrightarrow v $$ along $$\overrightarrow u = {{\overrightarrow v .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow v .\overrightarrow u } \over 2}$$
<br><br>projection of $$\overrightarrow w $$ along $$\overrightarrow u = {{\overrightarrow w .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow w .\overrightarrow u } \over 2}$$
<br><br>Given $${{\overrightarrow v .\overrightarrow u } \over 2} = {{\overrightarrow w .\overrightarrow u } \over 2}\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>Also, $$\overrightarrow v .\overrightarrow w = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>Now $${\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right|^2}$$
<br><br>$$ = {\left| {\overrightarrow u } \right|^2} + {\left| {\overrightarrow v } \right|^2} + {\left| {\overrightarrow w } \right|^2} - $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\overrightarrow u .\overrightarrow v - 2\overrightarrow v .\overrightarrow w + 2\overrightarrow u .\overrightarrow w $$
<br><br>$$ = 1 + 4 + 9 + 0$$ [ From $$(1)$$ and $$(2)$$ ] $$=14$$
<br><br>$$\therefore$$ $$\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right| = \sqrt {14} $$ | mcq | aieee-2004 |
R1ZCf7lzVetrtUe4 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | The values of a, for which the points $$A, B, C$$ with position vectors $$2\widehat i - \widehat j + \widehat k,\,\,\widehat i - 3\widehat j - 5\widehat k$$ and $$a\widehat i - 3\widehat j + \widehat k$$ respectively are the vertices of a right angled triangle with $$C = {\pi \over 2}$$ are : | [{"identifier": "A", "content": "$$2$$ and $$1$$ "}, {"identifier": "B", "content": "$$-2$$ and $$-1$$ "}, {"identifier": "C", "content": "$$-2$$ and $$1$$ "}, {"identifier": "D", "content": "$$2$$ and $$-1$$ "}] | ["A"] | null | $$\overrightarrow {CA} = \left( {2 - a} \right)\widehat i + 2\widehat j;$$
<br><br>$$\overrightarrow {CB} = \left( {1 - a} \right)\widehat i - 6\widehat k$$
<br><br>$$\overrightarrow {CA} .\overrightarrow {CB} = 0$$
<br><br>$$\,\,\,\,\,\,\,\, \Rightarrow \left( {2 - a} \right)\left( {1 - a} \right) = 0$$
<br><br>$$ \Rightarrow a = 2,1$$ | mcq | aieee-2006 |
DRDbNWOhlsTgTTBc | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If the vectors $$\overrightarrow a = \widehat i - \widehat j + 2\widehat k,\,\,\,\,\,\overrightarrow b = 2\widehat i + 4\widehat j + \widehat k\,\,\,$$ and $$\,\overrightarrow c = \lambda \widehat i + \widehat j + \mu \widehat k$$ are mutually orthogonal, then $$\,\left( {\lambda ,\mu } \right)$$ is equal to : | [{"identifier": "A", "content": "$$(2, -3)$$"}, {"identifier": "B", "content": "$$(-2, 3)$$"}, {"identifier": "C", "content": "$$(3, -2)$$"}, {"identifier": "D", "content": "$$(-3, 2)$$"}] | ["D"] | null | Since, $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ are mutually orthogonal
<br><br> $$\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow b .\overrightarrow c = 0,\,\,\overrightarrow c .\overrightarrow a = 0$$
<br><br>$$ \Rightarrow 2\lambda + 4 + \mu = 0\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$ \Rightarrow \lambda - 1 + 2\mu = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>On solving $$(i)$$ and $$(ii)$$, we get $$\lambda = - 3,\mu = 2$$ | mcq | aieee-2010 |
LYE5VbPrgMavFhGX | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two unit vectors. If the vectors $$\,\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$ are perpendicular to each other, then the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is : | [{"identifier": "A", "content": "$${\\pi \\over 6}$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["C"] | null | Let $$\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$
<br><br>Since $$\overrightarrow c $$ and $$\overrightarrow d $$ are perpendicular to each other
<br><br>$$\therefore$$ $$\overrightarrow c .\overrightarrow d = 0 \Rightarrow \left( {\widehat a + 2\widehat b} \right).\left( {5\widehat a - 4\widehat b} \right) = 0$$
<br><br>$$ \Rightarrow 5 + 6\widehat a.\widehat b - 8 = 0$$ $$\,\,\,\,\,\,$$ (as $$\widehat a.\widehat a = 1$$)
<br><br>$$ \Rightarrow \widehat a.\widehat b = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$$ | mcq | aieee-2012 |
J6fh5bJMpbtUOm1J | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $$ and $$\angle BAD$$ be an acute angle. If $$\overrightarrow r $$ is the vector that coincide with the altitude directed from the vertex $$B$$ to the side $$AD,$$ then $$\overrightarrow r $$ is given by : | [{"identifier": "A", "content": "$$\\overrightarrow r = 3\\overrightarrow q - {{3\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ "}, {"identifier": "B", "content": "$$\\overrightarrow r = - \\overrightarrow q + {{\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ "}, {"identifier": "C", "content": "$$\\vec r = \\vec q - {{\\left( {\\vec p.\\vec q} \\right)} \\over {\\left( {\\vec p.\\vec p} \\right)}}\\vec p$$ "}, {"identifier": "D", "content": "$$\\overrightarrow r = - 3\\overrightarrow q - {{3\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}$$ "}] | ["B"] | null | Let $$ABCD$$ be a parallelogram such that
<br><br>$$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $$ and $$\angle BAD$$ be an acute angle.
