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Question

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


A
8π2
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B
π4
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C
π2
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D
π
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Solution

The correct option is B $$\displaystyle \dfrac{\pi }{4}$$
Given $$\displaystyle \bar{a}\times \left ( \bar{b}\times \bar{c} \right )=\dfrac{\bar{b}+\bar{c}}{\sqrt{2}}$$
$$\displaystyle \Rightarrow \left ( \bar{a}\cdot \bar{c} \right )\bar{b}-\left ( \bar{a} \cdot \bar{b}\right )\bar{c}=\dfrac{\bar{b}+\bar{c}}{\sqrt{2}}$$
$$\displaystyle \Rightarrow \left ( \bar{a}\cdot \bar{c}-\dfrac{1}{\sqrt{2}} \right )\bar{b}-\left ( \bar{a}\cdot \bar{b}+\dfrac{1}{\sqrt{2}} \right )\bar{c}=\bar{0}$$
$$\displaystyle \Rightarrow \bar{a}\cdot \bar{c}=\dfrac{1}{\sqrt{2}}$$ and $$\displaystyle \bar{a}\cdot \bar{b}=-\dfrac{1}{\sqrt{2}}$$
($$\displaystyle \because \bar{a}, \bar{b}, \bar{c}$$ are non-coplanar)
$$\displaystyle \cos \theta =\dfrac{\bar{a}\cdot \bar{b}} {\left | \bar{a} \right |\left | \bar{b} \right |}=\dfrac{-1/\sqrt{2}}{1.1}$$
$$\displaystyle =\dfrac{-1}{\sqrt{2}}$$
$$\displaystyle \therefore \theta =\pi -\dfrac{\pi }{4}$$
$$\displaystyle  =\dfrac{3\pi }{4}$$
$$\displaystyle  \therefore $$ angle between $$\displaystyle  \bar{a} $$ and $$\displaystyle  \bar{b} $$ is $$\displaystyle \dfrac{\pi }{4}$$

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