## Problem Statement

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

$$
a = p - 2q
$$

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

$a \times b=(p-2 q) \times(2 p+q)=2 \cdot p \times p+p \times q-2 \cdot 2 \cdot q \times p-2 \cdot q \times q=$ $=p \times q-4 \cdot q \times p=p \times q+4 \cdot p \times q=(1+4) \cdot p \times q=5 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|5 \cdot p \times q|=5 \cdot|p \times q|=5 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =5 \cdot 2 \cdot 3 \cdot \sin \frac{3 \pi}{4}=30 \cdot \sin \frac{3 \pi}{4}=30 \cdot \frac{\sqrt{2}}{2}=15 \sqrt{2}
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $15 \sqrt{2}$.

## Problem Kuznetsov Analytic Geometry 5-5