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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
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B
a & b is π4
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C
a & c is 3π4
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D
a & c is π4
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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 $$\overrightarrow { a } .\overrightarrow { c } =\dfrac { 1 }{ \sqrt { 2 }  } $$
Hence
Angle 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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