<br><br>We have
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263647/exam_images/vxui8byefbrtoqgpopbv.webp" loading="lazy" alt="AIEEE 2012 Mathematics - Vector Algebra Question 196 English Explanation">
<br><br>$$\overrightarrow {AX} = \left( {{{\overrightarrow p .\overrightarrow q } \over {\left| {\overrightarrow p } \right|}}} \right)\left( {{{\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}} \right) = {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p $$
<br><br>Let $$\overrightarrow r = \overrightarrow {BX} = \overrightarrow {BA} + \overrightarrow {AX} $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \overrightarrow q + {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p $$ | mcq | aieee-2012 |
MyLIse0cRI3zW4o3T2Zji | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$, $$-$$$$\widehat i$$ + 3$$\widehat j$$ + p$$\widehat k$$ and 5$$\widehat i$$ + q$$\widehat j$$ $$-$$ 4$$\widehat k$$, then the point (p, q) lies
on a line : | [{"identifier": "A", "content": "parallel to x-axis. "}, {"identifier": "B", "content": "parallel to y-axis."}, {"identifier": "C", "content": "making an acute angle with the positive direction of x-axis."}, {"identifier": "D", "content": "making an obtuse angle with the positive direction of x-axis. "}] | ["C"] | null | Given,
<br><br>$$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$$
<br><br>$$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$$
<br><br>$$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$$
<br><br>$$ \therefore $$ $$\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k$$
<br><br> $$\overrightarrow {AC} = 2\widehat i + \left( {q - 1} \right)\widehat j - 3\widehat k$$
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265163/exam_images/odwfpjssxxiwn1znqs07.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Mathematics - Vector Algebra Question 184 English Explanation">
<br><br>$$\Delta $$ABC is a right angle triangle.
<br><br>Here $$\overrightarrow {AB} $$ perpendicular to $$\overrightarrow {AC} $$
<br><br>$$ \therefore $$ $$\overrightarrow {AB} $$ . $$\overrightarrow {AC} $$ = 0
<br><br>$$ \Rightarrow $$ $$-$$ 8 + 2(q $$-$$ 1) $$-$$ 3(p + 1) = 0
<br><br>$$ \Rightarrow $$ 3p $$-$$ 2q + 13 = 0
<br><br>$$ \therefore $$ (p, q) lies on the line
<br><br>3x $$-$$ 2y + 13 = 0
<br><br>And slope of the line = $${3 \over 2}$$
<br><br>$$ \therefore $$ line makes an angle less than 90<sup>o</sup> or acute angle with the positive direction of x-axis. | mcq | jee-main-2016-online-9th-april-morning-slot |
CknOj9CXeWNGkcOX | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow u $$ be a vector coplanar with the vectors $$\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$$ and $$\overrightarrow b = \widehat j + \widehat k$$. If $$\overrightarrow u $$ is perpendicular to $$\overrightarrow a $$ and $$\overrightarrow u .\overrightarrow b = 24$$, then $${\left| {\overrightarrow u } \right|^2}$$ is equal to | [{"identifier": "A", "content": "336"}, {"identifier": "B", "content": "315"}, {"identifier": "C", "content": "256"}, {"identifier": "D", "content": "84"}] | ["A"] | null | You should know that, when $$\overrightarrow u $$ is coplanar with $$\overrightarrow a $$ and $$\overrightarrow b $$ then we can write $$\overrightarrow u = x\overrightarrow a + y\overrightarrow b $$
<br><br>Here, $$\overrightarrow u $$ is perpendicular with $$\overrightarrow a $$ then,
<br><br>$$\overrightarrow u .\overrightarrow a = 0$$
<br><br>$$ \Rightarrow \,\,\,\,\left( {x\,\overrightarrow a + y\overrightarrow b } \right)\,.\overrightarrow a = 0$$
<br><br>$$ \Rightarrow \,\,\,\,x\,.\overrightarrow {\,a} \,.\,\overrightarrow a \, + \,y\,.\,\overrightarrow a \,.\,\overrightarrow b = 0$$
<br><br>$$ \Rightarrow \,\,\,\,x\,.\,{\left| {\overrightarrow a } \right|^2} + \,y\overrightarrow a \,.\,\overrightarrow b = 0$$
<br><br>[ As $$\left| {\overrightarrow a } \right| = \sqrt {{2^2} + {3^2} + {{\left( { - 1} \right)}^2}} = \sqrt {14} $$
<br><br>and $$\overrightarrow a .\overrightarrow b = \left( {2\widehat i + 3\widehat j - \widehat k} \right).\left( {\widehat j + \widehat k} \right)$$
<br><br>$$ = \,\,\,\,\left( {2.0 + 3.1 + \left( { - 1} \right).1} \right)$$
<br><br>$$ = \,\,\,\,2$$ ]
<br><br>$$ \Rightarrow \,\,\,\,x\,.\,\left( {14} \right) + y\,.\,2 = 0$$
<br><br>$$ \Rightarrow \,\,\,\,7x\, + \,y\, = 0........\left( 1 \right)$$
<br><br>Given, $$\overrightarrow u \,.\,\overrightarrow b = 24$$
<br><br>$$ \Rightarrow \,\,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow b = 24$$
<br><br>$$ \Rightarrow \,\,\,\,x\left( {\overrightarrow a .\overrightarrow b } \right) + y{\left| {\overrightarrow b } \right|^2} = 24$$
<br><br>$$ \Rightarrow \,\,\,\,x.2 + y.{\left( {\sqrt {{1^2} + {1^2}} } \right)^2} = 24$$
<br><br>$$ \Rightarrow \,\,\,\,2x + 2y = 24$$
<br><br>$$ \Rightarrow \,\,\,\,x + y = 12.......\left( 2 \right)$$
<br><br>By solvig (1) and (2) we get,
<br><br>x = - 2 and y = 14
<br><br>Now, $${\left| {\overrightarrow u } \right|^2} = \overrightarrow u .\overrightarrow u $$
<br><br>$$ = \,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow u $$
<br><br>$$ = \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u $$
<br><br>$$ = \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u $$
<br><br>$$ = \,\,\,\,0 + 14 \times 24$$ [as $$\overrightarrow a .\overrightarrow u = 0$$ and $$\overrightarrow u .\overrightarrow b = 24]$$
<br><br>$$=\,\,\,\,$$ 336 | mcq | jee-main-2018-offline |
