## Task Condition

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

$a=6 p-q$

\[
\begin{aligned}
& b=p+q \\
& |p|=3 \\
& |q|=4 \\
& (\widehat{p, q})=\frac{\pi}{4}
\end{aligned}
\]
## 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:

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

$=6 \cdot p \times q-q \times p=6 \cdot p \times q+p \times q=(6+1) \cdot p \times q=7 \cdot p \times q$

We compute the area:

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

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

Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is $42 \sqrt{2}$.

Problem Kuznetsov Analytical Geometry 5-19