## Task Condition

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

$a=7 p-2 q$

$b=p+3 q$

$|p|=\frac{1}{2}$

$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$
## 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:

$a \times b=(7 p-2 q) \times(p+3 q)=7 \cdot p \times p+7 \cdot 3 \cdot p \times q-2 \cdot q \times p-2 \cdot 3 \cdot q \times q=$ $=21 \cdot p \times q-2 \cdot q \times p=21 \cdot p \times q+2 \cdot p \times q=(21+2) \cdot p \times q=23 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|23 \cdot p \times q|=23 \cdot|p \times q|=23 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =23 \cdot \frac{1}{2} \cdot 2 \cdot \sin \frac{\pi}{2}=23 \cdot \sin \frac{\pi}{2}=23 \cdot 1=23
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 23.

## Problem Kuznetsov Analytical Geometry 5-18