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3D Geometry question

2025 · 23 Jan · Shift 2 · Q36
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  5. /2025 · 23 Jan · Shift 2 · Q36

3D Geometry question

2025 · 23 Jan · Shift 2 · Q36

JEE MainMathematics3D GeometryMCQ+4 / −1
The distance of the line x−22=y−63=z−34\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}2x−2​=3y−6​=4z−3​ from the point (1,4,0)(1,4,0)(1,4,0) along the line x1=y−22=z+33\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}1x​=2y−2​=3z+3​ is :
  1. A
    17\sqrt{17}17​
  2. B
    13\sqrt{13}13​
  3. C
    15\sqrt{15}15​
  4. D
    14\sqrt{14}14​
View written solutionFree

Correct answer: D

  1. Interpret the question

We need the distance from the point P(1,4,0)P(1,4,0)P(1,4,0) to the given line

L1:x−22=y−63=z−34L_1:\quad \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}L1​:2x−2​=3y−6​=4z−3​

along the line

L2:x1=y−22=z+33.L_2:\quad \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}.L2​:1x​=2y−2​=3z+3​.

This means: starting from P(1,4,0)P(1,4,0)P(1,4,0), move in the direction of L2L_2L2​ until we hit L1L_1L1​. The required distance is the length of that segment.

  1. Parametric equations of the two lines

For L1L_1L1​, let the parameter be λ\lambdaλ:

x−22=y−63=z−34=λ\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}=\lambda2x−2​=3y−6​=4z−3​=λ

So,

x=2+2λ,y=6+3λ,z=3+4λ.x=2+2\lambda,\quad y=6+3\lambda,\quad z=3+4\lambda.x=2+2λ,y=6+3λ,z=3+4λ.

For L2L_2L2​, let the parameter be ttt:

x1=y−22=z+33=t\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}=t1x​=2y−2​=3z+3​=t

So,

x=t,y=2+2t,z=−3+3t.x=t,\quad y=2+2t,\quad z=-3+3t.x=t,y=2+2t,z=−3+3t.

Since the required line passes through P(1,4,0)P(1,4,0)P(1,4,0) and is parallel to L2L_2L2​, its direction ratios are (1,2,3)(1,2,3)(1,2,3). Hence its equation is

x−11=y−42=z−03=s.\frac{x-1}{1}=\frac{y-4}{2}=\frac{z-0}{3}=s.1x−1​=2y−4​=3z−0​=s.

Thus points on this line are

x=1+s,y=4+2s,z=3s.x=1+s,\quad y=4+2s,\quad z=3s.x=1+s,y=4+2s,z=3s.
  1. Find intersection of this line with L1L_1L1​

At the intersection,

1+s=2+2λ...(1)1+s=2+2\lambda \quad ...(1)1+s=2+2λ...(1) 4+2s=6+3λ...(2)4+2s=6+3\lambda \quad ...(2)4+2s=6+3λ...(2) 3s=3+4λ...(3)3s=3+4\lambda \quad ...(3)3s=3+4λ...(3)

From (1):

s=1+2λ.s=1+2\lambda.s=1+2λ.

From (3):

3(1+2λ)=3+4λ3(1+2\lambda)=3+4\lambda3(1+2λ)=3+4λ 3+6λ=3+4λ3+6\lambda=3+4\lambda3+6λ=3+4λ 2λ=0⇒λ=0.2\lambda=0 \Rightarrow \lambda=0.2λ=0⇒λ=0.

Then

s=1.s=1.s=1.

Check in (2):

4+2(1)=6+3(0)=6,4+2(1)=6+3(0)=6,4+2(1)=6+3(0)=6,

which is correct.

So the intersection point is obtained at s=1s=1s=1 on the line through PPP parallel to L2L_2L2​.

  1. Compute the distance

The direction vector is (1,2,3)(1,2,3)(1,2,3), so moving by parameter s=1s=1s=1 gives displacement

d⃗=(1,2,3).\vec{d}=(1,2,3).d=(1,2,3).

Hence distance is

∣d⃗∣=12+22+32=14.|\vec{d}|=\sqrt{1^2+2^2+3^2}=\sqrt{14}.∣d∣=12+22+32​=14​.
  1. Option check
  • A: 17\sqrt{17}17​
  • B: 13\sqrt{13}13​
  • C: 15\sqrt{15}15​
  • D: 14\sqrt{14}14​

Therefore the correct option is

14.\boxed{\sqrt{14}}.14​​.
  1. Comparison with stored answer

Stored correct answer: DDD

Our derived answer: DDD

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