The distance of the point (7, 10, 11) from the line along the line is [2025]

Question

The distance of the point (7, 10, 11) from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is [2025]

Smart Achievers · Accepted Answer

(1)

Let $P \equiv (7, 10, 11)$

Any point on the line

$L_{1} : \frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3} = λ$

has coordinates

$x = λ + 4, y = 4, z = 3 λ + 2$

i.e., $Q (λ + 4, 4, 3 λ + 2)$

$∵$ Line PQ is parallel to line

$L_{2} : \frac{x - 9}{2} = \frac{y - 3}{3} = \frac{z - 17}{6}$

$\Rightarrow \frac{λ + 4 - 7}{2} = \frac{4 - 10}{3} = \frac{3 λ + 2 - 11}{6} \Rightarrow λ = - 1$

$∴$ Q(3, 4, –1) is the point on the line $L_{1}$ .

Hence, $P Q = \sqrt{16 + 36 + 144} = 14$ units.

Solution

Q.
The distance of the point (7, 10, 11) from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is [2025]

1 14

2 18

3 16

4 12

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