Let a, b and c be three unit vectors such tha | vedclass

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(C) Given that $a, b$ and $c$ are unit vectors,so $|a| = |b| = |c| = 1$. Using the vector triple product formula,$a 	imes (b 	imes c) = (a \cdot c)b

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$\frac{\pi}{2}$

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(C) Given that $a, b$ and $c$ are unit vectors,so $|a| = |b| = |c| = 1$. Using the vector triple product formula,$a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$. Given $a 	imes (b 	imes c) = \frac{1}{\sqrt{2}}(b + c)$,we have $(a \cdot c)b - (a \cdot b)c = \frac{1}{\sqrt{2}}b + \frac{1}{\sqrt{2}}c$. Since $b$ and $c$ are not parallel,we can equate the coefficients of $b$ and $c$: $a \cdot c = \frac{1}{\sqrt{2}}$ and $-(a \cdot b) = \frac{1}{\sqrt{2}} \Rightarrow a \cdot b = -\frac{1}{\sqrt{2}}$. By definition of dot product,$a \cdot c = |a||c| \cos \beta = \cos \beta = \frac{1}{\sqrt{2}} \Rightarrow \beta = \frac{\pi}{4}$. Similarly,$a \cdot b = |a||b| \cos \alpha = \cos \alpha = -\frac{1}{\sqrt{2}} \Rightarrow \alpha = \frac{3 \pi}{4}$. Therefore,$\alpha - \beta = \frac{3 \pi}{4} - \frac{\pi}{4} = \frac{2 \pi}{4} = \frac{\pi}{2}$.

Let $a, b$ and $c$ be three unit vectors such that $a \times (b \times c) = \frac{1}{\sqrt{2}}(b + c)$ and $b$ is not parallel to $c$. If $\alpha$ and $\beta$ are the angles between $a, b$ and $a, c$ respectively,then $\alpha - \beta =$

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