SOLUTION: Given: The point A(x0, y0, z0) lies outside a plane Q with equation ax+by+cz+d=0. The point P(x, y, z) lies on the plane Q. AB is the distance between the point A and the plane Q (

Question 30815: Given: The point A(x0, y0, z0) lies outside a plane Q with equation ax+by+cz+d=0. The point P(x, y, z) lies on the plane Q. AB is the distance between the point A and the plane Q (B is a point ON plane Q).
Derive the equation: (AB)= | ax0+by0+cz0+d | / sq.root of (a^2+b^2+c^2).
Basically, this question is asking you to derive the generality of finding the distance from a point to a plane. Any help is GREATLY appreciated, I would like it by tomorrow if possible!!
Thanks,
-Stephanie
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Given: The point A(x0, y0, z0) lies outside a plane Q with equation ax+by+cz+d=0.
DIRECTION RATIOS OF NORMAL TO PLANE ARE A,B AND C.
The point P(x, y, z) lies on the plane Q. AB is the distance between the point A and the plane Q (B is a point ON plane Q).
SO AB IS PERPENDICULAR TO PLANE OR PARALLEL TO ITS NORMAL.
LET COORDINATES OF B BE P,Q,R.
SO DIRECTION RATIOS OF AB ARE
P-X0,Q-Y0,R-Z0.
THESE ARE IN SAME RATIO AS THAT OF NORMAL AS BOTH ARE PARALLEL.HENCE
(P-X0)/A = (Q-Y0)/B = (R-Z0)/C =K SAY
SO...
P=X0+AK
Q=Y0+BK
R=Z0+CK
BUT B(P,Q,R)IS ON PLANE.
HENCE
AP+BQ+CR+D=0
A(X0+AK)+B(Y0+BK)+C(Z0+CK)+D=0
K(A^2+B^2+C^2)=-AX0-BY0-CZ0-D
K*SQRT{A^2+B^2+C^2)=-(AX0+BY0+CZ0+D)/SQRT.{(A^2+B^2+C^2)}
AB=DISTANCE OF POINT FROM PLANE =SQRT.{(P-X0)^2+(Q-Y0)^2+(R-Z0)^2}
=SQRT.{(AK)^2+(BK)^2+(CK)^2}=K*SQRT{A^2+B^2+C^2}
=-(AX0+BY0+CZ0+D)/SQRT.{(A^2+B^2+C^2)}
IT IS USUALLY PUT AS MODULUS OF THE ABOVE RATIO...

Derive the equation: (AB)= | ax0+by0+cz0+d | / sq.root of (a^2+b^2+c^2).
Basically, this question is asking you to derive the generality of finding the distance from a point to a plane. Any help is GREATLY appreciated, I would like it by tomorrow if possible!!
Thanks,
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