Question

Let the plane 'V' be defined as ax + by + cz + d = 0 with at least one of a, b or c different from 0 and d > and = 0.

Then the distance between V and the origin is d/ square root of a squared + b squared + c squared.

Prove this statement.

please assist.

Answer 1

Since at least one of a, b or c different from 0 and $d\ge 0$,the origin does not belong to the plane V.

The distance between the origin and the plane is the distance between the origin and a point Q(x,y,z) on the plane V, which can be obtained as a projection of vector OQ on the plane's normal vector n(a,b,c), and is given by

$L=\frac{\vec{OQ}\cdot \vec{n}}{|\vec{n}|}=\frac{|a(x_Q-x_0)+b(y_Q-y_0)+c(z_Q-z_0)+d|}{\sqrt{a^2+b^2+c^2}}=\frac{|d|}{\sqrt{a^2+b^2+c^2}}$