## Problem Statement

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

$a=3 p+q$

$b=p-3 q$
\[
\begin{aligned}
|p| & =7 \\
|q| & =2
\end{aligned}
\]

\[
(\widehat{p, q})=\frac{\pi}{4}
\]
## 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:

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

We compute the area:

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

$=10 \cdot 7 \cdot 2 \cdot \sin \frac{\pi}{4}=140 \cdot \sin \frac{\pi}{4}=140 \cdot \frac{\sqrt{2}}{2}=70 \sqrt{2}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $70 \sqrt{2}$.

## Problem Kuznetsov Analytic Geometry 5-25