Let a→=aii^+a2j^+a3k^ and b→=b1i^+b2j^+b3k^ be two vectors such that |a→|=1;a→·b→=2 and |b→|=4. If c→=2(a→×b→)-3b→, then the angle between b→ and c→ is equal to :

Question

Let

\vec{a} = a_{i} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}

and

\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}

be two vectors such that

| \vec{a} | = 1; \vec{a} \cdot \vec{b} = 2

and

| \vec{b} | = 4

. If

\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}

, then the angle between

\vec{b}

and

\vec{c}

is equal to :

\cos^{- 1} (\frac{2}{\sqrt{3}})

\cos^{- 1} (- \frac{1}{\sqrt{3}})

\cos^{- 1} (- \frac{\sqrt{3}}{2})

\cos^{- 1} (\frac{2}{3})

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