- 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]}