Logo
Jeeify
MCQMediumJEE 2025Equation of Line in 3D

JEE Mathematics 2025 Question with Solution

Let a straight line LL pass through the point P(2,−1,3)P(2, -1, 3) and be perpendicular to the lines x−12=y+11=z−3−2andx−31=y−23=z+24.\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \quad \text{and} \quad \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. If the line LL intersects the yzyz-plane at the point QQ, then the distance between the points PP and QQ is:

  • A

    22

  • B

    10\sqrt{10}

  • C

    33

  • D

    232\sqrt{3}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The line LL passes through P(2,−1,3)P(2, -1, 3) and is perpendicular to the two lines with direction ratios (2,1,−2)(2, 1, -2) and (1,3,4)(1, 3, 4).

Find: The distance PQPQ, where QQ is the point where LL intersects the yzyz-plane.

Since LL is perpendicular to both given lines, its direction vector is the cross product of their direction vectors.

dL⃗=(2,1,−2)×(1,3,4)\vec{d_L} = (2, 1, -2) \times (1, 3, 4)

Using determinant form,

dL⃗=∣i^j^k^21−2134∣=i^(1×4−(−2)×3)−j^(2×4−(−2)×1)+k^(2×3−1×1)\vec{d_L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = \hat{i}(1 \times 4 - (-2) \times 3) - \hat{j}(2 \times 4 - (-2) \times 1) + \hat{k}(2 \times 3 - 1 \times 1) dL⃗=i^(4+6)−j^(8+2)+k^(6−1)\vec{d_L} = \hat{i}(4 + 6) - \hat{j}(8 + 2) + \hat{k}(6 - 1) dL⃗=10i^−10j^+5k^\vec{d_L} = 10\hat{i} - 10\hat{j} + 5\hat{k}

So a convenient set of direction ratios for LL is (2,−2,1)(2, -2, 1).

Therefore, the line through P(2,−1,3)P(2, -1, 3) is

x−22=y+1−2=z−31=λ\frac{x - 2}{2} = \frac{y + 1}{-2} = \frac{z - 3}{1} = \lambda

To find its intersection with the yzyz-plane, set x=0x = 0.

0−22=λ\frac{0 - 2}{2} = \lambda λ=−1\lambda = -1

Substituting λ=−1\lambda = -1 into the parametric equations,

y=−1−2(−1)=1,z=3+(−1)=2y = -1 - 2(-1) = 1, \qquad z = 3 + (-1) = 2

Hence,

Q(0,1,2)Q(0, 1, 2)

Now use the distance formula:

PQ=(2−0)2+(−1−1)2+(3−2)2PQ = \sqrt{(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2} PQ=4+4+1=9=3PQ = \sqrt{4 + 4 + 1} = \sqrt{9} = 3

Therefore, the distance between PP and QQ is 33. The correct option is C.

The first approach in the source gives Q(0,0,2)Q(0, 0, 2) and writes 4+3+2\sqrt{4 + 3 + 2}, but this is inconsistent with the line equation and the distance formula. The second approach correctly gives Q(0,1,2)Q(0, 1, 2) and hence PQ=3PQ = 3.

Common mistakes

  • Using the direction ratios of LL directly as one of the given lines is incorrect. Since LL is perpendicular to both lines, its direction vector must be obtained from the cross product of the two given direction vectors.

  • While finding the intersection with the yzyz-plane, students may forget that the condition is x=0x = 0. Setting y=0y = 0 or z=0z = 0 would correspond to different coordinate planes.

  • After getting the parameter value, substituting it incorrectly into yy and zz can lead to the wrong point QQ. Use the same parameter consistently in all three parametric equations.

Practice more Equation of Line in 3D questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Equation of Line in 3D

Related questions