Let the line passing through the points (–1, 2, 1) and parallel to the line intersect the line at the point P. Then the distance of P from the point Q(4, –5, 1) is [2025]

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Solution

Q.
Let the line passing through the points (–1, 2, 1) and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point P. Then the distance of P from the point Q(4, –5, 1) is [2025]

1 $5 \sqrt{5}$

2 $5 \sqrt{6}$

3 5

4 10

Ans.
(1)

Equation of line passing through (–1, 2, 1) and parallel to $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ is $\frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 1}{4}$ .

$\frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 1}{4} = λ (say)$

Coordinates of P are $(2 λ - 1, 3 λ + 2, 4 λ + 1)$ .

As point P also lies on

$\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} = μ (say)$

$∴$ Coordinates of point P are $(3 μ - 2, 2 μ + 3, μ + 4)$

$\Rightarrow 2 λ - 1 = 3 μ - 2, 3 λ + 2 = 2 μ + 3, 4 λ + 1 = μ + 4$

On solving, we get $λ = μ = 1$

$∴$ The point P is (1, 5, 5)

$∴$ Required distance $Q P = \sqrt{{(4 - 1)}^{2} + {(- 5 - 5)}^{2} + {(1 - 5)}^{2}}$

$= \sqrt{9 + 100 + 16} = 5 \sqrt{5}$

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