The shortest distance between the lines x - 4 4 = y + 2 5 = z + 3 3 and x - 1 3 = y - 3 4 = z - 4 2 is [2023]

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Solution

Q.
The shortest distance between the lines $\frac{x - 4}{4} = \frac{y + 2}{5} = \frac{z + 3}{3}$ and $\frac{x - 1}{3} = \frac{y - 3}{4} = \frac{z - 4}{2}$ is [2023]

1 $3 \sqrt{6}$

2 $6 \sqrt{2}$

3 $2 \sqrt{6}$

4 $6 \sqrt{3}$

Ans.
(1)

$We have,$ $l_{1} : \frac{x - 4}{4} = \frac{y + 2}{5} = \frac{z + 3}{3},$ $l_{2} : \frac{x - 1}{3} = \frac{y - 3}{4} = \frac{z - 4}{2}$

$Here, x_{1} = 4, y_{1} = - 2, z_{1} = - 3, a_{1} = 4, b_{1} = 5, c_{1} = 3$

$x_{2} = 1, y_{2} = 3, z_{2} = 4, a_{2} = 3, b_{2} = 4, c_{2} = 2$

$Shortest distance,$ $d =$ $| \frac{| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} |$

$= \frac{| \begin{matrix} 1 - 4 & 3 + 2 & 4 + 3 \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{matrix} |}{\sqrt{{(4 \times 4 - 3 \times 5)}^{2} + {(5 \times 2 - 4 \times 3)}^{2} + {(3 \times 3 - 4 \times 2)}^{2}}}$

$= \frac{18}{\sqrt{6}} = 3 \sqrt{6}$

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