Plane formula explanation and implicit form

According to wikipedia a plane can be expressed as: $$(p-p_{0})\cdot n=0$$ where $$n - normal$$$$p_{0} - point$$however nothing is told about p. What is it? As I understand to define a plane I only need a vector and a normal.

Also how to solve it to get implicit formula of the plane?:$$Ax+By+Cz+D=0$$

• $p$ is an arbitrary point in the plane. $p_0$ is a fixed point in the plane. So $p-p_0$ is parallel to the plane. $$(p-p_0)\cdot n = \big((x,y,z)-(x_0,y_0,z_0)\big)\cdot(A,B,C) = Ax+By+Cz -(Ax_0 +By_0 +Cz_0) = Ax+By+Cz+D = 0$$ – user137731 Nov 29 '16 at 18:18
• @Bye_World thanks! Can you explain where D comes from? – lukas.pukenis Nov 29 '16 at 18:23
• $-(Ax_0 + By_0 +Cz_0)$ is just a constant. So we can just rename it $D$. – user137731 Nov 29 '16 at 18:24
• @Bye_World thank you, that explains it for me – lukas.pukenis Nov 29 '16 at 18:41
• Just define $D := -Ax_0 -By_0 -Cz_0$. You can always define negative constants, that's no problem -- e.g. $D = -5$ is fine. – user137731 Nov 29 '16 at 19:04

$p$ is the point you want to test and see if it belongs to the plane.

Let $n = (n_x, n_y, n_z)$ be a vector perpendicular to the plane and let $P_0$ be some point on the plane. Then the point $P$ belongs to the plane iff the vector $P - P_0$ is perpendicular to $n$. In terms of dot product, that is

$$(P - P_0)\cdot n = 0$$

If you write out explicitly the coordinates of the vector $P - P_0$ and expand the dot product, then you get something of the form $Ax + By + Cz = D$:

$$\begin{cases}P_0 = (x_0, y_0, z_0)\\P - P_0 = (x - x_0, y - y_0, z - z_0)\end{cases}\\ (P - P_0)\cdot n = (x - x_0, y - y_0, z - z_0)\cdot(n_x, n_y, n_z) = \\ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0$$

Collecting all the terms with the subscript $0$ gives the constant $D$. That is the reason why if you have a plane $Ax + By + Cz + D = 0$, then $(A, B, C)$ is a normal vector to the plane.

Let $p_0=(a_0,b_0,c_0)$ and $p$ be any point on the plane $(x,y,z)$ Then $$\vec p-\vec p_0=(x-a)\hat i+(y-b) \hat j+(z-c)\hat k$$ And if the normal be $\vec n =n_x \hat i+n_y \hat j+n_z\hat k$ then you get an equation by $$(\vec p-\vec p_0)\cdot \vec n=0$$ i.e.

$$n_x(x-a)+n_y(y-a)+n_z(z-a)=0\\\implies A=n_x\ ;B=n_y\ ;C=n_z\ ;D=-(n_xa+n_yb+n_zc)$$