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Question

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


A
3π4
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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 D $$\displaystyle \dfrac{3\pi}{4}$$
$$\vec {a}\times(\vec {b}\times\vec {c})$$
$$=\vec {b}(\vec {a}.\vec {c})-\vec {c}\vec {a}.\vec {b})$$
$$=\dfrac{\vec {b}}{\sqrt{2}}+\dfrac{\vec {c}}{\sqrt{2}}$$
Comparing coefficients, we get
$$\vec {a}.\vec {c}=\dfrac{1}{\sqrt{2}}$$ and $$\vec {a}.\vec {b}=\dfrac{-1}{\sqrt{2}}$$
If $$\theta $$ is the angle between $$\vec {a} $$ and $$\vec {b}$$, then $$\cos\theta = \displaystyle \dfrac{-1}{2}$$
Hence angle between $$\vec {a}$$ and $$\vec {b}$$ is $$\dfrac{3\pi}{4}$$

Maths

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