The distance of the line from the point (1, 4, 0) along the line is: [2025]

Question

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1, 4, 0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is: [2025]

Smart Achievers · Accepted Answer

(4)

Let the parallel line is $\frac{x - 1}{1} = \frac{y - 4}{2} = \frac{z - 0}{3} = λ$

$\Rightarrow x = λ + 1, y = 2 λ + 4, z = 3 λ$

and $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4} = t$

$\Rightarrow x = 2 t + 2, y = 3 t + 6, z = 4 t + 3$

So, their point of intersection is

$(λ + 1, 2 λ + 4, 3 λ) = (2 t + 2, 3 t + 6, 4 t + 3)$

$\Rightarrow λ = 2 t + 1$

and $2 λ + 4 = 3 t + 6$ [ $∵ λ = 2 t + 1$ ]

$\Rightarrow t = 0, λ = 1$

So, point of intersection is (2, 6, 3).

So, distance $= \sqrt{{(2 - 1)}^{2} + {(6 - 4)}^{2} + {(3 - 0)}^{2}} = \sqrt{14}$ .

Solution

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

1 $\sqrt{15}$

2 $\sqrt{17}$

3 $\sqrt{13}$

4 $\sqrt{14}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. The distance of the line x–22=y–63=z–34 from the point (1, 4, 0) along the line x1=y–22=z+33 is: [2025]

1 15

2 17

3 13