U1UOPS1llBEOw9OxMNW7E | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\sqrt 3 \widehat i + \widehat j,$$ $$\widehat i + \sqrt 3 \widehat j$$ and $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $${3 \over {\sqrt 2 }}$$, then the sum of all possible values of $$\beta $$ is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["B"] | null | Angle bisector is x $$-$$ y = 0
<br><br>$$ \Rightarrow $$ $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow $$ $$\left| {2\beta - 1} \right| = 3$$
<br><br>$$ \Rightarrow $$ $$\beta $$ = 2 or $$-$$ 1 | mcq | jee-main-2019-online-11th-january-evening-slot |
wXePTqL6zttwwsSMnl3rsa0w2w9jwy0f6ft | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos ($$\angle $$GOA) (O being the origin) is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt {15} }}$$"}, {"identifier": "B", "content": "$${1 \\over {6\\sqrt {10} }}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt {30} }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt {15} }}$$"}] | ["A"] | null | G is the centroid of $$\Delta $$ABC<br><br>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266720/exam_images/loddnxhki4jx9azlbmpr.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265343/exam_images/jmidgraxtr6s8vtd6inv.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265158/exam_images/zxjfdnwd3egwntp3eayb.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266855/exam_images/lhy81ppnjqc6maus8j7k.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264469/exam_images/wc2vniyerbsrv1ihactm.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265983/exam_images/roek14f6jydhrinxpgwx.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264933/exam_images/jnj8x9uncciwtqldtb9p.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266031/exam_images/ube74chtyryryxq4ef6j.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Vector Algebra Question 171 English Explanation"></picture><br><br>
$$OG = \sqrt {4 + 16 + 4} $$, OA = $$\sqrt {9 + 1} $$<br><br>
$$AG = \sqrt {1 + 16 + 9} $$<br><br>
$$\cos \theta = {{24 + 10 - 26} \over {2\sqrt {24} \sqrt {10} }}$$<br><br>
$$ \Rightarrow {8 \over {2\sqrt {8 \times 3 \times 2 \times 5} }}$$<br><br>
$$ \Rightarrow {4 \over {4\sqrt {15} }} = {1 \over {\sqrt {15} }}$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
5UABRSZyRW07gN53tq18hoxe66ijvww95a3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If a unit vector $$\overrightarrow a $$ makes angles $$\pi $$/3 with $$\widehat i$$ , $$\pi $$/ 4
with $$\widehat j$$ and $$\theta $$$$ \in $$(0, $$\pi $$) with $$\widehat k$$, then a value of $$\theta $$
is :- | [{"identifier": "A", "content": "$${{5\\pi } \\over {6}}$$"}, {"identifier": "B", "content": "$${{5\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$${{2\\pi } \\over {3}}$$"}, {"identifier": "D", "content": "$${{\\pi } \\over {4}}$$"}] | ["C"] | null | A unit vector $$\overrightarrow a $$ makes angles $$\pi $$/3 with $$\widehat i$$
<br><br>$$ \therefore $$ $$\alpha $$ = $$\pi $$/3
<br><br> and $$\pi $$/ 4
with $$\widehat j$$
<br><br>$$ \therefore $$ $$\beta $$ = $$\pi $$/ 4
<br><br>and $$\theta $$$$ \in $$(0, $$\pi $$) with $$\widehat k$$
<br><br>$$ \therefore $$ $$\gamma $$ = $$\theta $$
<br><br>We also know,
<br><br>$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma $$ = 1
<br><br>$$ \Rightarrow $$ $${\cos ^2}{\pi \over 3} + {\cos ^2}{\pi \over 4} + {\cos ^2}\theta $$ = 1
<br><br>$$ \Rightarrow $$ $${1 \over 4} + {1 \over 2} + {\cos ^2}\theta $$ = 1
<br><br>$$ \Rightarrow $$ $${\cos ^2}\theta $$ = $$ \pm {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $$\theta $$ = $${\pi \over 3}$$ or $${{2\pi } \over {3}}$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
6C4xzfmrghEfZNGqxuFTE | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$$ $$\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$$ and $$\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$$ be three vectors such that $$\overrightarrow b = 2\overrightarrow a $$ and $$\overrightarrow a $$ is perpendicular to $$\overrightarrow c $$. Then a possible value of $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right)$$ is : | [{"identifier": "A", "content": "(1, 5, 1)"}, {"identifier": "B", "content": "(1, 3, 1)"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 2},4,0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},4, - 2} \\right)$$"}] | ["C"] | null | Given $$\overrightarrow b = 2\overrightarrow a $$
<br><br>$$ \therefore $$ $$4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$$
<br><br>$$ \Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$$
<br><br>Given $$\overrightarrow a $$ is perpendicular to $$\overrightarrow c $$ <br><br>$$ \therefore $$ $$\overrightarrow a .\overrightarrow c = 0$$
<br><br>$$ \Rightarrow 6 + 6{\lambda _1} + 3\left( {{\lambda _3} - 1} \right) = 0$$
<br><br>$$ \Rightarrow 2{\lambda _1} + {\lambda _3} = - 1\,\,\,\,\,\,\,\,\,\,...(2)$$
<br><br>Now $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right) = \left( {{\lambda _1},3 - 2{\lambda _1}, - 1 - 2{\lambda _1}} \right)$$
<br><br>By checking each option you can see,
<br><br>when $${\lambda _1}$$ = $$ - {1 \over 2}$$
<br><br>then $${\lambda _2}$$ = $$3 - 2{\lambda _1}$$ = 3 + 1 = 4
<br><br>and $${\lambda _3}$$ = $$-1 - 2{\lambda _1}$$ = - 1 + 1 = 0 | mcq | jee-main-2019-online-10th-january-morning-slot |
fzUXOCvAXq7qZfRqzTtn7 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$$ $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$$, $$\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$$ be three vectors such that the projection vector of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$.
