## problem statement

Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.

$a=3 p+2 q$

$$
\begin{aligned}
& b=p-q \\
& |p|=10 \\
& |q|=1 \\
& (\widehat{p, q})=\frac{\pi}{2}
\end{aligned}
$$
## 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:

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

$=-3 \cdot p \times q+2 \cdot q \times p=-3 \cdot p \times q-2 \cdot p \times q=(-3-2) \cdot p \times q=-5 \cdot p \times q$

We compute the area:

$S=|a \times b|=|-5 \cdot p \times q|=5 \cdot|p \times q|=5 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=5 \cdot 10 \cdot 1 \cdot \sin \frac{\pi}{2}=50 \cdot \sin \frac{\pi}{2}=50 \cdot 1=50$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 50.

## Problem Kuznetsov Analytical Geometry 5-11