If the shortest distance between the lines x – 2 = y – 4 3 = z – 3 4 and x – 2 4 = y – 4 6 = z – 7 8 is 13 29 , then a value of is [2024]

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Solution

Q.
If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$ , then a value of $λ$ is [2024]

1 $- \frac{13}{25}$

2 $\frac{13}{25}$

3 1

4 –1

Ans.
(3)

We have $\vec{a} = λ \hat{i} + 4 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + 4 \hat{j} + 7 \hat{k}$ and $\vec{p} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$

Shortest distance, $d =$ $| \frac{(\vec{b} - \vec{a}) \times \vec{p}}{| \vec{p} |} |$

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

$(\vec{b} - \vec{a}) \times \vec{p} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 - λ & 0 & 4 \\ 2 & 3 & 4 \end{matrix} |$

$= \hat{i} (- 12) - \hat{j} (8 - 4 λ - 8) + \hat{k} (6 - 3 λ)$

$= - 12 \hat{i} + 4 λ \hat{j} + 3 (2 - λ) \hat{k}$

$∴ d =$ $| \frac{\sqrt{144 + 16 λ^{2} + 9 {(2 - λ)}^{2}}}{\sqrt{4 + 9 + 16}} |$ $= \frac{13}{\sqrt{29}}$ (Given)

$\Rightarrow \sqrt{144 + 16 λ^{2} + 36 + 9 λ^{2} - 36 λ} = 13$

$\Rightarrow 25 λ^{2} - 36 λ + 180 = 169 \Rightarrow 25 λ^{2} - 36 λ + 11 = 0$

$\Rightarrow (25 λ - 11) (λ - 1) = 0 \Rightarrow λ = \frac{11}{25} or 1$ .

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