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

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

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 $\sqrt{14}$

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

Want Us to Email you About Special Offers & Updates?