Q72 / 120 · VITEEE 2017 · Mathematics
The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is
- Correct answerA1
- B2
- C4
- D$2 \sqrt{3}$
Solution
Equation of the line through $(1,-2,3)$ parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$ (say)
Then any point on (1) is $(2 r+1,3 r-2,-6 r$ $+3$ )
If this point lies on the plane $x-y+z=5$ then
$$
(2 r+1)-(3 r-2)+(-6 r+3)=5 \Rightarrow r=\frac{1}{7}
$$
Hence the point is $\left(\frac{9}{7},-\frac{11}{7}, \frac{15}{7}\right)$
Distance between $(1,-2,3)$ and
$$
\begin{array}{l}
\left(\frac{9}{7},-\frac{11}{7}, \frac{15}{7}\right) \\
=\sqrt{\left(\frac{4}{49}+\frac{9}{49}+\frac{36}{49}\right)}=\sqrt{\left(\frac{49}{49}\right)}=1
\end{array}
$$
Hence, the correct answer is option: 'A'.