The distance of the point (7, 10, 11) from the line x - 41 = y - 40 = z - 23 along the line x - 72 = y - 103 = z - 116 is

Given: Point is P(7,10,11). The line is L1: x-41= y-40= z-23 and the distance is to be measured along the line L2: x-72= y-103= z-116. Find: The distance of point P from line L1 along line L2. Take a general point on L1 by putting x-41= y-40= z-23= Then x=4+, y=4, z=2+3 So a general point on L1 is Q(4+,4,2+3). A general point on L2 is obtained by putting x-72= y-103= z-116= Thus x=7+2, y=10+3, z=11+6 At the point of intersection, coordinates from the two lines must be equal. Therefore, 4+ =7+2 4=10+3 2+3 =11+6 From 4=10+3 we get 3 =-6 =-2 Now substitute in 4+ =7+2 So, 4+ =7-4=3 =-1 Hence the intersection point is Q=(4-1,4,2+3(-1))=(3,4,-1) The required distance is PQ: PQ= (7-3)^2+(10-4)^2+(11-(-1))^2 = 4^2+6^2+12^2 = 16+36+144 = 196=14 Therefore, the required distance is 14, so the correct option is B. Given: P(7,10,11) and a variable point Q on L1. Find: Distance PQ when PQ is along the given line with direction ratios (2,3,6). Let Q=( +4,4,3 +2) Then direction ratios of PQ are ( +4-7, 4-10, 3 +2-11)=( -3,-6,3 -9) Since PQ is parallel to the line with direction ratios (2,3,6), -32= -63= 3 -96 Now -63=-2 So, -32=-2 -3=-4 =-1 Thus Q=(3,4,-1) Therefore, PQ= (7-3)^2+(10-4)^2+(11+1)^2= 196=14 Hence the correct option is B.

The distance of the point (7, 10, 11) from the line x - 41 = y -… — Solution | JEE Mathematics

MCQMediumJEE 2025Equation of Line in 3D

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 $\frac{x - 7}{2} = \frac{y - 10}{3} = \frac{z - 11}{6}$ is

A
$18$
B
$14$
C
$12$
D
$16$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Point is $P(7,10,11)$ . The line is $L_1: \frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ and the distance is to be measured along the line $L_2: \frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}$ .

Find: The distance of point $P$ from line $L_1$ along line $L_2$ .

Take a general point on $L_1$ by putting

\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}=\lambda

Then

x=4+\lambda,\quad y=4,\quad z=2+3\lambda

So a general point on $L_1$ is $Q(4+\lambda,4,2+3\lambda)$ .

A general point on $L_2$ is obtained by putting

\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\mu

Thus

x=7+2\mu,\quad y=10+3\mu,\quad z=11+6\mu

Using intersection of the two lines

At the point of intersection, coordinates from the two lines must be equal. Therefore,

4+\lambda=7+2\mu

4=10+3\mu

2+3\lambda=11+6\mu

From

4=10+3\mu

we get

3\mu=-6 \Rightarrow \mu=-2

Now substitute in

4+\lambda=7+2\mu

So,

4+\lambda=7-4=3

\lambda=-1

Hence the intersection point is

Q=(4-1,4,2+3(-1))=(3,4,-1)

The required distance is $PQ$ :

PQ=\sqrt{(7-3)^2+(10-4)^2+(11-(-1))^2}

=\sqrt{4^2+6^2+12^2}

=\sqrt{16+36+144}

=\sqrt{196}=14

Therefore, the required distance is $14$ , so the correct option is B.

Using parallel direction ratios

Given: $P(7,10,11)$ and a variable point $Q$ on $L_1$ .

Find: Distance $PQ$ when $PQ$ is along the given line with direction ratios $(2,3,6)$ .

Let

Q=(\lambda+4,4,3\lambda+2)

Then direction ratios of $PQ$ are

(\lambda+4-7,\;4-10,\;3\lambda+2-11)=(\lambda-3,-6,3\lambda-9)

Since $PQ$ is parallel to the line with direction ratios $(2,3,6)$ ,

\frac{\lambda-3}{2}=\frac{-6}{3}=\frac{3\lambda-9}{6}

Now

\frac{-6}{3}=-2

So,

\frac{\lambda-3}{2}=-2 \Rightarrow \lambda-3=-4 \Rightarrow \lambda=-1

Thus

Q=(3,4,-1)

Therefore,

PQ=\sqrt{(7-3)^2+(10-4)^2+(11+1)^2}=\sqrt{196}=14

Hence the correct option is B.

Common mistakes

Treating the problem as the shortest distance from a point to a line is incorrect. Here the distance is measured along a given line, so the joining line must be parallel to the second line, not perpendicular to the first line.
Misreading $\frac{y-4}{0}$ and trying to divide by zero is wrong. It means the direction ratio along $y$ is $0$ , so on $L_1$ the coordinate remains constant: $y=4$ .
While forming the point on $L_1$ , students may write $Q=(4+\lambda,\lambda+4,2+3\lambda)$ . This is incorrect because the middle direction ratio is $0$ . The correct point is $Q=(4+\lambda,4,2+3\lambda)$ .

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