## Problem Statement

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

\[
\begin{aligned}
& a=2 p+3 q \\
& b=p-2 q \\
& |p|=2 \\
& |q|=3 \\
& (\widehat{p, q})=\frac{\pi}{4}
\end{aligned}
\]
## 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:

$$
\begin{aligned}
& a \times b=(2 p+3 q) \times(p-2 q)=2 \cdot p \times p+2 \cdot(-2) \cdot p \times q+3 \cdot q \times p+3 \cdot(-2) \cdot q \times q= \\
& =-4 \cdot p \times q+3 \cdot q \times p=-4 \cdot p \times q-3 \cdot p \times q=(-4-3) \cdot p \times q=-7 \cdot p \times q
\end{aligned}
$$

We compute the area:

$S=|a \times b|=|-7 \cdot p \times q|=7 \cdot|p \times q|=7 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=7 \cdot 2 \cdot 3 \cdot \sin \frac{\pi}{4}=42 \cdot \sin \frac{\pi}{4}=42 \cdot \frac{\sqrt{2}}{2}=21 \sqrt{2}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $21 \sqrt{2}$.

## Problem Kuznetsov Analytic Geometry 5-15