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If $|\mathbf{a}|=8,|\mathbf{b}|=3$ and $|\mathbf{a} \times \mathbf{b}|=12$, then find the angle between $\mathbf{a}$ and $\mathbf{b}$.
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The correct answer is:
$\frac{\pi}{6}$
Given, $|\mathfrak{a}|=8,|\vec{b}|=3$ and $|\mathbf{a} \times \mathbf{b}|=12$
We know that, $\sin \theta=\frac{|a \times b|}{|a| b \mid}$
$\Rightarrow \quad \sin \theta=\frac{12}{8 \times 3}=\frac{1}{2} \Rightarrow \sin \theta=\sin \frac{\pi}{6}$
$\Rightarrow \quad \theta=\frac{\pi}{6} \quad\left[\sin \frac{\pi}{6}=\frac{1}{2}\right]$
Hence, angle between and b is $\frac{\pi}{6}$.
We know that, $\sin \theta=\frac{|a \times b|}{|a| b \mid}$
$\Rightarrow \quad \sin \theta=\frac{12}{8 \times 3}=\frac{1}{2} \Rightarrow \sin \theta=\sin \frac{\pi}{6}$
$\Rightarrow \quad \theta=\frac{\pi}{6} \quad\left[\sin \frac{\pi}{6}=\frac{1}{2}\right]$
Hence, angle between and b is $\frac{\pi}{6}$.
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