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

Question

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

Smart Achievers · Accepted Answer

(3)

$a + b, a - b$ are mutually perpendicular $\Rightarrow (a + b) \cdot (a - b) = 0 \Rightarrow a^{2} - b^{2} = 0$

$\Rightarrow$ $| \hat{i} + λ \hat{j} - 3 \hat{k} |^{2} - | 3 \hat{i} - \hat{j} + 2 \hat{k} |^{2} = 0 \Rightarrow 1 + λ^{2} + 9 = 9 + 1 + 4 \Rightarrow λ^{2} = 4 \Rightarrow λ = 2$ $[∵ λ > 0]$

$\cos θ = \frac{a \cdot b}{| a | | b |} = \frac{(\hat{i} + 2 \hat{j} - 3 \hat{k}) \cdot (3 \hat{i} - \hat{j} + 2 \hat{k})}{| \hat{i} + 2 \hat{j} - 3 \hat{k} | | 3 \hat{i} - \hat{j} + 2 \hat{k} |}$

$= \frac{3 - 2 - 6}{\sqrt{1 + 4 + 9} \sqrt{9 + 1 + 4}} = \frac{- 5}{14}$

$\Rightarrow 14 \cos θ = - 5 \Rightarrow {(14 \cos θ)}^{2} = 25$

Solution

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

1 40

2 100

3 25

4 20

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