What is the d in the formula of a plane in $ R^3$

Question

In algebra the formula for a line is $y=ax+b$ the $b$ moves the position of the line up and down the y axis.

The formula for a plane is given to me as $ax+by+cz+d=0$ the $d$ must move the position of the plane. what I am trying to find out is along what does ti shift the plane in $ R^3$

Answer 1

The minimum distance of the plane to the origin is $$ h = \frac{|d|}{\sqrt{a^2+b^2+c^2}}$$

So by changing $d$ you are moving the plane along it's normal direction, with less absolute value closer to the origin and and larger absolute value away from the origin.

Answer 2

The modulus of d is essentially the perpendicular distance of the plane from the origin. If $r_0$ is any point on the plane and n = (a, b, c) is a unit normal vector to the plane then -d = $r_0.n$ where . is dot product. If n is not a unit vector simply divide by its length first.