## Task Condition

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

$a=2 p+3 q$

$b=p-2 q$

$|p|=6$

$|q|=7$

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

$a \times b=(2 p+3 q) \times(p-2 q)=2 \cdot p \times p+2 \cdot(-2) \cdot p \times q+3 \cdot q \times p+3 \cdot(-2) \cdot q \times q=$ $=-4 \cdot p \times q+3 \cdot q \times p=-4 \cdot p \times q-3 \cdot p \times q=(-4-3) \cdot p \times q=-7 \cdot p \times q$

We compute the area:

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

$=7 \cdot 6 \cdot 7 \cdot \sin \frac{\pi}{3}=294 \cdot \sin \frac{\pi}{3}=294 \cdot \frac{\sqrt{3}}{2}=147 \sqrt{3}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $147 \sqrt{3}$.

## Problem Kuznetsov Analytic Geometry 5-13