## Task Condition

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

\[
\begin{aligned}
& a=p-4 q \\
& b=3 p+q \\
& |p|=1 \\
& |q|=2 \\
& (\widehat{p, q})=\frac{\pi}{6}
\end{aligned}
\]
## 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:

$a \times b=(p-4 q) \times(3 p+q)=3 \cdot p \times p+p \times q-4 \cdot 3 \cdot q \times p-4 \cdot q \times q=$ $=p \times q-12 \cdot q \times p=p \times q+12 \cdot p \times q=(1+12) \cdot p \times q=13 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|13 \cdot p \times q|=13 \cdot|p \times q|=13 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =13 \cdot 1 \cdot 2 \cdot \sin \frac{\pi}{6}=26 \cdot \sin \frac{\pi}{6}=26 \cdot \frac{1}{2}=13
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 13.

## Problem Kuznetsov Analytic Geometry 5-9