If a→ and b→ are two vectors such that |a→|=1,|b→|=4,a·→b→=2 and c→=(2a→×b→)−3b→ then the angle between b→ and c→ is

Given c→=(2a→×b→)−3b→b→⋅c→=b→{(2a→×b→)−3b→}b→⋅c→=−3|b→|2 Now |a→×b→|2=a→2b→2−(a→⋅b→)2=16−4=12 And c→2=((2a→×b→)−3b→)2|c→|2=4|a→×b→|2+9b→2−6b→.(2a→×b→)⏟zero =48+144=192|c→|=83 Now, cos⁡θ=b→⋅c→|b→||c→|=−3|b→|2|b→||c→|=−3|b→||c→|=−3(4)83=−32 Hence (b→,c→)=5π6

If a→ and b→ are two vectors such that |a→|=1,|b→|=4,a·→b→=2 and c→=(2a→×b→)−3b→ then the angle between b→ and c→ is

Given c→=(2a→×b→)−3b→b→⋅c→=b→{(2a→×b→)−3b→}b→⋅c→=−3|b→|2 Now |a→×b→|2=a→2b→2−(a→⋅b→)2=16−4=12 And c→2=((2a→×b→)−3b→)2|c→|2=4|a→×b→|2+9b→2−6b→.(2a→×b→)⏟zero =48+144=192|c→|=83 Now, cos⁡θ=b→⋅c→|b→||c→|=−3|b→|2|b→||c→|=−3|b→||c→|=−3(4)83=−32 Hence (b→,c→)=5π6

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$If \vec{a} and \vec{b} are two vectors such that | \vec{a} | = 1, | \vec{b} | = 4, \vec{a \cdot} \vec{b} = 2 and \vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b} then the$ $angle between \vec{b} and \vec{c} is$

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$\frac{3 π}{4}$

$\frac{5 π}{6}$

$\frac{2 π}{3}$

$\frac{π}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} Given \vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b} \\ \vec{b} \cdot \vec{c} = \vec{b} {(2 \vec{a} \times \vec{b}) - 3 \vec{b}} \\ \vec{b} \cdot \vec{c} = - 3 | \vec{b} |^{2} \\ Now | \vec{a} \times \vec{b} |^{2} = {\vec{a}}^{2} {\vec{b}}^{2} - (\vec{a} \cdot \vec{b})^{2} \end{array}$

$= 16 - 4 = 12$

$And \begin{array}{l} {\vec{c}}^{2} = ((2 \vec{a} \times \vec{b}) - 3 \vec{b})^{2} \\ | \vec{c} |^{2} = 4 | \vec{a} \times \vec{b} |^{2} + 9 {\vec{b}}^{2} - \underset{zero}{\underset{⏟}{6 \vec{b} . (2 \vec{a} \times \vec{b})}} \\ = 48 + 144 = 192 \\ | \vec{c} | = 8 \sqrt{3} \end{array}$

$\begin{array}{l} Now, \cos θ = \frac{\vec{b} \cdot \vec{c}}{| \vec{b} | | \vec{c} |} = \frac{- 3 | \vec{b} |^{2}}{| \vec{b} | | \vec{c} |} = \frac{- 3 | \vec{b} |}{| \vec{c} |} = \frac{- 3 (4)}{8 \sqrt{3}} \\ = \frac{- \sqrt{3}}{2} \\ Hence (\vec{b}, \vec{c}) = \frac{5 π}{6} \end{array}$