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MCQMediumJEE 2025Skew Lines & Shortest Distance

JEE Mathematics 2025 Question with Solution

Line L1L_1 passes through the point (1,2,3)(1, 2, 3) and is parallel to z-axis. Line L2L_2 passes through the point (λ,5,6)(\lambda, 5, 6) and is parallel to y-axis. Let for λ=λ1,λ2,λ2<λ1\lambda = \lambda_1, \lambda_2, \lambda_2 < \lambda_1, the shortest distance between the two lines be 33. Then the square of the distance of the point (λ1,λ2,7)(\lambda_1, \lambda_2, 7) from the line L1L_1 is

  • A

    4040

  • B

    3232

  • C

    2525

  • D

    3737

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Line L1L_1 passes through (1,2,3)(1,2,3) and is parallel to the z-axis, so its direction vector is  (0,0,1)\,(0,0,1). Line L2L_2 passes through (λ,5,6)(\lambda,5,6) and is parallel to the y-axis, so its direction vector is  (0,1,0)\,(0,1,0).

Find: The square of the distance of the point (λ1,λ2,7)(\lambda_1, \lambda_2, 7) from line L1L_1.

Using the shortest distance formula for two skew lines,

D=∣(b−a)⋅(d1×d2)∣∣d1×d2∣D = \frac{|(\mathbf{b-a}) \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|}

Here,

d1=(0,0,1),d2=(0,1,0)\mathbf{d_1} = (0,0,1), \qquad \mathbf{d_2} = (0,1,0)

so

d1×d2=(−1,0,0)\mathbf{d_1} \times \mathbf{d_2} = (-1,0,0)

and

∣d1×d2∣=1|\mathbf{d_1} \times \mathbf{d_2}| = 1

Take points A(1,2,3)A(1,2,3) on L1L_1 and B(λ,5,6)B(\lambda,5,6) on L2L_2. Then

B−A=(λ−1,3,3)\mathbf{B-A} = (\lambda-1, 3, 3)

Hence,

D=∣(λ−1,3,3)⋅(−1,0,0)∣=∣1−λ∣=∣λ−1∣D = |(\lambda-1,3,3) \cdot (-1,0,0)| = |1-\lambda| = |\lambda-1|

Given the shortest distance is 33,

∣λ−1∣=3|\lambda-1| = 3

So,

λ−1=±3\lambda - 1 = \pm 3

which gives

λ=4,−2\lambda = 4, -2

Since λ2<λ1\lambda_2 < \lambda_1, we get

λ1=4,λ2=−2\lambda_1 = 4, \qquad \lambda_2 = -2

Therefore, the point is

P(4,−2,7)P(4,-2,7)

Let QQ be the foot of the perpendicular from PP to line L1L_1. A general point on L1L_1 is

Q=(1,2,3+t)Q = (1,2,3+t)

Then

PQ→=(1−4,2−(−2),3+t−7)=(−3,4,t−4)\overrightarrow{PQ} = (1-4, 2-(-2), 3+t-7) = (-3,4,t-4)

Since PQ⊥L1PQ \perp L_1 and the direction vector of L1L_1 is (0,0,1)(0,0,1),

(−3,4,t−4)⋅(0,0,1)=0(-3,4,t-4) \cdot (0,0,1) = 0

Thus,

t−4=0⇒t=4t-4 = 0 \Rightarrow t = 4

So,

Q=(1,2,7)Q = (1,2,7)

Now,

PQ2=(4−1)2+(−2−2)2+(7−7)2PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 =32+(−4)2+02= 3^2 + (-4)^2 + 0^2 =9+16=25= 9 + 16 = 25

Therefore, the square of the distance is 2525. The correct option is C.

Shortcut Using Distance from z-axis Parallel Line

Given: λ1=4\lambda_1 = 4 and λ2=−2\lambda_2 = -2 from the condition ∣λ−1∣=3|\lambda-1|=3.

Find: The square of the distance of (4,−2,7)(4,-2,7) from L1L_1.

Since L1L_1 is parallel to the z-axis and passes through (1,2,3)(1,2,3), the distance of any point (x,y,z)(x,y,z) from L1L_1 depends only on the change in the xx and yy coordinates. The zz-coordinate does not affect the perpendicular distance.

So for point (4,−2,7)(4,-2,7),

distance2=(4−1)2+(−2−2)2=32+(−4)2=25\text{distance}^2 = (4-1)^2 + (-2-2)^2 = 3^2 + (-4)^2 = 25

Therefore, the square of the distance is 2525. The correct option is C.

Common mistakes

  • Using the wrong formula for the shortest distance between the two lines. These lines are skew lines with direction vectors (0,0,1)(0,0,1) and (0,1,0)(0,1,0), so the cross-product formula must be used. Do not treat them as intersecting or parallel lines.

  • Not ordering λ1\lambda_1 and λ2\lambda_2 correctly after solving ∣λ−1∣=3|\lambda-1|=3. The two values are 44 and −2-2, and since λ2<λ1\lambda_2 < \lambda_1, we must take λ1=4\lambda_1=4 and λ2=−2\lambda_2=-2.

  • While finding distance from P(4,−2,7)P(4,-2,7) to L1L_1, including the z-component in the perpendicular distance. Since L1L_1 is parallel to the z-axis, only the horizontal displacement in xx and yy matters for the perpendicular distance.

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