Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then
Distance of the point (x1,y1,z1) from the plane Ax + By + Cz = D is
⇒
The equation of a plane having intercepts a, b, c on the x-, y-, z- axis respectively is
⇒
Comparing with Ax + By + Cz = D, we get -
A = 1/a, B = 1/b, C = 1/c, D = 1
Given, the plane is at a distance of 'p' units from the origin.
So, The point is O(0,0,0)
∴ x1 = 0, y1 = 0, z1 = 0
Now,
Distance
Substituting all values, we get -
⇒
squaring both sides, we get -
⇒
Hence Proved.