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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Three Dimensional Geometry

Let the line L intersect the lines x-2=-y=z-1,2(x+1)=2(y-1)=z+1 and...

Let the line L\mathrm{L} intersect the lines x−2=−y=z−1,2(x+1)=2(y−1)=z+1x-2=-y=z-1,2(x+1)=2(y-1)=z+1 and be parallel to the line x−23=y−11=z−22\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}. Then which of the following points lies on L?Let the line L intersect the lines x-2=-y=z-1,2(x+1)=2(y-1)=z+1 and be parallel to the lineLet the line L intersect the lines x-2=-y=z-1,2(x+1)=2(y-1)=z+1 and be parallel to the line

Options

  1. A(−13,1,−1)\left(-\frac{1}{3}, 1,-1\right)
  2. B(−13,−1,1)\left(-\frac{1}{3},-1,1\right)
  3. C(−13,1,1)\left(-\frac{1}{3}, 1,1\right)
  4. D(−13,−1,−1)\left(-\frac{1}{3},-1,-1\right)

Correct answer

A. (−13,1,−1)\left(-\frac{1}{3}, 1,-1\right)

Step-by-step solution

Let the line L intersect the lines x-2=-y=z-1,2(x+1)=2(y-1)=z+1 and be parallel to the line
L1:x−21=y−1=z−11=λ\mathrm{L}_1: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\lambda L2:x+112=y−112=z+11=μ\mathrm{L}_2: \frac{\mathrm{x}+1}{\frac{1}{2}}=\frac{\mathrm{y}-1}{\frac{1}{2}}=\frac{\mathrm{z}+1}{1}=\mu dr of line MN will be <3+λ−μ2,−1−λ−μ2,2+λ−μ>&< 3+\lambda-\frac{\mu}{2},-1-\lambda-\frac{\mu}{2}, 2+\lambda-\mu>\& it will be proportional to <3,1,2>< 3,1,2> Let the line L intersect the lines x-2=-y=z-1,2(x+1)=2(y-1)=z+1 and be parallel to the line
⇒λ=−43&μ=−23\Rightarrow \lambda=-\frac{4}{3} \& \mu=-\frac{2}{3} ∴\therefore Coordinate of M\mathrm{M} will be <(23,43,−13)< \left(\frac{2}{3}, \frac{4}{3},-\frac{1}{3}\right) and equation of required line will be. x−233=y−431=z+132=k\frac{\mathrm{x}-\frac{2}{3}}{3}=\frac{\mathrm{y}-\frac{4}{3}}{1}=\frac{\mathrm{z}+\frac{1}{3}}{2}=\mathrm{k} So any point on this line will be (23+3k,43+k,−13+2k)∵23+3k=−13⇒k=−13\begin{aligned} & \left(\frac{2}{3}+3 \mathrm{k}, \frac{4}{3}+\mathrm{k},-\frac{1}{3}+2 \mathrm{k}\right) \\ & \because \frac{2}{3}+3 \mathrm{k}=-\frac{1}{3} \Rightarrow \mathrm{k}=-\frac{1}{3} \end{aligned} ∴\therefore Point lie on the line for k=−13 is (−13,1,−1)\mathrm{k}=-\frac{1}{3} \text{ is }\left(-\frac{1}{3}, 1,-1\right)
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