## Problem Statement

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

$a=3 p-2 q$

$b=p+5 q$

$|p|=4$

$|q|=\frac{1}{2}$

$(\widehat{p, q})=\frac{5 \pi}{6}$
## 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:

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

$=15 \cdot p \times q-2 \cdot q \times p=15 \cdot p \times q+2 \cdot p \times q=(15+2) \cdot p \times q=17 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|17 \cdot p \times q|=17 \cdot|p \times q|=17 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =17 \cdot 4 \cdot \frac{1}{2} \cdot \sin \frac{5 \pi}{6}=34 \cdot \sin \frac{5 \pi}{6}=34 \cdot \frac{1}{2}=17
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 17.

## Problem Kuznetsov Analytic Geometry 5-4