## Task Condition

Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.

\[
\begin{aligned}
& a=4 p-q \\
& b=p+2 q \\
& |p|=5 \\
& |q|=4 \\
& (\widehat{p, q})=\frac{\pi}{4}
\end{aligned}
\]
## 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:

$a \times b=(4 p-q) \times(p+2 q)=4 \cdot p \times p+4 \cdot 2 \cdot p \times q-q \times p-2 \cdot q \times q=$ $=8 \cdot p \times q-q \times p=8 \cdot p \times q+p \times q=(8+1) \cdot p \times q=9 \cdot p \times q$

We compute the area:

$S=|a \times b|=|9 \cdot p \times q|=9 \cdot|p \times q|=9 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=9 \cdot 5 \cdot 4 \cdot \sin \frac{\pi}{4}=180 \cdot \sin \frac{\pi}{4}=180 \cdot \frac{\sqrt{2}}{2}=90 \sqrt{2}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $90 \sqrt{2}$.

## Problem Kuznetsov Analytical Geometry 5-12