Question

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

• A. $1$
• B. $6/7$
• C. $7/6$
• D. None of these

Answer 1

The correct option is A $1$

Plane : $x-y+2=5---\left(1\right)$

Point : $(1,-2,3)$

line is parallel to: $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$

Direction ratio of line from point $(1,-2,3)$ to plane will be same as that of given line

Let line from $(1,-2,3)$ meet & on plane

equation of line passing through $(1,-2,3)$ & having direction ratio $(2,3,-6)$ is

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

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

This satisfy the equation of plane $\left(1\right)$

So, $(2\lambda +1)-(3\lambda -2)+(-6\lambda +3)=5$

$-7\lambda +6=5$

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

$Q=(\frac{2}{7}+1),(\frac{2}{7}-2),(\frac{-6}{7}+3)\Rightarrow (\frac{9}{7},\frac{-11}{7},\frac{15}{7}1)$

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

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

$=\sqrt{\frac{49}{49}}=1$

$\therefore$ Distance $=1$, option $A$ is correct