Given a plane equation (ax+by+cz=d), how does one find a point on the plane (x0, y0,z0)?

Question

Given a plane equation in the form ax+by+cz=d, how can one find the point (x0, y0, z0)?

I already know what the a,b,c,d coeffiecients are. I am not referring to the x,y,z intercepts

ie. d=-ax0-by0-cz0 See the example here on wolfram

http://mathworld.wolfram.com/Plane.html

Read further in that Wolfram article. It tells you how to find the perpendicular distance from the origin to the plane given the equation, so the nearest point on the plane to the origin is… ? — amd, Jul 13 '16 at 07:27
thank you for the reply. is this correct x0=(a*d)/(a^2+b^2+c^2) — math_Artist88, Jul 13 '16 at 14:34

score 2 · Answer 1 · answered Jul 29 '16 at 15:42

I don't think there's such thing as "The point" on a plane. There are many points and not a special one. However, you can preset 2 of the 3 coords of a particular point on the plane, and find the other one by substituting in the equation.

Any further explanation needs more detailing about what you're trying to accomplish.

score 1 · Answer 2 · edited Jul 29 '16 at 15:58

1

I would simply replace $(x,y)\,$ by $\,(x_0,y_0),\,$ i.e $\;ax_0 + by_0 + cz + d = 0,\;$ and solve the equation for $z_0.$

edited Jul 29 '16 at 15:58

amWhy

209,954

answered Jul 29 '16 at 14:30

Carlos Afonso

343

Given a plane equation (ax+by+cz=d), how does one find a point on the plane (x0, y0,z0)?

2 Answers2