If vec{a}, vec{b}, vec{c} are unit vectors su | vedclass

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(A) Given $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}$.Using the vector triple product formula $\vec{a} 	imes (\vec

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$\frac{3\pi}{4}$

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(A) Given $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}$.Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.So,$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{\sqrt{2}}\vec{b} + \frac{1}{\sqrt{2}}\vec{c}$.Rearranging the terms,we get: $(\vec{a} \cdot \vec{c} - \frac{1}{\sqrt{2}})\vec{b} - (\vec{a} \cdot \vec{b} + \frac{1}{\sqrt{2}})\vec{c} = \vec{0}$.Since $\vec{b}$ and $\vec{c}$ are non-collinear (as they are unit vectors and the equation involves their linear combination),the coefficients must be zero.Thus,$\vec{a} \cdot \vec{c} = \frac{1}{\sqrt{2}}$ and $\vec{a} \cdot \vec{b} = -\frac{1}{\sqrt{2}}$.We know $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.Since $\vec{a}$ and $\vec{b}$ are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.Therefore,$1 	imes 1 	imes \cos 	heta = -\frac{1}{\sqrt{2}}$.$\cos 	heta = -\frac{1}{\sqrt{2}} \implies 	heta = \frac{3\pi}{4}$.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}$,then the angle between $\vec{a}$ and $\vec{b}$ is:

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