If a→,b→,c→ are non-coplanar unit vector such that a→×(b→×c→)=12(b→+c→) then the angle between the vectors a→,b→ is

(a→⋅c→)b→−(a→⋅b→)c→=12b→+12c→∴a→.c→=12 and a→⋅b→=−12⇒ angle between a→ and c→=π4 and angle between a→ and b→=3π4

If a→,b→,c→ are non-coplanar unit vector such that a→×(b→×c→)=12(b→+c→) then the angle between the vectors a→,b→ is

(a→⋅c→)b→−(a→⋅b→)c→=12b→+12c→∴a→.c→=12 and a→⋅b→=−12⇒ angle between a→ and c→=π4 and angle between a→ and b→=3π4

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$If \vec{a}, \vec{b}, \vec{c} are non-coplanar unit vector such that$ $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{\sqrt{2}} (\vec{b} + \vec{c}) then the angle between the vectors$ $\vec{a}, \vec{b} is$

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$\frac{3 π}{4}$

$\frac{π}{4}$

$\frac{π}{8}$

$\frac{π}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{\sqrt{2}} \vec{b} + \frac{1}{\sqrt{2}} \vec{c} \\ ∴ \vec{a} . \vec{c} = \frac{1}{\sqrt{2}} and \vec{a} \cdot \vec{b} = - \frac{1}{\sqrt{2}} \\ \Rightarrow angle between \vec{a} and \vec{c} = \frac{π}{4} and angle between \vec{a} and \vec{b} = \frac{3 π}{4} \end{array}$