Let $|\vec{a}|=2,|\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}$ is equal to :
Solution
$$
\begin{aligned}
& |\vec{a}|=2 \\\\
& |\vec{b}|=3 \\\\
& \vec{a} \cdot \vec{b}=\frac{\pi}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\
& = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\
& = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2 \\\\
& = |-7 \vec{a} \times \vec{b}|^2 \\\\
& = \left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2
\end{aligned}
$$
<br/><br/>= $49 \times 4 \times 9 \times \frac{1}{2}=882$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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