The shortest distance between the lines x - 2 3 = y + 1 2 = z - 6 2 and x - 6 3 = 1 - y 2 = z + 8 0 is equal to ______ . [2023]

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Solution

Q.
The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______ . [2023]

Ans.
(14)

$L_{1} : \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$

$L_{2} : \frac{x - 6}{3} = \frac{y - 1}{- 2} = \frac{z + 8}{0}$

$Shortest distance =$ $\frac{| | \begin{matrix} 6 - 2 & 1 - (- 1) & - 8 - 6 \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}{| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}$

$=$ $\frac{| | \begin{matrix} 4 & 2 & - 14 \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}{| 4 \hat{i} + 6 \hat{j} - 12 \hat{k} |}$ $= \frac{4 (0 + 4) - 2 (0 - 6) - 14 (- 6 - 6)}{\sqrt{4^{2} + 6^{2} + 12^{2}}}$

$= \frac{16 + 12 + 14 \times 12}{14} = 14 units.$

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