<br/>If $$\overrightarrow a + \overrightarrow b $$ is perpendicular to $$\overrightarrow c $$ , then $$\left| {\overrightarrow b } \right|$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt {32} $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$\\sqrt {22} $$"}, {"identifier": "D", "content": "4"}] | ["B"] | null | Projection of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$
<br><br>$$ \therefore $$ $${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$$
<br><br>$$ \Rightarrow $$ $${{{b_1} + {b_2} + 2} \over 2} = 2$$
<br><br>$$ \Rightarrow $$ $${b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)$$
<br><br>and $$\overrightarrow a $$ + $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$
<br><br>$$ \Rightarrow $$ $$\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \overrightarrow c = 0$$
<br><br>$$ \Rightarrow $$ $$5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0$$
<br><br>$$ \Rightarrow $$ $$5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)$$
<br><br>solving (1) & (2)
<br><br>b<sub>1</sub> $$=$$ $$-$$ 3 and b<sub>2</sub> $$=$$ 5
<br><br>$$ \Rightarrow $$ $$\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
gBxihkaJRK9XmHPQAQ7k9k2k5e2n780 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)$$ lies in the plane of the vectors, $$\overrightarrow b = \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j + 4\widehat k$$. If $$\overrightarrow a $$ bisects the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$, then: | [{"identifier": "A", "content": "$$\\overrightarrow a .\\widehat i + 3 = 0$$"}, {"identifier": "B", "content": "$$\\overrightarrow a .\\widehat k - 4 = 0$$"}, {"identifier": "C", "content": "$$\\overrightarrow a .\\widehat i + 1 = 0$$"}, {"identifier": "D", "content": "$$\\overrightarrow a .\\widehat k + 2 = 0$$"}] | ["B"] | null | Angle bisector $$\overrightarrow a = \lambda \left( {\widehat b + \widehat c} \right)$$
<br><br>= $$\lambda \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} + {{\widehat i - \widehat j + 4\widehat k} \over {3\sqrt 2 }}} \right)$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$$
<br><br>comparing with $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$$
<br><br>$${{2\lambda } \over {3\sqrt 2 }}$$ = 2
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${3\sqrt 2 }$$
<br><br>$$ \therefore $$ $$\overrightarrow a = \left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$$
<br><br>Then $$\overrightarrow a .\widehat k - 4 $$
<br><br>= 4 - 4 = 0 | mcq | jee-main-2020-online-7th-january-morning-slot |
wxnynWl5nmUW9g7btjjgy2xukewt1roa | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors such that
<br/>$${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$$ = 8.
<br/><br/>Then $${\left| {\overrightarrow a + 2\overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a + 2\overrightarrow c } \right|^2}$$ is equal to ______. | [] | null | 2 | Given, $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = 1$$
<br><br>$${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$$ = 8
<br><br>$$ \Rightarrow $$ $${\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} - 2\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow c } \right|^2} - 2\overrightarrow a .\overrightarrow c $$ = 8
<br><br>$$ \Rightarrow $$ $$\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c $$ = -2
<br><br>Now, $${\left| {\overrightarrow a + 2\overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a + 2\overrightarrow c } \right|^2}$$
<br><br>= $${\left| {\overrightarrow a } \right|^2} + 4{\left| {\overrightarrow b } \right|^2} + 4\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow a } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 4\overrightarrow a .\overrightarrow c $$
<br><br>= 10 + 4$$\left( {\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c } \right)$$
<br><br>= 10 + 4(-2)
<br><br>= 2 | integer | jee-main-2020-online-2nd-september-morning-slot |
qU27zRsydVCHN6lGyOjgy2xukf3zyzqz | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let a, b c $$ \in $$ R be such that a<sup>2</sup>
+ b<sup>2</sup>
+ c<sup>2</sup>
= 1. If <br/>$$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$$,
<br/>where
$${\theta = {\pi \over 9}}$$, then the angle between the vectors
$$a\widehat i + b\widehat j + c\widehat k$$ and $$b\widehat i + c\widehat j + a\widehat k$$ is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${{\\pi \\over 9}}$$"}, {"identifier": "C", "content": "$${{{2\\pi } \\over 3}}$$"}, {"identifier": "D", "content": "$${{\\pi \\over 2}}$$"}] | ["D"] | null | Let, $$\overrightarrow {{a_1}} = a\widehat i + b\widehat j + c\widehat k$$<br><br>and $$\overrightarrow {{a_2}} = b\widehat i + c\widehat j + a\widehat k$$<br><br>We know, Angle between two vectors<br><br>$$\cos \alpha = {{\overrightarrow {{a_1}} \,.\,\overrightarrow {{a_2}} } \over {|\overrightarrow {{a_1}} \,|.|\,\overrightarrow {{a_2}} |}}$$<br><br>$$ = {{ab + bc + ac} \over {\sqrt {{a^2} + {b^2} + {c^2}} .\sqrt {{a^2} + {b^2} + {c^2}} }}$$<br><br>$$ = {{ab + bc + ac} \over {({a^2} + {b^2} + {c^2})}}$$<br><br>Given, $${a^2} + {b^2} + {c^2} = 1$$<br><br>$$ \therefore $$ $$\cos \alpha = ab + bc + ac$$<br><br>$$ = ab\left( {{1 \over a} + {1 \over b} + {1 \over c}} \right)$$ .....