Question

The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line. $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6},is:$

• A. 1
• B. 6/7
• C. 7/6
• D. 1/6

Answer 1

The correct option is B 6/7

Given,

Let $P=(1,-2,3)$

given plane $x-y+z=5$

to find the distance of point $P=(1,-2,3)$ from the plane $x-y+z=5$ measured along the parallel line to

$\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$

equation of line passing through $(1,-2,3)$and having DR's $(2,3,-6)$

let $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda$

$x=2\lambda +1,y=3\lambda -2,z=-6\lambda +3$

$\Rightarrow 2\lambda +1-3\lambda +2-6\lambda +3=2+3-6$

$-7\lambda =5-6$

$\therefore \lambda =\frac{1}{7}$

Therefore the coordinates of Q are

$(\frac{2}{7}+1,\frac{3}{7}-2,-\frac{6}{7}+3)$

$=(\frac{9}{7},-\frac{11}{7},\frac{15}{7})$

$PQ=\sqrt{{(\frac{9}{7}-1)}^2+{(-\frac{11}{7}+2)}^2+{(\frac{15}{7}-3)}^2}$

$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}$

$=\sqrt{\frac{49}{49}}$

$=1$