## Task Condition

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

$a=2 p-3 q$

$b=5 p+q$

$|p|=2$

$|q|=3$

$(\widehat{p, q})=\frac{\pi}{2}$
## 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:

$a \times b=(2 p-3 q) \times(5 p+q)=2 \cdot 5 \cdot p \times p+2 \cdot p \times q-3 \cdot 5 \cdot q \times p-3 \cdot q \times q=$ $=2 \cdot p \times q-15 \cdot q \times p=2 \cdot p \times q+15 \cdot p \times q=(2+15) \cdot p \times q=17 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|17 \cdot p \times q|=17 \cdot|p \times q|=17 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =17 \cdot 2 \cdot 3 \cdot \sin \frac{\pi}{2}=102 \cdot \sin \frac{\pi}{2}=102 \cdot 1=102
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 102.

## Problem Kuznetsov Analytic Geometry 5-30