## Task Condition

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

$a=7 p+q$

$b=p-3 q$

$|p|=3$

$|q|=1$

$(\widehat{p, q})=\frac{3 \pi}{4}$
## 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:

$a \times b=(7 p+q) \times(p-3 q)=7 \cdot p \times p+7 \cdot(-3) \cdot p \times q+q \times p-3 \cdot q \times q=$ $=-21 \cdot p \times q+q \times p=-21 \cdot p \times q-\cdot p \times q=(-21-1) \cdot p \times q=-22 \cdot p \times q$

We compute the area:

$S=|a \times b|=|-22 \cdot p \times q|=22 \cdot|p \times q|=22 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=22 \cdot 3 \cdot 1 \cdot \sin \frac{3 \pi}{4}=66 \cdot \sin \frac{3 \pi}{4}=66 \cdot \frac{\sqrt{2}}{2}=33 \sqrt{2}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $33 \sqrt{2}$.

## Problem Kuznetsov Analytic Geometry 5-23