## Task Condition

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

$a=2 p-3 q$

$b=3 p+q$

$|p|=4$

$|q|=1$

$(\widehat{p, q})=\frac{\pi}{6}$
## 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:

$a \times b=(2 p-3 q) \times(3 p+q)=2 \cdot 3 \cdot p \times p+2 \cdot p \times q-3 \cdot 3 \cdot q \times p-3 \cdot q \times q=$ $=2 \cdot p \times q-9 \cdot q \times p=2 \cdot p \times q+9 \cdot p \times q=(2+9) \cdot p \times q=11 \cdot p \times q$

We compute the area:

$S=|a \times b|=|11 \cdot p \times q|=11 \cdot|p \times q|=11 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$
$=11 \cdot 4 \cdot 1 \cdot \sin \frac{\pi}{6}=44 \cdot \sin \frac{\pi}{6}=44 \cdot \frac{1}{2}=22$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 22.

## Problem Kuznetsov Analytic Geometry 5-16