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

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(A) 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

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

Vedclass · Answer

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