Question

Find the distance of point $(1,-5,9)$ from plane $x-y+z=5$ measured along line $x=y=z$.

Answer 1

The equation of line parallel to line $x=y=z$ and passing through $P$ is

$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=r$

Let point A be intersection of this line with plane.

$\therefore (x,y,z)=(r+1,r-5,r+9)$ is general point on line.

for $(x,y,z)$ to be point of intersection of line & plane it must satisfy the plane equation, $x-y+z=5\therefore (r+1)-(r-5)+(r+9)=5\Rightarrow r=-10\therefore pointA=(-9,-15,-1)\therefore distanceofP(1,-5,9)$ from plane along required direction$=APAP=\sqrt{(-9-1{)}^2+(-15-(-5){)}^2+(-1-9{)}^2}=10\sqrt{3}units$