• # question_answer If a, b, c are non-coplanar unit vectors such that $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=\frac{\mathbf{b}+\mathbf{c}}{\sqrt{2}}$, then the angle between a and b is [IIT 1995] A) $\frac{\pi }{4}$ B) $\frac{\pi }{2}$ C) $\frac{3\pi }{4}$ D) $\pi$

• $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=\frac{\mathbf{b}+\mathbf{c}}{\sqrt{2}}\Rightarrow (\mathbf{a}\,.\,\mathbf{c})\mathbf{b}-(\mathbf{a}\,.\,\mathbf{b})\,\mathbf{c}=\frac{\mathbf{b}+\mathbf{c}}{\sqrt{2}}$
• $\Rightarrow \left[ (\mathbf{a}\,.\,\mathbf{c})-\frac{1}{\sqrt{2}} \right]\mathbf{b}-\left[ (\mathbf{a}\,.\,\mathbf{b})+\frac{1}{\sqrt{2}} \right]\,\mathbf{c}=0$
• $\Rightarrow \mathbf{a}\,.\,\mathbf{c}=\frac{1}{\sqrt{2}},$ $\mathbf{a}\,.\,\mathbf{b}=-\frac{1}{\sqrt{2}}$
• $\Rightarrow \,|\mathbf{a}|\,|\mathbf{c}|\cos \theta =\frac{1}{\sqrt{2}},$ $|\mathbf{a}|\,|\mathbf{b}|\cos \varphi =-\frac{1}{\sqrt{2}}$
• $\Rightarrow \cos \theta =\frac{1}{\sqrt{2}},$ $\cos \varphi =-\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4},$ $\varphi =\frac{3\pi }{4}.$