## Task Condition

Calculate the area of the parallelogram constructed on vectors $a$ and $b$.

$a=3 p+q$
$b=p-2 q$

$|p|=4$

$|q|=1$

$\widehat{(\widehat{p}, q})=\frac{\pi}{4}$
## Solution

The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:

$S=|a \times b|$

We compute $a \times b$ using the properties of the vector product:

$$
\begin{aligned}
& a \times b=(3 p+q) \times(p-2 q)=3 \cdot p \times p+3 \cdot(-2) \cdot p \times q+q \times p-2 q \times q= \\
& =-6 \cdot p \times q+1 \cdot q \times p=-6 \cdot p \times q-1 \cdot p \times q=(-6-1) \cdot p \times q=-7 \cdot p \times q
\end{aligned}
$$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|-7 \cdot p \times q|=7 \cdot|p \times q|=7 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =7 \cdot 4 \cdot 1 \cdot \sin \frac{\pi}{4}=28 \cdot \sin \frac{\pi}{4}=28 \cdot \frac{\sqrt{2}}{2}=14 \sqrt{2}
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $14 \sqrt{2}$.

## Problem Kuznetsov Analytic Geometry 5-2