(1)<br><br>Given, $$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right) = \lambda $$ (Assume)<br><br>$$ \therefore $$ $${1 \over a} = {{\cos \theta } \over \lambda }$$<br><br>$${1 \over b} = {{\cos \left( {\theta + {{2\pi } \over 3}} \right)} \over \lambda }$$<br><br>$${1 \over c} = {{\cos \left( {\theta + {{4\pi } \over 3}} \right)} \over \lambda }$$<br><br>$$ \therefore $$ $${1 \over a} + {1 \over b} + {1 \over c} = {1 \over \lambda }\left[ {\cos \theta + \cos \left( {\theta + {{2\pi } \over 3}} \right) + \cos \left( {\theta + {{4\pi } \over 3}} \right)} \right]$$<br><br>$$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos \left( {{{2\theta + 2\pi } \over 2}} \right)\cos \left( {{{{{2\pi } \over 3}} \over 2}} \right)} \right]$$<br><br>$$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos (\theta + \pi )\cos \left( {{\pi \over 3}} \right)} \right]$$<br><br>$$ = {1 \over \lambda }\left[ {\cos \theta + 2( - \cos \theta ) \times {1 \over 2}} \right]$$<br><br>$$ = {1 \over \lambda } \times 0$$<br><br>$$ = 0$$<br><br>Putting value of $$\lambda $$ in equation (1),<br><br>cos$$\alpha $$ = ab(0) = 0<br><br>$$ \Rightarrow $$ $$\alpha = {\pi \over 2}$$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
IjCSOdBC2AtfLTHw6Cjgy2xukfqgbdp5 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let the vectors $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$
be such that
<br/>$$\left| {\overrightarrow a } \right| = 2$$, $$\left| {\overrightarrow b } \right| = 4$$
and $$\left| {\overrightarrow c } \right| = 4$$. If the projection of
<br/>$$\overrightarrow b $$
on $$\overrightarrow a $$
is equal to the projection of $$\overrightarrow c $$
on $$\overrightarrow a $$
<br/>and $$\overrightarrow b $$
is perpendicular to $$\overrightarrow c $$,
then the value of
<br/>$$\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$$
is ___________. | [] | null | 6 | Projection of $$\overrightarrow b $$
on $$\overrightarrow a $$
= Projection of $$\overrightarrow c $$
on $$\overrightarrow a $$
<br><br>$$ \Rightarrow $$ $${{\overrightarrow b .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = {{\overrightarrow c .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}}$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow b .\overrightarrow a = \overrightarrow c .\overrightarrow a $$
<br><br>$$ \because $$ $$\overrightarrow b $$
is perpendicular to $$\overrightarrow c $$
<br><br>$$ \therefore $$ $$\overrightarrow b .\overrightarrow c = 0$$
<br><br>Let $$\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$$ = k
<br><br>Square both sides
<br><br>k<sup>2</sup> = $${{{\left( {\overrightarrow a } \right)}^2}}$$ + $${{{\left( {\overrightarrow b } \right)}^2}}$$ + $${{{\left( {\overrightarrow c } \right)}^2}}$$ + $$2\overrightarrow a .\overrightarrow b $$ - $$2\overrightarrow b .\overrightarrow c $$ - $$2\overrightarrow a .\overrightarrow c $$
<br><br>$$ \Rightarrow $$ k<sup>2</sup> = $${{{\left( {\overrightarrow a } \right)}^2}}$$ + $${{{\left( {\overrightarrow b } \right)}^2}}$$ + $${{{\left( {\overrightarrow c } \right)}^2}}$$
<br><br>$$ \Rightarrow $$ k<sup>2</sup> = 2<sup>2</sup> + 4<sup>2</sup> + 4<sup>2</sup> = 36
<br><br>$$ \Rightarrow $$ k = 6 | integer | jee-main-2020-online-5th-september-evening-slot |
oBUoyZ4YjQ6r5Z8sqVjgy2xukg4n5m60 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\overrightarrow x $$ and $$\overrightarrow y $$ be two non-zero vectors such that
$$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$ and $${2\overrightarrow x + \lambda \overrightarrow y }$$ is perpendicular to $${\overrightarrow y }$$,
then the value of $$\lambda $$ is _________ .
| [] | null | 1 | $$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$
<br>Squaring both sides we get
<br><br>$${\left| {\overrightarrow x } \right|^2} + 2\overrightarrow x .\overrightarrow y + {\left| {\overrightarrow y } \right|^2} = {\left| {\overrightarrow x } \right|^2}$$
<br><br>$$ \Rightarrow $$ $$2\overrightarrow x .\overrightarrow y + \overrightarrow y .\overrightarrow y $$ = 0 ....(1)
<br><br>Given $${2\overrightarrow x + \lambda \overrightarrow y }$$ is perpendicular to $${\overrightarrow y }$$
<br><br>$$ \therefore $$ $$\left( {2\overrightarrow x + \lambda \overrightarrow y } \right).\overrightarrow y $$ = 0
<br><br>$$ \Rightarrow $$ $$2\overrightarrow x .\overrightarrow y + \lambda \overrightarrow y .\overrightarrow y $$ = 0 ....(2)
<br><br>Comparing (1) & (2) we get, $$\lambda $$ = 1 | integer | jee-main-2020-online-6th-september-evening-slot |
