## Task Condition

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

$$
\begin{aligned}
& a=5 p-q \\
& b=p+q \\
& |p|=5 \\
& |q|=3 \\
& (\widehat{p, q})=\frac{5 \pi}{6}
\end{aligned}
$$
## 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:

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

$=5 \cdot p \times q-q \times p=5 \cdot p \times q+p \times q=(5+1) \cdot p \times q=6 \cdot p \times q$

We compute the area:

$S=|a \times b|=|6 \cdot p \times q|=6 \cdot|p \times q|=6 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=6 \cdot 5 \cdot 3 \cdot \sin \frac{5 \pi}{6}=90 \cdot \sin \frac{5 \pi}{6}=90 \cdot \frac{1}{2}=45$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 45.

## Problem Kuznetsov Analytical Geometry 5-26