## Task Condition

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

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

$$
\begin{aligned}
& a \times b=(p+2 q) \times(3 p-q)=3 \cdot p \times p-p \times q+2 \cdot 3 \cdot q \times p+2 \cdot(-1) q \times q= \\
& =-1 \cdot p \times q+6 \cdot q \times p=-1 \cdot p \times q-6 \cdot p \times q=(-1-6) \cdot p \times q=-7 \cdot p \times q
\end{aligned}
$$

$$
\begin{aligned}
& S=|a \times b|=|-7 \cdot p \times q|=7 \cdot|p \times q|=7 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =7 \cdot 1 \cdot 2 \cdot \sin \frac{\pi}{6}=14 \cdot \sin \frac{\pi}{6}=14 \cdot \frac{1}{2}=7
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 7.

## Problem Kuznetsov Analytic Geometry 5-1