hnJwMJTUDbFeYVoTKR1kmknwag0 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow x $$ be a vector in the plane containing vectors $$\overrightarrow a = 2\widehat i - \widehat j + \widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$$. If the vector $$\overrightarrow x $$ is perpendicular to $$\left( {3\widehat i + 2\widehat j - \widehat k} \right)$$ and its projection on $$\overrightarrow a $$ is $${{17\sqrt 6 } \over 2}$$, then the value of $$|\overrightarrow x {|^2}$$ is equal to __________. | [] | null | 486 | Let, $$\overrightarrow x = k(\overrightarrow a + \lambda \overrightarrow b )$$<br><br>$$\overrightarrow x$$ is perpendicular to $$3\widehat i + 2\widehat j - \widehat k$$<br><br><b>I.</b> k{(2 + $$\lambda$$)3 + (2$$\lambda$$ $$-$$ 1)2 + (1 $$-$$ $$\lambda$$)($$-$$1) = 0<br><br>$$ \Rightarrow $$ 8$$\lambda$$ + 3 = 0<br><br>$$\lambda = {{ - 3} \over 8}$$<br><br><b>II.</b> Also projection of $$\overrightarrow x $$ on $$\overrightarrow a $$ is $${{17\sqrt 6 } \over 2}$$ therefore<br><br>$${{\overrightarrow x .\overrightarrow a } \over {|\overrightarrow a |}} = {{17\sqrt 6 } \over 2}$$<br><br>$$ \Rightarrow k\left\{ {{{(\overrightarrow a + \lambda \overrightarrow b ).\overrightarrow a } \over {\sqrt 6 }}} \right\} = {{17\sqrt 6 } \over 2}$$<br><br>$$ \Rightarrow k\left\{ {6 + \left( {{3 \over 8}} \right)} \right\} = {{17 \times 6} \over 2}$$<br><br>$$ \Rightarrow k = {{51} \over {51}} \times 8$$<br><br>k = 8<br><br>$$ \therefore $$ $$\overrightarrow x = 8\left( {{{13} \over 8}\widehat i - {{14} \over 8}\widehat j + {{11} \over 8}\widehat k} \right)$$<br><br>$$ = 13\widehat i - 14\widehat j + 11\widehat k$$<br><br>$$|\overrightarrow x {|^2} = 169 + 196 + 121 = 486$$ | integer | jee-main-2021-online-17th-march-evening-shift |
uSK38CUUVnLZXt4bdl1kmm3d8s3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, if $$|\overrightarrow {BC} | = 8,|\overrightarrow {CA} | = 7,|\overrightarrow {AB} | = 10$$, then the projection of the vector $$\overrightarrow {AB} $$ on $$\overrightarrow {AC} $$ is equal to : | [{"identifier": "A", "content": "$${{25} \\over 4}$$"}, {"identifier": "B", "content": "$${{127} \\over 20}$$"}, {"identifier": "C", "content": "$${{85} \\over 14}$$"}, {"identifier": "D", "content": "$${{115} \\over 16}$$"}] | ["C"] | null | <picture><source media="(max-width: 1728px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266979/exam_images/tvnba426ygdio3l2ckwo.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264573/exam_images/vlfqrjsuvhpnf9m6eq5m.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263571/exam_images/xgrresfpuuu91hsf42u2.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263714/exam_images/pcdamphydbyvfwc6litj.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265714/exam_images/qkbkis1tmtf62kwtxc59.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266498/exam_images/ejsakidefctwtpaxykbn.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266487/exam_images/hf9slgyute6ci2cnjkmc.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263298/exam_images/rovhevula1wlt2hu6mar.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - Vector Algebra Question 137 English Explanation"></picture>
<br>$$|\overrightarrow a | = 8,|\overrightarrow b | = 7,|\overrightarrow c | = 10$$<br><br>Projection of $$\overrightarrow {AB} $$ on $$\overrightarrow {AC} $$<br><br>= $$|\overrightarrow {AB} |cos\theta $$<br><br>$$ = 10\left( {{{|\overrightarrow {AB} {|^2} + |\overrightarrow {CA} {|^2} - |\overrightarrow {BC} {|^2}} \over {2.|\overrightarrow {CA} ||\overrightarrow {AB} {|}}}} \right)$$<br><br>$$ = 10\left( {{{{{10}^2} + {7^2} - {8^2}} \over {2(10)(7)}}} \right)$$<br><br>$$ = 10\left( {{{85} \over {140}}} \right)$$<br><br>$$ = {{85} \over {14}}$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1krq02yea | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $$\theta$$, with the vector $$\overrightarrow a $$ + $$\overrightarrow b $$ + $$\overrightarrow c $$. Then 36cos<sup>2</sup>2$$\theta$$ is equal to ___________. | [] | null | 4 | $${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2(\overrightarrow a \,.\,\overrightarrow b + \overrightarrow a \,.\,\overrightarrow c + \overrightarrow b \,.\,\overrightarrow c ) = 3$$<br><br>$$ \Rightarrow \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 \overrightarrow a .(\overrightarrow a + \overrightarrow b + \overrightarrow c ) = \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|\cos \theta $$<br><br>$$ \Rightarrow 1 = \sqrt 3 \cos \theta $$<br><br>$$ \Rightarrow \cos 2\theta = - {1 \over 3}$$<br><br>$$ \Rightarrow 36{\cos ^2}2\theta = 4$$ | integer | jee-main-2021-online-20th-july-morning-shift |
1krrv1lee | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, if $$\left| {\overrightarrow {BC} } \right| = 3$$, $$\left| {\overrightarrow {CA} } \right| = 5$$ and $$\left| {\overrightarrow {BA} } \right| = 7$$, then the projection of the vector $$\overrightarrow {BA} $$ on $$\overrightarrow {BC} $$ is equal to : | [{"identifier": "A", "content": "$${{19} \\over 2}$$"}, {"identifier": "B", "content": "$${{13} \\over 2}$$"}, {"identifier": "C", "content": "$${{11} \\over 2}$$"}, {"identifier": "D", "content": "$${{15} \\over 2}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264070/exam_images/ep6a24u2iigywhjyduzo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Vector Algebra Question 131 English Explanation"> <br><br>Projection of $$\overrightarrow {BA} $$<br><br>on $${\overrightarrow {BC} }$$ is equal to <br><br>$$ = \left| {\overrightarrow {BA} } \right|\cos \angle ABC$$<br><br>$$ = 7\left| {{{{7^2} + {3^2} - {5^2}} \over {2 \times 7 \times 3}}} \right| = {{11} \over 2}$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1krrw91t0 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | For p > 0, a vector $${\overrightarrow v _2} = 2\widehat i + (p + 1)\widehat j$$ is obtained by rotating the vector $${\overrightarrow v _1} = \sqrt 3 p\widehat i + \widehat j$$ by an angle $$\theta$$ about origin in counter clockwise direction. If $$\tan \theta = {{\left( {\alpha \sqrt 3 - 2} \right)} \over {\left( {4\sqrt 3 + 3} \right)}}$$, then the value of $$\alpha$$ is equal to _____________. | [] | null | 6 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265929/exam_images/pfsd2oapdy3p6fqplgh1.