The shortest distance, between lines and , where and is the line, passing through the points A(–4, 4, 3), B(–1, 6, 3) and perpendicular to the line , is [2024]

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Solution

Q.
The shortest distance, between lines $L_{1}$ and $L_{2}$ , where $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$ and $L_{2}$ is the line, passing through the points A(–4, 4, 3), B(–1, 6, 3) and perpendicular to the line $\frac{x - 3}{- 2} = \frac{y}{3} = \frac{z - 1}{1}$ , is [2024]

1 $\frac{141}{\sqrt{221}}$

2 $\frac{42}{\sqrt{117}}$

3 $\frac{121}{\sqrt{221}}$

4 $\frac{24}{\sqrt{117}}$

Ans.
(1)

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

Equation of the line passing through the points A(–4, 4, 3) and B(–1, 6, 3) is

$L_{2} = \frac{x + 4}{3} = \frac{y - 4}{2} = \frac{z - 3}{0}$

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

$a_{2} = 3, b_{2} = 2, c_{2} = 0; x_{2} = - 4, y_{2} = 4, z_{2} = 3$

$∴$ Required shortest distance

$d =$ $| \frac{| \begin{matrix} - 5 & 5 & 7 \\ 2 & - 3 & 2 \\ 3 & 2 & 0 \end{matrix} |}{\sqrt{{(4 + 9)}^{2} + {(0 - 4)}^{2} + {(6 - 0)}^{2}}} |$

$=$ $| \frac{- 5 (0 - 4) - 5 (0 - 6) + 7 (4 + 9)}{\sqrt{169 + 16 + 36}} |$ $=$ $| \frac{20 + 30 + 91}{\sqrt{221}} |$ $= \frac{141}{\sqrt{221}}$ .

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