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MCQMediumJEE 2023Equation of Line in 3D

JEE Mathematics 2023 Question with Solution

If the lines x−11=y−22=z+31\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1} and x−a2=y+23=z−31\frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1} intersect at the point PP, then the distance of the point PP from the plane z=az = a is:

  • A

    1616

  • B

    2828

  • C

    1010

  • D

    2222

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The lines x−11=y−22=z+31\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1} and x−a2=y+23=z−31\frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1} intersect at point PP.

Find: The distance of PP from the plane z=az=a.

Write the lines in parametric form:

x=1+λ,y=2+2λ,z=−3+λx = 1 + \lambda, \quad y = 2 + 2\lambda, \quad z = -3 + \lambda

and

x=a+2μ,y=−2+3μ,z=3+μx = a + 2\mu, \quad y = -2 + 3\mu, \quad z = 3 + \mu

At the point of intersection, corresponding coordinates are equal:

1+λ=a+2μ1 + \lambda = a + 2\mu 2+2λ=−2+3μ2 + 2\lambda = -2 + 3\mu −3+λ=3+μ-3 + \lambda = 3 + \mu

From the third equation,

λ−μ=6\lambda - \mu = 6

Using μ=λ−6\mu = \lambda - 6 in the second equation:

2+2λ=−2+3(λ−6)2 + 2\lambda = -2 + 3(\lambda - 6) 2+2λ=3λ−202 + 2\lambda = 3\lambda - 20 λ=22\lambda = 22

Hence,

μ=16\mu = 16

Now use the first equation:

1+22=a+2(16)1 + 22 = a + 2(16) 23=a+3223 = a + 32 a=−9a = -9

Substitute λ=22\lambda = 22 in the first line:

P=(1+22, 2+44, −3+22)=(23,46,19)P = (1+22,\, 2+44,\, -3+22) = (23, 46, 19)

The plane is z=a=−9z=a = -9. Therefore the perpendicular distance of PP from this plane is

∣z−a∣=∣19−(−9)∣=28|z-a| = |19-(-9)| = 28

Therefore, the correct option is B.

Discrepancy Check from the solution

The provided the solution concludes with 2828 and also marks Option C, but the listed options show Option C = 1010 and Option B = 2828. Using the working from the equations, the defensible answer is 2828, so the correct option must be B.

Common mistakes

  • Using the second line incorrectly as z=−3+μz=-3+\mu instead of reading z−31\frac{z-3}{1}, which gives z=3+μz=3+\mu. This changes the intersection equations. Always convert each symmetric-form term carefully.

  • Assuming the distance from point PP to the plane z=az=a requires the full point-to-plane formula with normalization. For a plane of the form z=az=a, the distance is directly ∣z−a∣|z-a|.

  • Trusting the marked option letter without checking whether it matches the computed value. Here the solution text and option list disagree, so the numerical result must be matched back to the correct option.

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