Q213 / 270 · NDA 2016 (Phase 2) · Mathematics

Paragraph: A line L passes through the point $\mathrm{P}(5,-6,7)$ and is parallel to the planes $\mathrm{x}+\mathrm{y}+\mathrm{z}=1$ and $2 \mathrm{x}-\mathrm{y}-2 \mathrm{z}=3$.
Question: What are the direction ratios of the line of intersection of the given planes?

  • A
    $\lt 1,4,3 \lt$
  • B
    $\lt -1,-4,3 \gt$
  • Correct answer
    C
    $\lt 1,-4,3 \gt$
  • D
    $\lt 1,-4,-3 \gt$

Solution

Let $a, b, c$ be the direction ratios of the line. Then its equation is $\frac{x-5}{a}=\frac{y+6}{b}=\frac{z-7}{c}$...(i) Since (i) is parallel to the planes $x+y+z=1$ and $2 x-y-2 z=3$ then $a(1)+b(1)+c(1)=0 \text { and } a(2)+b(-1)+c(-2)=0$ By cross multiplication $\begin{aligned} & \frac{a}{-1}=\frac{b}{4}=\frac{c}{-3}=\lambda \\ & \Rightarrow a=-\lambda, b=4 \lambda, c=-3 \lambda \end{aligned}$ $\Rightarrow$ Direction ratios of the line are $ \lt -1,4,-3 \gt = \lt 1,-4,3 \gt $ Substituting a, $\mathrm{b}, \mathrm{c}$ in (i), we get $\frac{x-5}{-1}=\frac{y+6}{4}=\frac{z-7}{-3}$ ...(ii) Hence equation of the line is $\frac{x-5}{-1}=\frac{y+6}{4}=\frac{z-7}{-3}$

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

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