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264150/exam_images/yosxjcnh7ins9tydvyvk.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263795/exam_images/vuqsv6zh5ecsiacx3auh.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Vector Algebra Question 132 English Explanation"></picture> <br><br>$$\left| {\overrightarrow {{V_1}} } \right| = \left| {\overrightarrow {{V_2}} } \right|$$<br><br>$$3{P^2} + 1 = 4 + {(P + 1)^2}$$<br><br>$$2{P^2} - 2P - 4 = 0 \Rightarrow {P^2} - P - 2 = 0$$<br><br>$$P = 2, - 1$$ (rejected)<br><br>$$\cos \theta = {{\overrightarrow {{V_1}} .\overrightarrow {{V_2}} } \over {\left| {\overrightarrow {{V_1}} } \right|\left| {\overrightarrow {{V_2}} } \right|}} = {{2\sqrt 3 P + (P + 1)} \over {\sqrt {{{(P + 1)}^2} + 4} \sqrt {3{P^2} + 1} }}$$<br><br>$$\cos \theta = {{4\sqrt 3 + 3} \over {\sqrt {13} \sqrt {13} }} = {{4\sqrt 3 + 3} \over {13}}$$<br><br>$$\tan \theta = {{\sqrt {112 - 24\sqrt 3 } } \over {4\sqrt 3 + 3}} = {{6\sqrt 3 - 2} \over {4\sqrt 3 + 3}} = {{\alpha \sqrt 3 - 2} \over {4\sqrt 3 + 3}}$$<br><br>$$ \Rightarrow \alpha = 6$$ | integer | jee-main-2021-online-20th-july-evening-shift |
1krzrafms | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\left( {\overrightarrow a + 3\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 5\overrightarrow b } \right)$$ and $$\left( {\overrightarrow a - 4\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 2\overrightarrow b } \right)$$, then the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ (in degrees) is _______________. | [] | null | 60 | $$\left( {\overrightarrow a + 3\overrightarrow b } \right) \bot \left( {7\overrightarrow a - 5\overrightarrow b } \right)$$<br><br>$$ \therefore $$ $$\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$$<br><br>$$ \Rightarrow $$ $$7{\left| {\overrightarrow a } \right|^2} - 15{\left| {\overrightarrow b } \right|^2} + 16\overrightarrow a \,.\,\overrightarrow b = 0$$ ....(1)<br><br>Also, $$\left( {\overrightarrow a - 4\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 2\overrightarrow b } \right) = 0$$<br><br>$$ \Rightarrow $$ $$7{\left| {\overrightarrow a } \right|^2} + 8{\left| {\overrightarrow b } \right|^2} - 30\overrightarrow a \,.\,\overrightarrow b = 0$$ .....(2)
<br><br>Equation (1) × 30
<br><br>$$210{\left| {\overrightarrow a } \right|^2} - 450{\left| {\overrightarrow b } \right|^2} + 480\overrightarrow a \,.\,\overrightarrow b = 0$$ ....(3)
<br><br>Equation (2) × 16
<br><br>$$112{\left| {\overrightarrow a } \right|^2} + 128{\left| {\overrightarrow b } \right|^2} - 480\overrightarrow a \,.\,\overrightarrow b = 0$$ .....(4)
<br><br>from (3) & (4)<br><br>$$322{\left| {\overrightarrow a } \right|^2} = 322{\left| {\overrightarrow b } \right|^2}$$
<br><br>$$ \Rightarrow $$ $${\left| {\overrightarrow a } \right|^2} = {\left| {\overrightarrow b } \right|^2}$$
<br><br>$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$$
<br><br>From equation (2),
<br><br>$$15\left| {\overrightarrow a } \right| = 30\overrightarrow a .\overrightarrow b $$
<br><br>$$ \Rightarrow $$ $$15{\left| {\overrightarrow a } \right|^2} = 30\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta $$
<br><br>$$\cos \theta = {{15} \over {30}} = {1 \over 2}$$<br><br>$$\therefore$$ $$\theta = 60^\circ $$ | integer | jee-main-2021-online-25th-july-evening-shift |
1ktd1se61 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A hall has a square floor of dimension 10 m $$\times$$ 10 m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is $${\cos ^{ - 1}}{1 \over 5}$$, then the height of the hall (in meters) is :<br/><br/><img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "2$$\\sqrt {10} $$"}, {"identifier": "C", "content": "5$$\\sqrt {3} $$"}, {"identifier": "D", "content": "5$$\\sqrt {2} $$"}] | ["D"] | null | $$A(\widehat j)\,.\,B(10\widehat i)$$<br><br>$$H(h\widehat j + 10\widehat k)$$<br><br>$$G(10\widehat i + h\widehat j + 10\widehat k)$$<br><br>$$\overrightarrow {AG} = 10\widehat i + h\widehat j + 10\widehat k$$<br><br>$$\overrightarrow {BH} = - 10\widehat i + h\widehat j + 10\widehat k$$<br><br>$$\cos \theta = {{\overrightarrow {AG} \overrightarrow {BH} } \over {\left| {\overrightarrow {AG} } \right|\left| {\overrightarrow {BH} } \right|}}$$<br><br>$${1 \over 5} = {{{h^2}} \over {{h^2} + 200}}$$<br><br>$$4{h^2} = 200 \Rightarrow h = 5\sqrt 2 $$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktd3i9x4 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If the projection of the vector $$\widehat i + 2\widehat j + \widehat k$$ on the sum of the two vectors $$2\widehat i + 4\widehat j - 5\widehat k$$ and $$ - \lambda \widehat i + 2\widehat j + 3\widehat k$$ is 1, then $$\lambda$$ is equal to __________. | [] | null | 5 | $$\overrightarrow a = \widehat i + 2\widehat j + \widehat k$$<br><br>$$\overrightarrow b = (2 - \lambda )\widehat i + 6\widehat j - 2\widehat k$$<br><br>$${{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = 1,\overrightarrow a \,.