If bar{a}, bar{b}, bar{c} are non-coplanar un | vedclass

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

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

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(A) Given $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = \frac{\bar{b} + \bar{c}}{\sqrt{2}}$.Using the vector triple product formula $\bar{a} 	imes (\bar{b} 	imes \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$.So,$(\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} = \frac{1}{\sqrt{2}}\bar{b} + \frac{1}{\sqrt{2}}\bar{c}$.Rearranging terms: $(\bar{a} \cdot \bar{c} - \frac{1}{\sqrt{2}})\bar{b} - (\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}})\bar{c} = 0$.Since $\bar{b}$ and $\bar{c}$ are non-coplanar (and thus linearly independent),their coefficients must be zero.Therefore,$\bar{a} \cdot \bar{c} = \frac{1}{\sqrt{2}}$ and $\bar{a} \cdot \bar{b} = -\frac{1}{\sqrt{2}}$.Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.The angle $	heta$ between $\bar{a}$ and $\bar{b}$ is given by $\cos 	heta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \frac{-1/\sqrt{2}}{1 	imes 1} = -\frac{1}{\sqrt{2}}$.Thus,$	heta = \cos^{-1}(-\frac{1}{\sqrt{2}}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

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

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