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


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+

which, amazingly, simplifies to


So the d=|Ah+Bk+Cm+D|/sqrt(A^2+B^2+C^2)

[improvement needed: add math tags]

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