Question

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

A
a & b is 3π4
B
a & b is π4
C
a & c is 3π4
D
a & c is π4

Solution

## The correct options are A $$\displaystyle \vec{a}$$ & $$\displaystyle \vec{b}$$ is $$\displaystyle \frac{3\pi}{4}$$ D $$\displaystyle \vec{a}$$ & $$\displaystyle \vec{c}$$ is $$\displaystyle \frac{\pi}{4}$$$$\overrightarrow { a } \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) =\dfrac { \overrightarrow { b } +\overrightarrow { c } }{ \sqrt { 2 } }$$$$\displaystyle\Rightarrow \left( \overrightarrow { a } .\overrightarrow { c } \right) \overrightarrow { b } -\left( \overrightarrow { a } .\overrightarrow { b } \right) \overrightarrow { c } =\frac { \overrightarrow { b } }{ \sqrt { 2 } } +\frac { \overrightarrow { c } }{ \sqrt { 2 } }$$$$\displaystyle \Rightarrow \left( \overrightarrow { a } .\overrightarrow { c } -\frac { 1 }{ \sqrt { 2 } } \right) \overrightarrow { b } -\left( \overrightarrow { a } .\overrightarrow { b } +\frac { 1 }{ \sqrt { 2 } } \right) \overrightarrow { c } =0$$$$\displaystyle\therefore \overrightarrow { a } .\overrightarrow { b } =\frac { -1 }{ \sqrt { 2 } }$$ and $$\overrightarrow { a } .\overrightarrow { c } =\dfrac { 1 }{ \sqrt { 2 } }$$HenceAngle between $$\overrightarrow { a }$$ and $$\overrightarrow { b }$$ is $$\displaystyle\cos ^{ -1 } \left( -\frac { 1 }{ \sqrt { 2 } } \right) =\frac { 3\pi }{ 4 }$$And angle between $$\overrightarrow { a }$$ and $$\overrightarrow { c }$$ is $$\displaystyle\cos ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) =\frac { \pi }{ 4 }$$Maths

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