## Problem Statement

Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=p-3q$

$b=p+2q$

$|p|=\frac{1}{5}$

$|q|=1$

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

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

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|5 \cdot p \times q|=5 \cdot|p \times q|=5 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =5 \cdot \frac{1}{5} \cdot 1 \cdot \sin \frac{\pi}{2}=\sin \frac{\pi}{2}=1
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 1.

## Problem Kuznetsov Analytic Geometry 5-3