- A student asks,
- "What is the distance, d, from plane Ax+By+Cz+D=0 to the point (h, k, m)?"
- MathHelp replies
- The shortest distance is along the line normal, i.e. perpendicular, to the plane, that passes through the given point. A normal vector to the given plane is (A, B, C). So the parametric equation of the line in question is
- x = h+tA
- y = k+tB
- z = m+tC
Then, the intersection of the given plane and this line is the point correstponding to the value of t that solves
- A(h+tA)+B(k+tB)+C(m+tC)+D=0
which is
- t = -(Ah+Bk+Cm+D)/(A^2+B^2+C^2)
plugging this expression for t into the equations of x, y, and z,
- x = h-A(Ah+Bk+Cm+D)/(A^2+B^2+C^2)
- y = k-B(Ah+Bk+Cm+D)/(A^2+B^2+C^2)
- z = m-C(Ah+Bk+Cm+D)/(A^2+B^2+C^2)
Now, d^2 = (x-h)^2+(y-k)^2+(z-m)^2, or
- d^2 = (A(Ah+Bk+Cm+D)/(A^2+B^2+C^2))^2+
- (B(Ah+Bk+Cm+D)/(A^2+B^2+C^2))^2+
- (C(Ah+Bk+Cm+D)/(A^2+B^2+C^2))^2,
which, amazingly, simplifies to
- (Ah+Bk+Cm+D)^2/(A^2+B^2+C^2)
So the d=|Ah+Bk+Cm+D|/sqrt(A^2+B^2+C^2)
[improvement needed: add math tags]