Let the angle between two unit vectors and be . If the vector , then the value of is [2025]

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Solution

Q.
Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin^{- 1} (\frac{\sqrt{65}}{9})$ . If the vector $\vec{c} = 3 \hat{a} + 6 \hat{b} + 9 (\hat{a} \times \hat{b})$ , then the value of $9 (\vec{c} \cdot \hat{a}) - 3 (\vec{c} \cdot \hat{b})$ is [2025]

1 24

2 29

3 27

4 31

Ans.
(2)

Given, $\vec{c} = 3 \hat{a} + 6 \hat{b} + 9 (\hat{a} \times \hat{b})$ and

$θ = \sin^{- 1} (\frac{\sqrt{65}}{9}) \Rightarrow \sin θ = \frac{\sqrt{65}}{9} \Rightarrow \cos θ = \frac{4}{9}$ $[∵ 0 < θ < \frac{π}{2}]$

Now, $\vec{c} \cdot \hat{a} = 3 | \hat{a} |^{2} + 6 \hat{a} \cdot \hat{b} + 9 (\hat{a} \times \hat{b}) \cdot \hat{a}$

$= 3 + \frac{6 \times 4}{9} + 0$ [ $∵ \hat{a} \cdot \hat{b} = \cos θ$ ]

$= \frac{51}{9}$

and $\vec{c} \cdot \hat{b} = 3 \hat{a} \cdot \hat{b} + 6 | \hat{b} |^{2} + 9 (\hat{a} \times \hat{b}) \cdot \hat{b}$

$= \frac{3 \times 4}{9} + 6 = \frac{22}{3}$

Now, $9 (\vec{c} \cdot \hat{a}) - 3 (\vec{c} \cdot \hat{b}) = 51 - 22 = 29$ .

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