## Task Condition

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

$a=p+4q$

$b=2p-q$

$|p|=7$

$|q|=2$

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

$a \times b=(p+4 q) \times(2 p-q)=2 \cdot p \times p-p \times q+4 \cdot 2 \cdot q \times p-4 \cdot q \times q=$ $=-p \times q+8 \cdot q \times p=-\cdot p \times q-8 \cdot p \times q=(-1-8) \cdot p \times q=-9 \cdot p \times q$

We compute the area:

$S=|a \times b|=|-9 \cdot p \times q|=9 \cdot|p \times q|=9 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$

$=9 \cdot 7 \cdot 2 \cdot \sin \frac{\pi}{3}=126 \cdot \sin \frac{\pi}{3}=126 \cdot \frac{\sqrt{3}}{2}=63 \sqrt{3}$

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $63 \sqrt{3}$.

## Problem Kuznetsov Analytical Geometry 5-10