2.12. Calculate the area of the parallelogram constructed on vectors $\vec{a}$ and $\vec{b}$, the lengths of the diagonals of the parallelogram, $\cos (\widehat{\vec{a}, \vec{p}})$ and $r_{\dot{a}} \ddot{b}$, if

$$
\vec{a}=5 \vec{p}+2 \vec{q}, \vec{b}=\vec{p}-3 \vec{q},|\vec{p}|=4 \sqrt{2},|\vec{q}|=6,(\widehat{\vec{p}, \vec{q}})=\pi / 4
$$
S o l u t i o n. The area $S$ of the parallelogram constructed on vectors $\vec{a}$ and $\vec{b}$ is equal to the magnitude of the vector product of these vectors. Using the algebraic properties of the vector product, we have

$$
\begin{aligned}
\vec{a} \times \vec{b}=(5 \vec{p}+2 \vec{q}) & \times(\vec{p}-3 \vec{q})=5 \cdot \vec{p} \times \vec{p}-15 \cdot \vec{p} \times \vec{q}+2 \cdot \vec{q} \times \vec{p}-6 \cdot \vec{q} \times \vec{q}= \\
& =-15 \cdot \vec{p} \times \vec{q}-2 \cdot \vec{p} \times \vec{q}=-17 \cdot \vec{p} \times \vec{q}
\end{aligned}
$$

Here we have taken into account that $\vec{p} \times \vec{p}=\vec{q} \times \vec{q}=\overrightarrow{0}$ according to the algebraic property 4 of the vector product, as well as the algebraic property 1. Now, in accordance with the definition of the magnitude of the vector product (formula $(2.31)$),

$$
\begin{gathered}
S=|\vec{a} \times \vec{b}|=17 \cdot|\vec{p} \times \vec{q}|=17 \cdot|\vec{p}| \cdot|\vec{q}| \cdot \sin (\widehat{\vec{p}, \vec{q}})=17 \cdot 4 \sqrt{2} \cdot 6 \cdot \sin \frac{\pi}{4}= \\
=17 \cdot 4 \cdot 6 \cdot \sqrt{2} \cdot \frac{\sqrt{2}}{2}=17 \cdot 6 \cdot 4=408
\end{gathered}
$$

Thus, $S=408$ (sq. units). To determine the lengths of the diagonals of the parallelogram, note that one of the diagonals, in accordance with the parallelogram rule for vector addition, is equal to $\vec{a}+\vec{b}$, and the other, in accordance with the rule of subtraction, is $-\vec{a}-\vec{b}$ (see fig. $2.2,2,3$). Adding and subtracting the vectors $\vec{a}$ and $\vec{b}$, we get

$$
\vec{a}+\vec{b}=6 \vec{p}-\vec{q}, \vec{a}-\vec{b}=4 \vec{p}+5 \vec{q}
$$

To determine the lengths of the diagonals of the parallelogram, i.e., $|\vec{a}+\vec{b}|$ and $|\vec{a}-\vec{b}|$, we will use formula (2.27). We get:

$$
\begin{gathered}
|\vec{a}+\vec{b}|=\sqrt{(\vec{a}+\vec{b})(\vec{a}+\vec{b})}=\sqrt{(6 \vec{p}-\vec{q})^{2}}=\sqrt{36 \vec{p}^{2}-12 \vec{p} \cdot \vec{q}+\vec{q}^{2}}= \\
=\sqrt{36|\vec{p}|^{2}-12|\vec{p}| \cdot|\vec{q}| \cos \frac{\pi}{4}+|\vec{q}|^{2}}=\sqrt{36 \cdot 32-12 \cdot 4 \cdot \sqrt{2} \cdot 6 \cdot \frac{\sqrt{2}}{2}+36}= \\
=\sqrt{36 \cdot 32-36 \cdot 8+36}=6 \sqrt{32-8+1}=6 \cdot 5=30 .
\end{gathered}
$$

Similarly,

$$
|\vec{a}-\vec{b}|=\sqrt{(4 \vec{p}+5 \vec{q})^{2}}=\sqrt{16 \cdot 32+40 \cdot 4 \cdot \sqrt{2} \cdot 6 \cdot \frac{\sqrt{2}}{2}+25 \cdot 36}=2 \sqrt{593}
$$

To determine $\cos (\widehat{\vec{a}, \vec{p}})$, we will use formula (2.22), according to which

$$
\cos (\widehat{\vec{a}, \vec{p}})=\frac{\vec{a} \cdot \vec{p}}{|\vec{a}| \cdot|\vec{p}|}
$$

We calculate the quantities entering the right-hand side of this equality:

$$
\begin{gathered}
\vec{a} \cdot \vec{p}=(5 \vec{p}+2 \vec{q}) \cdot \vec{p}=5 \vec{p}^{2}+2 \vec{q} \cdot \vec{p}=5|\vec{p}|^{2}+2|\vec{q}| \cdot|\vec{p}| \cdot \cos \frac{\pi}{4}= \\
=5 \cdot 32+2 \cdot 6 \cdot 4 \sqrt{2} \cdot \frac{\sqrt{2}}{2}=160+48=208 \\
|\vec{a}|=\sqrt{(5 \vec{p}+2 \vec{q})^{2}}=\sqrt{25|\vec{p}|^{2}+20|\vec{p}| \cdot|\vec{q}| \cdot \cos \frac{\pi}{4}+4|\vec{q}|^{2}}= \\
=\sqrt{25 \cdot 32+20 \cdot 4 \sqrt{2} \cdot 6 \frac{\sqrt{2}}{2}+4 \cdot 36}=2 \sqrt{356}
\end{gathered}
$$

Substituting the obtained results into the formula for $\cos (\widehat{\vec{a}, \vec{p}})$, we get

$$
\cos (\widehat{\vec{a}, \vec{p}})=\frac{208}{2 \sqrt{356} \cdot \sqrt{2}}=\frac{13}{\sqrt{178}}
$$

Now let's find $\Pi_{\vec{a}} \vec{b}$, i.e., the projection of vector $\vec{b}$ onto vector $\vec{a}$. For this, we need to calculate the quantities $\vec{a} \cdot \vec{b}$ and $|\vec{a}|$, entering the formula

$$
\Pi_{\vec{a}} \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}
$$

We have

$$
\begin{gathered}
\vec{a} \cdot \vec{b}=(5 \vec{p}+2 \vec{q})(\vec{p}-3 \vec{q})=5 \vec{p}^{2}-15 \vec{p} \cdot \vec{q}+2 \vec{p} \cdot \vec{q}-6 \vec{q}^{2}= \\
=5|\vec{p}|^{2}-13 \vec{p} \cdot \vec{q}-6|\vec{q}|^{2}=5 \cdot 32-13 \cdot 4 \sqrt{2} \cdot 6 \frac{1}{\sqrt{2}}-6 \cdot 36= \\
=160-312-216=-368 . \\
|\vec{a}|=\sqrt{\vec{a} \cdot \vec{a}}=\sqrt{(5 \vec{p}+2 \vec{q})(5 \vec{p}+2 \vec{q})}= \\
=\sqrt{25|\vec{p}|^{2}+20|\vec{p}| \cdot|\vec{q}| \cos \frac{\pi}{4}+4|\vec{q}|^{2}}=4 \sqrt{89}
\end{gathered}
$$

Thus,

$$
\Pi \mathrm{p}_{\vec{a}} \vec{b}=-\frac{368}{4 \sqrt{89}}=-\frac{92}{\sqrt{89}}
$$