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If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar unit vectors such that $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{\mathbf{b}+\mathbf{c}}{\sqrt{2}}$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is
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Verified Answer
The correct answer is:
$\frac{3 \pi}{4}$
Since, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$
$$
=(\mathbf{a} \cdot \mathbf{c}) \cdot \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}=\frac{\mathbf{b}+\mathbf{c}}{\sqrt{2}} \quad (given)
$$
So, $\quad \mathbf{a} \cdot \mathbf{b}=\frac{-1}{\sqrt{2}} \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}}$
$\{\because \mathbf{a}$ and $\mathbf{b}$ are unit vector where $\theta$ is angle between vectors $\mathbf{a}$ and $\mathbf{b}\}$
$$
\Rightarrow \quad \theta=\frac{3 \pi}{4} \text {. }
$$
$$
=(\mathbf{a} \cdot \mathbf{c}) \cdot \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}=\frac{\mathbf{b}+\mathbf{c}}{\sqrt{2}} \quad (given)
$$
So, $\quad \mathbf{a} \cdot \mathbf{b}=\frac{-1}{\sqrt{2}} \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}}$
$\{\because \mathbf{a}$ and $\mathbf{b}$ are unit vector where $\theta$ is angle between vectors $\mathbf{a}$ and $\mathbf{b}\}$
$$
\Rightarrow \quad \theta=\frac{3 \pi}{4} \text {. }
$$
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