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(A) Using the vector triple product formula,$\overline{a} 	imes (\overline{b} 	imes \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} -

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Using the vector triple product formula,$\overline{a} 	imes (\overline{b} 	imes \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}$.Given $\overline{a} 	imes (\overline{b} 	imes \overline{c}) = \frac{\overline{b} + \overline{c}}{\sqrt{2}}$,we have $(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c} = \frac{1}{\sqrt{2}} \overline{b} + \frac{1}{\sqrt{2}} \overline{c}$.Rearranging the terms,we get $(\overline{a} \cdot \overline{c} - \frac{1}{\sqrt{2}}) \overline{b} - (\overline{a} \cdot \overline{b} + \frac{1}{\sqrt{2}}) \overline{c} = 0$.Since $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar,$\overline{b}$ and $\overline{c}$ are linearly independent.Therefore,the coefficients must be zero: $\overline{a} \cdot \overline{b} + \frac{1}{\sqrt{2}} = 0$.This implies $\overline{a} \cdot \overline{b} = -\frac{1}{\sqrt{2}}$.Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| |\overline{b}| \cos 	heta = -\frac{1}{\sqrt{2}}$,where $	heta$ is the angle between $\overline{a}$ and $\overline{b}$.Thus,$\cos 	heta = -\frac{1}{\sqrt{2}}$,which gives $	heta = \frac{3 \pi}{4}$.

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

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