Q67 / 120 · NDA 2024 (Phase 2) · Mathematics

Paragraph: Let S be the line of intersection of two planes $x+y+z=1$ and $2 x+3 y-4 z=8$.
Question: Which of the following are the direction ratios of S ?

  • A
    $\langle-7,-6,1\rangle$
  • Correct answer
    B
    $ < -7,6,1>$
  • C
    $\langle-6,5,1\rangle$
  • D
    $ < 6,5,1>$

Solution

2 planes are $x+y+z=1$ and $2 x+3 y-4 z=8$. Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are $\mathrm{dr}^{\prime} \mathrm{s}$ is of lines of formed by intersection of planes then the lines is perpendicular to normal of both planes. $a+b+c=0$ ....(i) $\text { and } 2 a+3 b-4 c=0$ .....(ii) $\frac{\mathrm{a}}{-4-3}=\frac{\mathrm{b}}{2+\mathrm{a}}=\frac{\mathrm{c}}{3-2}=\lambda$ (Solving (i) and (ii) $\mathrm{a}=-7 \lambda, \mathrm{~b}=6 \lambda, \mathrm{c}=\lambda$ Dr's or $\langle-7,6,1\rangle$.

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

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