## Task Condition

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

$a=3 p+2 q$

$b=2 p-q$

$|p|=4$

$|q|=3$

$(\widehat{p, q})=\frac{3 \pi}{4}$
## 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:

$$
\begin{aligned}
& a \times b=(3 p+2 q) \times(2 p-q)=3 \cdot 2 \cdot p \times p+3 \cdot(-1) \cdot p \times q+2 \cdot 2 \cdot q \times p+2 \cdot(-1) \cdot q \times q= \\
& =-3 \cdot p \times q+4 \cdot q \times p=-3 \cdot p \times q-4 \cdot p \times q=(-3-4) \cdot p \times q=-7 \cdot p \times q
\end{aligned}
$$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|-7 \cdot p \times q|=7 \cdot|p \times q|=7 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =7 \cdot 4 \cdot 3 \cdot \sin \frac{3 \pi}{4}=84 \cdot \sin \frac{3 \pi}{4}=84 \cdot \frac{\sqrt{2}}{2}=42 \sqrt{2}
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $42 \sqrt{2}$.

## Problem Kuznetsov Analytic Geometry 5-31