\,\overrightarrow b = 12 - \lambda $$<br><br>$$\left( {\overrightarrow a \,.\,\overrightarrow b } \right) = |\overrightarrow b {|^2}$$<br><br>$$\lambda$$<sup>2</sup> $$-$$ 24$$\lambda$$ + 144 = $$\lambda$$<sup>2</sup> $$-$$ 4$$\lambda$$ + 4 + 40<br><br>20$$\lambda$$ = 100 $$\Rightarrow$$ $$\lambda$$ = 5 | integer | jee-main-2021-online-26th-august-evening-shift |
1ktip5iva | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two vectors <br/>such that $$\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$$ and the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is 60$$^\circ$$. If $${1 \over 8}\overrightarrow a $$ is a unit vector, then $$\left| {\overrightarrow b } \right|$$ is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "8"}] | ["C"] | null | $${\left| {3\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {2\overrightarrow a + 3\overrightarrow b } \right|^2}$$<br><br>$$\left( {3\overrightarrow a + \overrightarrow b } \right).\left( {3\overrightarrow a + \overrightarrow b } \right) = \left( {2\overrightarrow a + 3\overrightarrow b } \right).\left( {2\overrightarrow a + 3\overrightarrow b } \right)$$<br><br>$$9\overrightarrow a .\,\overrightarrow a + 6\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow b = 4\overrightarrow a \,.\,\overrightarrow a + 12\overrightarrow a \,.\,\overrightarrow b + 9\overrightarrow b \,.\,\overrightarrow b $$<br><br>$$5{\left| {\overrightarrow a } \right|^2} - 6\overrightarrow a \,.\,\overrightarrow b = 8{\left| {\overrightarrow b } \right|^2}$$<br><br>$$5{(8)^2} - 6.8\,.\,\left| {\overrightarrow b } \right|\cos 60^\circ = 8{\left| {\overrightarrow b } \right|^2}$$ $$\because$$ $$\left( \matrix{
{1 \over 8}\left| {\overrightarrow a } \right| = 1 \hfill \cr
\Rightarrow \left| {\overrightarrow a } \right| = 8 \hfill \cr} \right)$$<br><br>$$40 - 3\left| {\overrightarrow b } \right| = {\left| {\overrightarrow b } \right|^2}$$<br><br>$$ \Rightarrow {\left| {\overrightarrow b } \right|^2} + 3\left| {\overrightarrow b } \right| - 40 = 0$$<br><br>$$\left| {\overrightarrow b } \right| = - 8$$, $$\left| {\overrightarrow b } \right| = 5$$<br><br>(rejected) | mcq | jee-main-2021-online-31st-august-morning-shift |
1l5ainfwa | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$ $${a_i} > 0$$, $$i = 1,2,3$$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $$\overrightarrow a $$ on the vector $$3\widehat i + 4\widehat j$$ be 7. Let $$\overrightarrow b $$ be a vector obtained by rotating $$\overrightarrow a $$ with 90$$^\circ$$. If $$\overrightarrow a $$, $$\overrightarrow b $$ and x-axis are coplanar, then projection of a vector $$\overrightarrow b $$ on $$3\widehat i + 4\widehat j$$ is equal to:</p> | [{"identifier": "A", "content": "$$\\sqrt 7 $$"}, {"identifier": "B", "content": "$$\\sqrt 2 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "7"}] | ["B"] | null | <p>$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \Rightarrow {\cos ^2}\alpha = {1 \over 3} \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$$</p>
<p>$$\overrightarrow a = {\lambda \over 3}(\widehat i + \widehat j + \widehat k),\,\lambda > 0$$</p>
<p>$${\lambda \over {\sqrt 3 }}{{(\widehat i + \widehat j + \widehat k)\,.\,(3\widehat i + 4\widehat j)} \over {\sqrt {{3^2} + {4^2}} }} = 7$$</p>
<p>$$ \Rightarrow {\lambda \over {\sqrt 3 }}(3 + 4) = 7 \times 5$$</p>
<p>$$\therefore$$ $$\lambda = 5\sqrt 3 $$</p>
<p>$$\overrightarrow a = 5(\widehat i + \widehat j + \widehat k)$$</p>
<p>Let $$\overrightarrow b = p\widehat i + q\widehat j + r\widehat k$$</p>
<p>$$\overrightarrow a \,.\,\overrightarrow b = 0$$ and $$[\overrightarrow a \,\overrightarrow b \,\widehat i] = 0$$</p>
<p>$$ \Rightarrow p + q + r = 0$$ ..... (i)</p>
<p>& $$\left| {\matrix{
p & q & r \cr
1 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right| = 0 \Rightarrow \matrix{
{q = r} \cr
{p = - 2r} \cr
} $$</p>
<p>$$\overrightarrow b = - 2r\widehat i + r\widehat j + r\widehat k$$</p>
<p>$$\overrightarrow b = r( - 2\widehat i + \widehat j + \widehat k)$$</p>
<p>Now $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$$</p>
<p>$$5\sqrt 3 = \left| r \right|\sqrt b \Rightarrow \left| r \right| = {5 \over {\sqrt 2 }}$$</p>
<p>$$\Rightarrow$$ Projection of $$\overrightarrow b $$ on $$3\widehat i + 4\widehat j = \left| {{{\overrightarrow b \,.\,\left( {3\widehat i + 4\widehat j} \right)} \over {\sqrt {{3^2} + {4^2}} }}} \right|$$</p>
<p>$$ = \left| r \right|{{( - 6 + 4)} \over 5} = \left| {{{ - 2r} \over 5}} \right|$$</p>
<p>Projection $$ = {2 \over 5} \times {5 \over {\sqrt 2 }} = \sqrt 2 $$</p>
<p>$$\therefore$$ B is correct.</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
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