Let , , and be a unit vector such that and . If is perpendicular to , then is equal to __________. [2025]

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Solution

Q.
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 3 \hat{i} + 2 \hat{j} - \hat{k}$ , $\vec{c} = μ \hat{j} + u \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d} = 1$ . If $\vec{c}$ is perpendicular to $\vec{a}$ , then $| 3 λ \hat{d} + μ \vec{c} |^{2}$ is equal to __________. [2025]

Ans.
(5)

We have, $\vec{a} \times \hat{d} - \vec{b} \times \hat{d} = 0$

$\Rightarrow (\vec{a} - \vec{b}) \times \hat{d} = 0$

$∴ \hat{d} = t (\vec{a} - \vec{b})$ , for any scalar t.

$\Rightarrow \hat{d} = t (- 2 \hat{i} - \hat{j} + 2 \hat{k})$

Since, $| \hat{d} |$ $= 1$

$∴$ $| t |$ $= \frac{1}{3}$

Now, $\vec{c} \cdot \vec{a} = 0$ [ $∵ \vec{c}$ is perpendicular to $\vec{a}$ ]

$\Rightarrow λ + μ = 0 \Rightarrow μ = - λ$

$∴ \vec{c} = λ (\hat{j} - \hat{k}) \Rightarrow | \vec{c} |^{2} = 2 λ^{2}$

Also, $\vec{c} \cdot \hat{d} = 1$

$\Rightarrow λ (\hat{j} - \hat{k}) \cdot t (- 2 \hat{i} - \hat{j} + 2 \hat{k}) = 1$

$\Rightarrow λ t = \frac{- 1}{3} \Rightarrow λ^{2} = 1$ $[∵ | t | = \frac{1}{3}]$

$∴ | 3 λ \hat{d} + μ \vec{c} |^{2} = 9 λ^{2} | \hat{d} |^{2} + μ^{2} | \vec{c} |^{2} + 6 λ μ (\hat{d} \cdot \vec{c})$

$= 3 λ^{2} + 2 λ^{4} = 5$ [ $∵ λ^{2} = 1$ ]

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