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
B
π4
C
π2
D
π

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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