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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let (α, β, γ) be the co-ordinates of the foot of the perpendicular...

Let (α,β,γ)(\alpha, \beta, \gamma) be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2)(5,4,2) on the line r→=(−i^+3j^+k^)+λ(2i^+3j^−k^)\overrightarrow{\mathrm{r}}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k}).
Then the length of the projection of the vector αi^+βj^+γk^\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k} on the vector 6i^+2j^+3k^6 \hat{i}+2 \hat{j}+3 \hat{k} is :

Options

  1. A157\frac{15}{7}
  2. B4
  3. C187\frac{18}{7}
  4. D3

Correct answer

C. 187\frac{18}{7}

Step-by-step solution

Line: r⃗=(−1,3,1)+λ(2,3,−1)\vec{r} = (-1, 3, 1) + \lambda(2, 3, -1), Point P = (5, 4, 2).
Any point on line: Q=(−1+2λ,3+3λ,1−λ)Q = (-1 + 2\lambda, 3 + 3\lambda, 1 - \lambda).
PQ⃗=(−6+2λ,−1+3λ,−1−λ)\vec{PQ} = (-6 + 2\lambda, -1 + 3\lambda, -1 - \lambda).
For perpendicular: PQ⃗⋅(2,3,−1)=0\vec{PQ} \cdot (2, 3, -1) = 0 gives 14λ−14=014\lambda - 14 = 0, so λ=1\lambda = 1.
Foot (α,β,γ)=(1,6,0)(\alpha, \beta, \gamma) = (1, 6, 0).
Projection of (1,6,0)(1, 6, 0) on (6,2,3)(6, 2, 3): ∣6+12+0∣49=187\frac{|6 + 12 + 0|}{\sqrt{49}} = \frac{18}{7}.
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