## Problem Statement

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

$a=6 p-q$

$b=p+2 q$

$|p|=8$

$|q|=\frac{1}{2}$

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

$a \times b=(6 p-q) \times(p+2 q)=6 \cdot p \times p+6 \cdot 2 \cdot p \times q-q \times p-2 \cdot q \times q=$

$=12 \cdot p \times q-q \times p=12 \cdot p \times q+1 \cdot p \times q=(12+1) \cdot p \times q=13 \cdot p \times q$

We compute the area:

$$
\begin{aligned}
& S=|a \times b|=|13 \cdot p \times q|=13 \cdot|p \times q|=13 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =13 \cdot 8 \cdot \frac{1}{2} \cdot \sin \frac{\pi}{3}=52 \cdot \sin \frac{\pi}{3}=52 \cdot \frac{\sqrt{3}}{2}=26 \sqrt{3}
\end{aligned}
$$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $26 \sqrt{3}$.

## Problem Kuznetsov Analytical Geometry 5-21