## problem statement

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

$a=3 p-4 q$
$b=p+3 q$

$|p|=2$

$|q|=3$

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

$a \times b=(3 p-4 q) \times(p+3 q)=3 \cdot p \times p+3 \cdot 3 \cdot p \times q-4 \cdot q \times p-4 \cdot 3 q \times q=$

$=9 \cdot p \times q-4 \cdot q \times p=9 \cdot p \times q+4 \cdot p \times q=(9+4) \cdot p \times q=13 \cdot p \times q$

We compute the area:

$S=|a \times b|=|13 \cdot p \times q|=13 \cdot|p \times q|=13 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=13 \cdot 2 \cdot 3 \cdot \sin \frac{\pi}{4}=78 \cdot \sin \frac{\pi}{4}=78 \cdot \frac{\sqrt{2}}{2}=39 \sqrt{2}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $39 \sqrt{2}$.

## Problem Kuznetsov Analytical Geometry 5-27