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If $|\mathbf{a} \cdot \mathbf{b}|=3$ and $|\mathbf{a} \times \mathbf{b}|=4$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is
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Verified Answer
The correct answer is:
$\cos ^{-1} \frac{3}{5}$
| a.b $\mid=a b \cos \theta=3$ ...(ii)
and
$|\mathbf{a} \times \mathbf{b}|=a b \sin \theta=4$ ...(i)
Dividing (ii) by (i),
we get
$\tan \theta=\frac{4}{3} \Rightarrow \cos \theta=\frac{3}{5} \Rightarrow \theta=\cos ^{-1} \frac{3}{5} \text {. }$
and
$|\mathbf{a} \times \mathbf{b}|=a b \sin \theta=4$ ...(i)
Dividing (ii) by (i),
we get
$\tan \theta=\frac{4}{3} \Rightarrow \cos \theta=\frac{3}{5} \Rightarrow \theta=\cos ^{-1} \frac{3}{5} \text {. }$
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