Distance from a point to a given plane

 

Question

 

     Find the distance from a point (a, b, c) to a plane Ax + By + Cz + D = 0

 

 

Quick method

 

     If you know Cauchy inequality:

              (a12 + a22 + a32) (b12 + b22 + b32) ³ (a1b1 + a2b2 + a3b3)2,

     you can derive the result very easily.

 

Solution

     The distance from the point (a, b, c) to a point (x, y, z) on the plane Ax + By + Cz + D = 0 is:

              d = [(x – a)2 + (y – b)2 + (z – c)2]1/2

     We like to find the minimum of d, which is the distance required.

 

     By the Cauchy inequality,

              [(x – a)2 + (y – b)2 + (z – c)2]1/2[A2 + B2 + C2]1/2 ³ |A(x-a) + B(y-b) + C(z-c)|

 

\ min. of d = |A(x - a) + B(y - b) + C(z - c)| / [A2 + B2 + C2]1/2

                       = |Aa + Bb + Cc + D| / [A2 + B2 + C2]1/2.

                                (note : Ax + By + Cz + D = 0 since point (x, y, z) is on the plane)

Hurray!