## Task Condition

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

$a=10 p+q$

$b=3 p-2 q$

$|p|=4$

$|q|=1$
$(\widehat{p, q})=\frac{\pi}{6}$
## 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:

$a \times b=(10 p+q) \times(3 p-2 q)=10 \cdot 3 \cdot p \times p+10 \cdot(-2) \cdot p \times q+3 \cdot q \times p-2 q \times q=$

$=-20 \cdot p \times q+3 \cdot q \times p=-20 \cdot p \times q-3 \cdot p \times q=(-20-3) \cdot p \times q=-23 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|-23 \cdot p \times q|=23 \cdot|p \times q|=23 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =23 \cdot 4 \cdot 1 \cdot \sin \frac{\pi}{6}=92 \cdot \sin \frac{\pi}{6}=92 \cdot \frac{1}{2}=46
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 46.

## Problem Kuznetsov Analytical Geometry 5-20