## Task Condition

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

$a=p+3 q$

$b=3 p-q$

$|p|=3$

$|q|=5$

$(\widehat{p, q})=\frac{2 \pi}{3}$
## 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:

$a \times b=(p+3 q) \times(3 p-q)=3 \cdot p \times p-p \times q+3 \cdot 3 \cdot q \times p+3 \cdot(-1) \cdot q \times q=$ $=-p \times q+9 \cdot q \times p=-1 \cdot p \times q-9 \cdot p \times q=(-1-9) \cdot p \times q=-10 \cdot p \times q$

We compute the area:

$S=|a \times b|=|-10 \cdot p \times q|=10 \cdot|p \times q|=10 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=10 \cdot 3 \cdot 5 \cdot \sin \frac{2 \pi}{3}=150 \cdot \sin \frac{2 \pi}{3}=150 \cdot \frac{\sqrt{3}}{2}=75 \sqrt{3}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $75 \sqrt{3}$.

## Problem Kuznetsov Analytic Geometry 5-24