Logo
Jeeify
MCQMediumJEE 2024Equation of Line in 3D

JEE Mathematics 2024 Question with Solution

The square of the distance of the image of the point (6,1,5)\left(6, 1, 5\right) in the line x−13=y2=z−24\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}, from the origin is:

  • A

    6262

  • B

    5858

  • C

    6565

  • D

    6060

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The point is P(6,1,5)P(6,1,5) and the line is

x−13=y2=z−24\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}

Find: The square of the distance of the image of PP in the given line from the origin.

The direction ratios of the line are ⟨3,2,4⟩\langle 3,2,4 \rangle and a point on the line is ⟨1,0,2⟩\langle 1,0,2 \rangle. So the line in vector form is

r=⟨1,0,2⟩+λ⟨3,2,4⟩\mathbf{r} = \langle 1,0,2 \rangle + \lambda \langle 3,2,4 \rangle

Let the foot of the perpendicular from PP to the line be

Q=⟨1+3λ,2λ,2+4λ⟩Q = \langle 1+3\lambda, 2\lambda, 2+4\lambda \rangle

Then

PQ→=⟨3λ−5,2λ−1,4λ−3⟩\overrightarrow{PQ} = \langle 3\lambda-5, 2\lambda-1, 4\lambda-3 \rangle

Since PQ→\overrightarrow{PQ} is perpendicular to the line direction,

PQ→⋅⟨3,2,4⟩=0\overrightarrow{PQ} \cdot \langle 3,2,4 \rangle = 0

Therefore,

(3λ−5)⋅3+(2λ−1)⋅2+(4λ−3)⋅4=0(3\lambda-5)\cdot 3 + (2\lambda-1)\cdot 2 + (4\lambda-3)\cdot 4 = 0 9λ−15+4λ−2+16λ−12=09\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0 29λ=2929\lambda = 29 λ=1\lambda = 1

Substituting λ=1\lambda=1,

Q=(4,2,6)Q = (4,2,6)

The image point is the reflection of PP in QQ, so

Q′=2Q−PQ' = 2Q - P Q′=2(4,2,6)−(6,1,5)=(2,3,7)Q' = 2(4,2,6) - (6,1,5) = (2,3,7)

Now the square of the distance of Q′Q' from the origin is

(2)2+(3)2+(7)2=4+9+49=62(2)^2 + (3)^2 + (7)^2 = 4 + 9 + 49 = 62

Therefore, the square of the distance is 6262. Hence, the correct option is A.

Coordinate Reflection Method

Given: M(3λ+1,2λ,4λ+2)M(3\lambda+1,2\lambda,4\lambda+2) is a point on the line. Find: The image of A(6,1,5)A(6,1,5) in the line and its squared distance from the origin.

Using the perpendicular condition from the solution,

AM→⋅b=0\overrightarrow{AM} \cdot \mathbf{b} = 0

where b=⟨3,2,4⟩\mathbf{b} = \langle 3,2,4 \rangle. So,

9λ−15+4λ−2+16λ−12=09\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0 29λ=29  ⟹  λ=129\lambda = 29 \implies \lambda = 1

Hence,

M=(4,2,6)M = (4,2,6)

and the image point is

I=(2,3,7)I = (2,3,7)

So the distance from the origin is

22+32+72=62\sqrt{2^2+3^2+7^2} = \sqrt{62}

Therefore, the square of the distance is 6262.

Common mistakes

  • Students often treat the given line as a plane mirror without first finding the foot of the perpendicular from the point to the line. This is wrong because reflection in a line in 3D requires the midpoint relation through the perpendicular foot. First find the projection point on the line, then reflect the point.

  • A common error is writing the point on the line incorrectly from the symmetric form. In x−13=y2=z−24\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}, the point is (1,0,2)\left(1,0,2\right) and direction ratios are ⟨3,2,4⟩\langle 3,2,4 \rangle. Misreading these gives the wrong projection point.

  • Some students compute the distance from the origin to the foot of the perpendicular instead of to the image point. This is incorrect because the question asks for the image of the point in the line. After finding the foot QQ, calculate the reflected point Q′=2Q−PQ' = 2Q - P and then its distance from the origin.

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