The shortest distance between the lines x – 3 2 = y + 15 – 7 = z – 9 5 and x + 1 2 = y – 1 1 = z – 9 – 3 is [2024]

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Solution

Q.
The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is [2024]

1 $8 \sqrt{3}$

2 $5 \sqrt{3}$

3 $4 \sqrt{3}$

4 $6 \sqrt{3}$

Ans.
(3)

$L_{1} : \frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$

$L_{2} : \frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$

$\vec{a_{1}} = 3 \hat{i} - 15 \hat{j} + 9 \hat{k}, \vec{a_{2}} = - \hat{i} + \hat{j} + 9 \hat{k}$

$\vec{b_{1}} = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}, \vec{b_{2}} = 2 \hat{i} + \hat{j} - 3 \hat{k}$

Shortest distance = $= \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$

$= \frac{| (- 4 \hat{i} + 16 \hat{j}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} | |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$

$= \frac{| (- 4 \hat{i} + 16 \hat{j}) \cdot (16 \hat{i} + 16 \hat{j} + 16 \hat{k}) |}{| 16 \hat{i} + 16 \hat{j} + 16 \hat{k} |} = \frac{192}{16 \sqrt{3}} = 4 \sqrt{3}$ .

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