Q246 / 270 · NDA 2023 (Phase 2) · Mathematics
If $L$ is the line with DIRECTIONS ratios $ \lt 3,-2,6 \gt $ and passing through $(1,-1,1)$, then what are the coordinates of the points on L whose distance from $(1,-1,1)$ is 2 units?
- A$\left(-\frac{11}{7}, \frac{13}{7}, \frac{19}{7}\right)$ and $\left(\frac{1}{7}, \frac{3}{7}, \frac{5}{7}\right)$
- B$\left(\frac{19}{7},-\frac{11}{7}, \frac{13}{7}\right)$ and $\left(-\frac{1}{7}, \frac{3}{7},-\frac{5}{7}\right)$
- C$\left(\frac{13}{7}, \frac{11}{7}, \frac{19}{7}\right)$ and $\left(-\frac{1}{7},-\frac{3}{7}, \frac{5}{7}\right)$
- Correct answerD$\left(\frac{13}{7},-\frac{11}{7}, \frac{19}{7}\right)$ and $\left(\frac{1}{7},-\frac{3}{7},-\frac{5}{7}\right)$
Solution
Since equation of line is
$\begin{aligned}
& \frac{x-1}{3}=\frac{y+1}{-2}=\frac{z-1}{6}=\lambda \text { (let) } \\
& \Rightarrow(x, y, z)=(3 \lambda+1,-2 \lambda-1,6 \lambda+1)
\end{aligned}$
Now,
$\begin{aligned}
& \sqrt{(3 \lambda+1-1)^2+(-2 \lambda-1+1)^2+(6 \lambda+1-1)^2}=2 \\
& =\sqrt{9 \lambda^2+4 \lambda^2+36 \lambda^2}=2 \\
& \Rightarrow \sqrt{49 \lambda^2}=2 \Rightarrow \pm 7 \lambda=2 \Rightarrow \lambda= \pm \frac{2}{7}
\end{aligned}$
So, $(x, y, z)=\left(\frac{13}{7}, \frac{11}{7}, \frac{19}{7}\right)$ or $\left(\frac{1}{7}, \frac{-3}{7}, \frac{-5}{7}\right)$
Hence, the correct answer is option: 'D'.