## Task Condition

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

$a=5 p+q$

$b=p-3 q$

$|p|=1$

$|q|=2$

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

$a \times b=(5 p+q) \times(p-3 q)=5 \cdot p \times p+5 \cdot(-3) \cdot p \times q+q \times p-3 \cdot q \times q=$ $=-15 \cdot p \times q+q \times p=-15 \cdot p \times q-p \times q=(-15-1) \cdot p \times q=-16 \cdot p \times q$

We compute the area:

$S=|a \times b|=|-16 \cdot p \times q|=16 \cdot|p \times q|=16 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=16 \cdot 1 \cdot 2 \cdot \sin \frac{\pi}{3}=32 \cdot \sin \frac{\pi}{3}=32 \cdot \frac{\sqrt{3}}{2}=16 \sqrt{3}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $16 \sqrt{3}$.

## Problem Kuznetsov Analytic Geometry 5-17