Question

Find the distance of point $(1,-2,3)$ from the plane $x-y+z=5$, measured parallel to the line whose direction cosines are proportional to $2,3,-6.$

Answer 1

Here, we are not required to find the perpendicular distance of the point $(1,-2,3)$ from the plane, but the distance measured parallel to line whose D.R.'s are $2,3,-6$.

Equation of the line passing through $(1,-2,3)$ and whose D.R.'s are $2,3,-6$,

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r,$ say.

Any point on it is $(2r+1,3r-2,-6r+3)$.

If it lies on the plane $x-y+z=5,$ then

$(2r+1)-(3r-2)+(-6r+3)=5.$

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

$\therefore$ The point is $(\frac{9}{7},\frac{-11}{7},\frac{15}{7})$ and its distance from $(1,-2,3)$

is $\frac{1}{7}\sqrt{(4+9+36)}=\frac{7}{7}=1.$