Let a ? = a 1 i ^ + a 2 j ^ + a 3 k ^ and b ? = b 1 i ^ + b 2 j ^ + b 3 k ^ be two vectors such that, | a ? | = 1 , a ? · b ? = 2 and | b ? | = 4 . If c ? = 2 ( a ? × b ? ) – 3 b ? then the angle between b ? and c ? is equal to [2

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Solution

Q.
Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ 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 [2024]

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

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

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

4 $\cos^{- 1} (\frac{2}{3})$

Ans.
(1)

We have, $| \vec{a} | = 1$ , $\vec{a} \cdot \vec{b} = 2$ and $| \vec{b} | = 4$

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

$\Rightarrow \vec{b} \cdot \vec{c} = 2 \vec{b} \cdot (\vec{a} \times \vec{b}) - 3 \vec{b} \cdot \vec{b} = 2 (0) - 3 | \vec{b} |^{2}$

$\Rightarrow \vec{b} \cdot \vec{c} = - 48 \Rightarrow | \vec{b} | | \vec{c} | \cos θ = - 48$

$\Rightarrow | \vec{c} | \cos θ = - 12$ ... (i)

Consider, $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b} \Rightarrow | \vec{c} |^{2} = 4 | \vec{a} \times \vec{b} |^{2} + 9 | \vec{b} |^{2}$

$\Rightarrow | \vec{c} |^{2} = 4 (| \vec{a} |^{2} | \vec{b} |^{2} - {(\vec{a} \cdot \vec{b})}^{2}) + 9 | \vec{b} |^{2}$

$\Rightarrow | \vec{c} |^{2} = 4 (16 - 4) + 144 = 48 + 144$

$\Rightarrow | \vec{c} |^{2} = 192 \Rightarrow \vec{c} = 8 \sqrt{3}$

From (i), $\cos θ = \frac{- 12}{8 \sqrt{3}}$ [Using (i)]

$\Rightarrow \cos θ = \frac{- 3}{2 \sqrt{3}} \Rightarrow \cos θ = \frac{- \sqrt{3}}{2} \Rightarrow θ = \cos^{- 1} (- \frac{\sqrt{3}}{2})$

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