## problem statement

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

$a=3 p-q$

$b=p+2 q$

$|p|=3$

$|q|=4$

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

$a \times b=(3 p-q) \times(p+2 q)=3 \cdot p \times p+3 \cdot 2 \cdot p \times q-q \times p-2 \cdot q \times q=$ $=6 \cdot p \times q-q \times p=6 \cdot p \times q+p \times q=(6+1) \cdot p \times q=7 \cdot p \times q$

We compute the area:

$S=|a \times b|=|7 \cdot p \times q|=7 \cdot|p \times q|=7 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=7 \cdot 3 \cdot 4 \cdot \sin \frac{\pi}{3}=84 \cdot \sin \frac{\pi}{3}=84 \cdot \frac{\sqrt{3}}{2}=42 \sqrt{3}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $42 \sqrt{3}$.

## Problem Kuznetsov Analytical Geometry 5-14