The shortest distance between the lines x – 3 4 = y + 7 – 11 = z – 1 5 and x – 5 3 = y – 9 – 6 = z + 2 1 is [2024]

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Solution

Q.
The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is [2024]

1 $\frac{179}{\sqrt{563}}$

2 $\frac{185}{\sqrt{563}}$

3 $\frac{187}{\sqrt{563}}$

4 $\frac{178}{\sqrt{563}}$

Ans.
(3)

Given lines can be written as

$\frac{x - 3}{4} = \frac{y - (- 7)}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z - (- 2)}{1}$

${\vec{a}}_{1} = 3 \hat{i} - 7 \hat{j} + \hat{k}, {\vec{a}}_{2} = 5 \hat{i} + 9 \hat{j} - 2 \hat{k}, {\vec{b}}_{1} = 4 \hat{i} - 11 \hat{j} + 5 \hat{k}$

and ${\vec{b}}_{2} = 3 \hat{i} - 6 \hat{j} + \hat{k}$

$∴ {\vec{a}}_{2} - {\vec{a}}_{1} = 2 \hat{i} + 16 \hat{j} - 3 \hat{k}$ and ${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 11 & 5 \\ 3 & - 6 & 1 \end{matrix} |$ $= 19 \hat{i} + 11 \hat{j} + 9 \hat{k}$

Shortest distance between two lines,

$d = \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| ({\vec{b}}_{1} \times {\vec{b}}_{2}) |} \Rightarrow d = \frac{| 38 + 176 - 27 |}{\sqrt{361 + 121 + 81}} = \frac{187}{\sqrt{563}}$ .

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