## Task Condition

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

$a=4 p+q$

$b=p-q$

$|p|=7$

$|q|=2$

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

$$
\begin{aligned}
& a \times b=(4 p+q) \times(p-q)=4 \cdot p \times p-4 \cdot p \times q+q \times p-q \times q= \\
& =-4 \cdot p \times q+q \times p=-4 \cdot p \times q-p \times q=(-4-1) \cdot p \times q=-5 \cdot p \times q
\end{aligned}
$$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|-5 \cdot p \times q|=5 \cdot|p \times q|=5 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =5 \cdot 7 \cdot 2 \cdot \sin \frac{\pi}{4}=70 \cdot \sin \frac{\pi}{4}=70 \cdot \frac{\sqrt{2}}{2}=35 \sqrt{2}
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $35 \sqrt{2}$.

## Problem Kuznetsov Analytic Geometry 5-8