Start free

Mathematics · 3D Geometry

JEE Main 2024 — 4 April, Shift 2 — Question 20

Let PP the point of intersection of the lines x−21=y−45=z−21\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \quad and x−32=y−23=z−32\quad \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}.

Then, the shortest distance of P from the line 4x=2y=z4 \mathrm{x}=2 \mathrm{y}=\mathrm{z} is

  1. Option A:

    5147\frac{5 \sqrt{14}}{7}

  2. Option B:

    147\frac{\sqrt{14}}{7}

  3. Option C:

    3147\frac{3 \sqrt{14}}{7}

    Correct
  4. Option D:

    6147\frac{6 \sqrt{14}}{7}

Answer: C

Step-by-step solution

L1≡x−21=y−45=z−21=λ\mathrm{L}_{1} \equiv \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}-2}{1}=\lambda

P(λ+2,5λ+4,λ+2)\mathrm{P}(\lambda+2,5 \lambda+4, \lambda+2)

L2≡x−32=y−23=z−32=μ⇒P(2μ+3, 3μ+2, 2μ+3)L_{2} \equiv \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}=\mu \quad\Rightarrow\quad P(2\mu+3,\,3\mu+2,\,2\mu+3) λ+2=2μ+3,3μ+2=5λ+4\lambda+2 = 2\mu+3,\qquad 3\mu+2 = 5\lambda+4 λ=2μ+1,3μ=5λ+2\lambda = 2\mu+1,\qquad 3\mu = 5\lambda + 2 3μ=5(2μ+1)+23\mu = 5(2\mu+1) + 2 3μ=10μ+7⇒μ=−1, λ=−13\mu = 10\mu + 7 \quad\Rightarrow\quad \mu = -1,\ \lambda = -1 Both satisfy P.P(1,−1,1)\text{Both satisfy P.} \qquad P(1,-1,1) L3≡x1/4=y1/2=z1L_{3} \equiv \frac{x}{1/4}=\frac{y}{1/2}=\frac{z}{1} L3=x1=y2=z4=k⇒Q(k, 2k, 4k)L_{3} = \frac{x}{1} = \frac{y}{2} = \frac{z}{4} = k \qquad\Rightarrow\qquad Q(k,\,2k,\,4k) DRs of PQ=⟨k−1, 2k+1, 4k−1⟩\text{DRs of PQ} = \langle k-1,\ 2k+1,\ 4k-1\rangle PQ⊥L3PQ \perp L_{3} (k−1)+2(2k+1)+4(4k−1)=0(k-1) + 2(2k+1) + 4(4k-1) = 0 k−1+4k+2+16k−4=0k - 1 + 4k + 2 + 16k - 4 = 0 k=17k = \frac{1}{7} Q(17, 27, 47)Q\left(\frac{1}{7},\,\frac{2}{7},\,\frac{4}{7}\right) PQ=(1−17)2+(−1−27)2+(1−47)2PQ = \sqrt{\left(1-\frac{1}{7}\right)^{2} + \left(-1-\frac{2}{7}\right)^{2} + \left(1-\frac{4}{7}\right)^{2}} =3649+8149+949=1267=3147= \sqrt{\frac{36}{49} + \frac{81}{49} + \frac{9}{49}} = \frac{\sqrt{126}}{7} = \frac{3\sqrt{14}}{7}
Solution figure

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
4 April, Shift 2
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them

More 3D Geometry questions from this paper

Let P the point of intersection of the lines x-2/1=y-4/5=z-2/1 and… | JEE Main 2024 PYQ with Solution · DhiX AI