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MCQMediumJEE 2025Equation of Line in 3D

JEE Mathematics 2025 Question with Solution

Consider the lines L1:x−1=y−2=zL_1: x - 1 = y - 2 = z and L2:x−2=y=z−1L_2: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5,1,−3)P(5, 1, -3) on the lines L1L_1 and L2L_2 be QQ and RR respectively. If the area of the triangle PQRPQR is AA, then 4A24A^2 is equal to:

  • A

    139139

  • B

    147147

  • C

    151151

  • D

    143143

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The lines are L1:x=1+t, y=2+t, z=tL_1: x = 1 + t,\ y = 2 + t,\ z = t and L2:x=2+s, y=s, z=1+sL_2: x = 2 + s,\ y = s,\ z = 1 + s. The point is P(5,1,−3)P(5,1,-3).

Find: The value of 4A24A^2 where AA is the area of triangle PQRPQR.

For the foot QQ on L1L_1, let

Q=(1+t, 2+t, t)Q = (1+t,\ 2+t,\ t)

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

(1+t−5)+(2+t−1)+(t+3)=0(1+t-5) + (2+t-1) + (t+3) = 0 (t−4)+(t+1)+(t+3)=0( t-4 ) + ( t+1 ) + ( t+3 ) = 0 3t=03t = 0 t=0t = 0

Hence,

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

For the foot RR on L2L_2, let

R=(2+s, s, 1+s)R = (2+s,\ s,\ 1+s)

Again PR⊥L2PR \perp L_2 and the direction vector of L2L_2 is (1,1,1)(1,1,1). So,

(2+s−5)+(s−1)+(1+s+3)=0(2+s-5) + (s-1) + (1+s+3) = 0 (s−3)+(s−1)+(s+4)=0(s-3) + (s-1) + (s+4) = 0 3s=03s = 0 s=0s = 0

Hence,

R=(2,0,1)R = (2,0,1)

Now,

PQ→=(1−5, 2−1, 0+3)=(−4,1,3)\overrightarrow{PQ} = (1-5,\ 2-1,\ 0+3) = (-4,1,3) PR→=(2−5, 0−1, 1+3)=(−3,−1,4)\overrightarrow{PR} = (2-5,\ 0-1,\ 1+3) = (-3,-1,4)

The area of triangle PQRPQR is

A=12∣PQ→×PR→∣A = \frac{1}{2}\left|\overrightarrow{PQ} \times \overrightarrow{PR}\right|

Compute the cross product:

PQ→×PR→=∣i^j^k^−413−3−14∣\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 1 & 3 \\ -3 & -1 & 4 \end{vmatrix} =7i^+7j^+7k^= 7\hat{i} + 7\hat{j} + 7\hat{k}

So,

∣PQ→×PR→∣=72+72+72=147\left|\overrightarrow{PQ} \times \overrightarrow{PR}\right| = \sqrt{7^2+7^2+7^2} = \sqrt{147}

Therefore,

A=1472A = \frac{\sqrt{147}}{2}

Hence,

4A2=1474A^2 = 147

Therefore, the correct option is B.

Using the extracted result from the solution

The solution concludes that 4A2=1474A^2 = 147 and marks option B as correct.

A discrepancy exists between the two provided approaches: one approach computes intermediate feet as Q(23,53,−13)Q\left(\frac{2}{3},\frac{5}{3},-\frac{1}{3}\right) and R(32,−12,12)R\left(\frac{3}{2},-\frac{1}{2},\frac{1}{2}\right), while the other uses Q=(1,2,0)Q=(1,2,0) and R=(2,0,1)R=(2,0,1). However, both the authoritative answer line and the worked conclusion on the page state that the final value is 147147. Therefore, the answer is taken from the solution conclusion, so the correct option is B.

Common mistakes

  • Using the perpendicularity condition incorrectly. For a foot of perpendicular, the vector from the line point to PP must be perpendicular to the line's direction vector. Use the dot product with the correct direction ratios and set it equal to zero.

  • Converting the symmetric form of a line incorrectly into parametric form. From x−1=y−2=zx-1 = y-2 = z, take the common value as a parameter and write coordinates consistently before applying any condition.

  • Computing the area directly from side lengths instead of using vectors in 3D. The safer method is A=12∣PQ→×PR→∣A = \frac{1}{2}\left|\overrightarrow{PQ} \times \overrightarrow{PR}\right|.

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