## Task Condition

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

$a=6 p-q$

$b=5 q+p$

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

$|q|=4$
$(\widehat{p, q})=\frac{5 \pi}{6}$
## 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:

$$
\begin{aligned}
& a \times b=(6 p-q) \times(5 q+p)=6 \cdot 5 \cdot p \times q+6 \cdot p \times p-5 \cdot q \times q-q \times p= \\
& =30 \cdot p \times q-q \times p=30 \cdot p \times q+p \times q=(30+1) \cdot p \times q=31 \cdot p \times q
\end{aligned}
$$

We compute the area:

$S=|a \times b|=|31 \cdot p \times q|=31 \cdot|p \times q|=31 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=31 \cdot \frac{1}{2} \cdot 4 \cdot \sin \frac{5 \pi}{6}=62 \cdot \sin \frac{5 \pi}{6}=62 \cdot \frac{1}{2}=31$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 31.

## Problem Kuznetsov Analytical Geometry 5-28