For , let be the angle between the vectors and . If the vectors and are mutually perpendicular, then the value of is equal to [2024]

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Solution

Q.
For $λ > 0$ , let $θ$ be the angle between the vectors $\vec{a} = \hat{i} + λ \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ . If the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular, then the value of ${(14 \cos θ)}^{2}$ is equal to [2024]

1 20

2 25

3 50

4 40

Ans.
(2)

$\vec{a} + \vec{b} = 4 \hat{i} + (λ - 1) \hat{j} - \hat{k}$ and $\vec{a} - \vec{b} = - 2 \hat{i} + (λ + 1) \hat{j} - 5 \hat{k}$

Now, given that $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular.

$\Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0$

$\Rightarrow - 8 + λ^{2} - 1 + 5 = 0 \Rightarrow λ = \pm 2$

$∴ λ = 2 (∵ λ > 0)$

Now, $\cos θ \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | \cdot | \vec{b} |} = \frac{(\hat{i} + 2 \hat{j} - 3 \hat{k}) \cdot (3 \hat{i} - \hat{j} + 2 \hat{k})}{\sqrt{{(1)}^{2} + {(2)}^{2} + {(3)}^{2}} \sqrt{{(3)}^{2} + {(- 1)}^{2} + {(2)}^{2}}}$

$\Rightarrow \cos θ = \frac{- 5}{14}$

$∴ {(14 \cos θ)}^{2} = {(14 \times \frac{- 5}{14})}^{2} = 25$ .

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