Find the perpendicular distance from the point $P(4,-1,2)$ to the plane $3x+4y-2z=12$.

Step-by-step solution

1. Substitute the coordinates of $P$ into the plane equation: $3(4)+4(-1)-2(2)-12=12-4-4-12=-8$.
2. Take the absolute value of the numerator: $|-8|=8$.
3. Compute the magnitude of the normal vector: $\sqrt{3^2+4^2+(-2{)}^2}=\sqrt{29}$.
4. The distance is the absolute value of the numerator divided by the magnitude of the normal: $8/\sqrt{29}$.
5. Hence the perpendicular distance is $8/\sqrt{29}$.

Concept tested: Distance from point to plane

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked