Let\vec{a},\vec{b},and\: \vec{c}  be three unit vectors such that

\vec{a}\times \left ( \vec{b}\: \: \times\: \: \vec{c} \right )= \frac{\sqrt{3}}{2}\left ( \vec{b} +\vec{c}\right )\: if \: \vec{b}

is not parallel  to  \vec{c} then the angle between \vec{a}\: and\: \vec{b}\: is

  • Option 1)

    \frac{3\pi }{4}

  • Option 2)

    \frac{\pi }{2}

  • Option 3)

    \frac{2\pi }{3}

  • Option 4)

    \frac{5\pi }{6}

 

Answers (1)

As we learnt in 

Vector Triple Product (VTP) -

\vec{a}\times \left ( \vec{b} \times \vec{c}\right )= \left ( \vec{a}.\vec{c} \right )\vec{b}-\left ( \vec{a}.\vec{b}\right )\vec{c}

\left ( \vec{a}\times \vec{b} \right )\times \vec{c}= \left ( \vec{a}.\vec{c} \right )\vec{b}-\left ( \vec{b}.\vec{c}\right )\vec{a}

- wherein

\vec{a}, \vec{b}, \vec{c}are three vectors.

 

 \vec{a}\times \left (\vec{b}\times \vec{c} \right )= \left (\vec{a}\cdot \vec{c} \right )\vec{b}- \left (\vec{a}\cdot \vec{b} \right )\vec{c}

\left (\vec{a}\cdot \vec{c} \right )\vec{b}- \left (\vec{a}\cdot \vec{b} \right )\vec{c}=\frac{\sqrt{3}}{1}\, \vec{b}+\frac{\sqrt{3}}{1}\, \vec{c}

\vec{a}\cdot \vec{b}=-\frac{\sqrt{3}}{2}=\left | \vec{a} \right |\left | \vec{b} \right |cos\theta

Hence, cos\theta =- \frac{\sqrt{3}}{2}

\theta =\frac{5\pi }{6}

 


Option 1)

\frac{3\pi }{4}

This option is incorrect.

Option 2)

\frac{\pi }{2}

This option is incorrect.

Option 3)

\frac{2\pi }{3}

This option is incorrect.

Option 4)

\frac{5\pi }{6}

This option is correct.

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