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MCQMediumJEE 2023Equation of Line in 3D

JEE Mathematics 2023 Question with Solution

Consider the lines L1L_1 and L2L_2 given by

L1:x−12=y−32=z−22,L2:x−21=y−22=z−33.L_1: \frac{x-1}{2} = \frac{y-3}{2} = \frac{z-2}{2}, \quad L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}.

A line L3L_3 having direction ratios 1,−1,−21, -1, -2 intersects L1L_1 and L2L_2 at the points PP and QQ respectively. Then the length of line segment PQPQ is:

  • A

    262\sqrt{6}

  • B

    323\sqrt{2}

  • C

    434\sqrt{3}

  • D

    44

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

L1:x−12=y−32=z−22,L2:x−21=y−22=z−33L_1: \frac{x-1}{2} = \frac{y-3}{2} = \frac{z-2}{2}, \quad L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}

A line L3L_3 has direction ratios 1,−1,−21, -1, -2 and intersects L1L_1 at PP and L2L_2 at QQ.

Find: The length of segment PQPQ.

Take a general point on L1L_1 as

P=(2λ+1,λ+3,2λ+2)P = (2\lambda + 1, \lambda + 3, 2\lambda + 2)

and a general point on L2L_2 as

Q=(μ+2,2μ+2,3μ+3)Q = (\mu + 2, 2\mu + 2, 3\mu + 3)

Since PQ→\overrightarrow{PQ} is along L3L_3, its direction ratios are proportional to 1,−1,−21, -1, -2. Hence

(μ+2)−(2λ+1)1=(2μ+2)−(λ+3)−1=(3μ+3)−(2λ+2)−2\frac{(\mu + 2) - (2\lambda + 1)}{1} = \frac{(2\mu + 2) - (\lambda + 3)}{-1} = \frac{(3\mu + 3) - (2\lambda + 2)}{-2}

Using the extracted working,

λ=μ=3\lambda = \mu = 3

Therefore,

P=(7,6,8),Q=(5,8,12)P = (7, 6, 8), \quad Q = (5, 8, 12)

Now apply the distance formula:

PQ=(7−5)2+(6−8)2+(8−12)2PQ = \sqrt{(7-5)^2 + (6-8)^2 + (8-12)^2} =22+(−2)2+(−4)2= \sqrt{2^2 + (-2)^2 + (-4)^2} =4+4+16=24=26= \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}

Therefore, the length of the line segment is 262\sqrt{6}. So the correct option is A.

The solution states option C, but the actual working gives PQ=26PQ = 2\sqrt{6}, which matches option A. The worked solution is taken as authoritative.

Distance Formula Expansion

Given: The intersection points obtained from the working are

P(7,6,8),Q(5,8,12)P(7, 6, 8), \quad Q(5, 8, 12)

Find: PQPQ.

Use the three-dimensional distance formula:

PQ=(x1−x2)2+(y1−y2)2+(z1−z2)2PQ = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}

Substitute the coordinates:

PQ=(7−5)2+(6−8)2+(8−12)2PQ = \sqrt{(7-5)^2 + (6-8)^2 + (8-12)^2} =4+4+16= \sqrt{4 + 4 + 16} =24=26= \sqrt{24} = 2\sqrt{6}

Hence, the correct option is A.

Common mistakes

  • Students often take the direction ratios of L3L_3 as the coordinates of a point on the line. This is wrong because direction ratios describe the direction of PQ→\overrightarrow{PQ}, not the location of PP or QQ. First write general points on L1L_1 and L2L_2, then use their difference vector parallel to ⟨1,−1,−2⟩\langle 1,-1,-2 \rangle.

  • A common mistake is writing incorrect parametric coordinates for L1L_1 or L2L_2. For example, from x−12=y−32=z−22\frac{x-1}{2} = \frac{y-3}{2} = \frac{z-2}{2}, the point on L1L_1 must be written consistently using one parameter. Any mismatch in coefficients gives the wrong points PP and QQ.

  • Some students use the distance formula with sign errors, especially in terms like (6−8)2(6-8)^2 and (8−12)2(8-12)^2. The subtraction may be negative, but the square must be taken afterward. Compute each coordinate difference carefully before squaring.

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