Q214 / 270 · NDA 2017 (Phase 1) · Mathematics
The line passing through the points $(1,2,-1)$ and $(3,-1,2)$ meets the yz-plane at which one of the following points?
- A$\left(0,-\frac{7}{2}, \frac{5}{2}\right)$
- B$\left(0, \frac{7}{2}, \frac{1}{2}\right)$
- C$\left(0,-\frac{7}{2},-\frac{5}{2}\right)$
- Correct answerD$\left(0,-\frac{7}{2},-\frac{5}{2}\right)$
Solution
Eqn. of line
$\Rightarrow \frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{z}_2-\mathrm{z}_1}$
$\begin{aligned}
& \text { i.e, } \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=K(\text { say }) \\
& \Rightarrow x-1=2 K ; y-2=-3 K ; z+1=3 K \\
& \Rightarrow x=2 K+1 ; y=-3 K+2 ; z=3 K-1
\end{aligned}$
Since the line meets $y z$ plane, $x=0$
$\begin{aligned}
& \therefore 2 \mathrm{~K}+1=0 \Rightarrow \mathrm{~K}=\frac{-1}{2} \\
& \therefore \mathrm{y}=-3\left(\frac{-1}{2}\right)+2=\frac{3}{2}+2=\frac{7}{2} \\
& \mathrm{z}=3\left(\frac{-1}{2}\right)-1=\frac{-3}{2}-1=\frac{-5}{2}
\end{aligned}$
Hence, the correct answer is option: 'D'.