Start free

Mathematics · 3D Geometry

JEE Main 2024 — 30 January, Shift 1 — Question 10

Let (α,β,γ)(\alpha, \beta, \gamma) be the foot of perpendicular from the point (1,2,3)(1,2,3) on the line x+35=y−12=z+43\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}. then 19(α+β+γ)19(\alpha+\beta+\gamma) is equal to :

  1. Option A:

    102

  2. Option B:

    101

    Correct
  3. Option C:

    99

  4. Option D:

    100

Answer: B

Step-by-step solution

figure

Let foot P(5k−3,2k+1,3k−4)\mathrm{P}(5 \mathrm{k}-3,2 \mathrm{k}+1,3 \mathrm{k}-4)

DR's →\rightarrow AP:5k−4,2k−1,3k−7AP: 5k-4, 2k-1, 3k-7

DR's →\rightarrow Line: 5, 2, 3 Condition of perpendicular lines

(25k−20)+(4k−2)+(9k−21)=0(25 \mathrm{k}-20)+(4 \mathrm{k}-2)+(9 \mathrm{k}-21)=0

Then k=4338\mathrm{k}=\frac{43}{38}

Then 19(α+β+γ)=10119(\alpha+\beta+\gamma)=101

Answer key and solution verified before publishing.

Practise 3D Geometry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
30 January, Shift 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Vector & Cartesian forms of lines and planes

More 3D Geometry questions from this paper

Let (α, β, γ) be the foot of perpendicular from the point (1,2,3) on… | JEE Main 2024 PYQ with Solution · DhiX AI