Question

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

Answer 1

Consider the given equation of the line.

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

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

Co-ordinate of any point on the line are $2\lambda +1,3\lambda -2,-6\lambda +3$. $Q$ lies on the line, therefore, co-ordinate of $Q$ is $(2\lambda +1,3\lambda -2,-6\lambda +3)$ for some$\lambda$. But $Q$ also lies in the plane so, it's the point of intersection of the line and the plane.

From the equation of the plane, we have

$2\lambda +1-3\lambda +2-6\lambda +3=5$

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

$\therefore$ Co-ordinate of $Q$ is,

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

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

By using the distance formula, we have

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

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

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

$=1$

Hence, the required distance is $1unit$.