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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2017 · Mathematics · Three Dimensional Geometry

The line passing through the points (1,2,-1) and (3,-1,2) meets the...

The line passing through the points (1,2,−1)(1,2,-1) and (3,−1,2)(3,-1,2) meets the yz-plane at which one of the following points?

Options

  1. A(0,−72,52)\left(0,-\frac{7}{2}, \frac{5}{2}\right)
  2. B(0,72,12)\left(0, \frac{7}{2}, \frac{1}{2}\right)
  3. C(0,−72,−52)\left(0,-\frac{7}{2},-\frac{5}{2}\right)
  4. D(0,−72,−52)\left(0,-\frac{7}{2},-\frac{5}{2}\right)

Correct answer

D. (0,−72,−52)\left(0,-\frac{7}{2},-\frac{5}{2}\right)

Step-by-step solution

Eqn. of line ⇒x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\Rightarrow \frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{z}_2-\mathrm{z}_1} i.e, x−12=y−2−3=z+13=K( say )⇒x−1=2K;y−2=−3K;z+1=3K⇒x=2K+1;y=−3K+2;z=3K−1\begin{aligned} & \text{ i.e, } \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=K(\text{ say }) \\ & \Rightarrow x-1=2 K ; y-2=-3 K ; z+1=3 K \\ & \Rightarrow x=2 K+1 ; y=-3 K+2 ; z=3 K-1 \end{aligned}Since the line meets $y z$ plane, $x=0\begin{aligned} & \therefore 2 \mathrm{ K}+1=0 \Rightarrow \mathrm{ K}=\frac{-1}{2} \\ & \therefore \mathrm{y}=-3\left(\frac{-1}{2}\right)+2=\frac{3}{2}+2=\frac{7}{2} \\ & \mathrm{z}=3\left(\frac{-1}{2}\right)-1=\frac{-3}{2}-1=\frac{-5}{2} \end{aligned}$$$
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