## Task Condition

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

$a=3 p+4 q$

$b=q-p$

$|p|=2.5$

$|q|=2$

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

$a \times b=(3 p+4 q) \times(q-p)=3 \cdot p \times q+3 \cdot(-1) \cdot p \times p+4 \cdot q \times q+4 \cdot(-1) \cdot q \times p=$ $=3 \cdot p \times q-4 \cdot q \times p=3 \cdot p \times q+4 \cdot p \times q=(3+4) \cdot p \times q=7 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|7 \cdot p \times q|=7 \cdot|p \times q|=7 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =7 \cdot 2.5 \cdot 2 \cdot \sin \frac{\pi}{2}=35 \cdot \sin \frac{\pi}{2}=35 \cdot 1=35
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 35.

## Problem Kuznetsov Analytic Geometry 5-22