If a, b, c are non-coplanar unit vectors such | vedclass

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(C) Given the vector triple product formula: $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$.Equating this to the given expression: $(a \cdot c)

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

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(C) Given the vector triple product formula: $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$.Equating this to the given expression: $(a \cdot c)b - (a \cdot b)c = \frac{1}{\sqrt{2}}b + \frac{1}{\sqrt{2}}c$.Since $b$ and $c$ are non-coplanar,they are linearly independent. We can compare the coefficients of $b$ and $c$:$a \cdot c = \frac{1}{\sqrt{2}}$$-(a \cdot b) = \frac{1}{\sqrt{2}} \Rightarrow a \cdot b = -\frac{1}{\sqrt{2}}$.Since $a$ and $b$ are unit vectors,$a \cdot b = |a||b| \cos \phi$,where $\phi$ is the angle between $a$ and $b$.$1 \cdot 1 \cdot \cos \phi = -\frac{1}{\sqrt{2}}$.Therefore,$\cos \phi = -\frac{1}{\sqrt{2}}$,which implies $\phi = \frac{3\pi}{4}$.

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

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