Line passes through the point (1, 2, 3) and is parallel to z-axis. Line passes through the point (, 5, 6) and is parallel to y-axis. Let for , , the shortest distance between the two lines be 3. Then the square of the distance of the point from the lin

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Solution

Q.
Line $L_{1}$ passes through the point (1, 2, 3) and is parallel to z-axis. Line $L_{2}$ passes through the point ( $λ$ , 5, 6) and is parallel to y-axis. Let for $λ = λ_{1}, λ_{2}$ , $λ_{2} < λ_{1}$ , the shortest distance between the two lines be 3. Then the square of the distance of the point $(λ_{1}, λ_{2}, 7)$ from the line $L_{1}$ is [2025]

1 40

2 32

3 25

4 37

Ans.
(3)

$L_{1} \equiv \frac{x - 1}{0} = \frac{y - 2}{0} = \frac{z - 3}{1}$

[ $∵ L_{1}$ is parallel to z-axis and passing through (1, 2, 3)]

and $L_{2} \equiv \frac{x - λ}{0} = \frac{y - 5}{1} = \frac{z - 6}{0}$

[ $∵ L_{2}$ is parallel to y-axis and passing through $(λ, 5, 6)$ ]

Now, shortest distance between $L_{1}$ and $L_{2}$ is given by

$\frac{| \begin{matrix} λ - 1 & 3 & 3 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} |}{\sqrt{{(- 1)}^{2} + 0 + 0}}$ $=$ $| λ - 1 |$

$\Rightarrow | λ - 1 | = 3 \Rightarrow λ = 4, - 2$

$\Rightarrow λ_{1} = 4, λ_{2} = - 2$ [ $∵ λ_{2} < λ_{1}$ ]

Let the foot the perpendicular from point P(4, –2, 7) to the line $L_{1}$ is $Q (1, 2, μ + 3)$ .

Direction ratios of the QP are (3, –4, 4 – $μ$ )

$∵$ QP is perpendicular to $L_{1}$

So, $3 \times 0 - 4 \times 0 + (4 - μ) \times 1 = 0 \Rightarrow μ = 4$

The coordinates of point Q are (1, 2, 7).

$∴ P Q^{2} = 9 + 16 + 0 \Rightarrow P Q^{2} = 25$

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