Q69 / 120 · NDA 2024 (Phase 2) · Mathematics

Paragraph: Let $\mathrm{L}: x+y+z+4=0=2 x-y-z+8$ be a line and P : $x+2 y+3 z+1=0$ be a plane
Question: What are the direction ratios of the line?

  • A
    $ < 2,1,-1>$
  • B
    $ < 0,-1,2>$
  • Correct answer
    C
    $ < 0,1,-1>$
  • D
    $ < 2,3,-3>$

Solution

We have $\mathrm{L}: \mathrm{x}+\mathrm{y}+\mathrm{z}+4=0=2 \mathrm{x}-\mathrm{y}-\mathrm{z}-8$ and $P: x+2 y+3 z+1=0$ is a plane. Let Dr's of line $\langle$ a, b, c $\rangle$ Since, line $L$ obtained by intersection of planes $x+y+z$ $+4=0 \text { and } 2 x-y-z-8=0$ So, $\mathrm{a}+\mathrm{b}+\mathrm{c}=0$ .....(i) $2 a-b-c=0$ ....(ii) (line is perpendicular to normal of both plane) $\begin{aligned} & \frac{\mathrm{a}}{-1+1}=\frac{\mathrm{b}}{2+1}=\frac{\mathrm{c}}{-1-2}=\lambda \\ & \Rightarrow \mathrm{a}=0, \mathrm{~b}=3 \lambda, \mathrm{c}=-3 \lambda \end{aligned}$ (solving (i) and (ii) Dr's of line is $\langle 0,3,-3\rangle$ or $\langle 0,1,-1\rangle$

Hence, the correct answer is option: 'C'.

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