calculate arbitrary points from a plane equation

I understand how one can calculate a plane equation (ax+by+cz=d) from three points but how can you go in reverse?

How can you calculate arbitrary points from a plane equation?

• Do you mean tell whether or not a point lies in the plane? Just plug in its coordinates into the equation; if ax+by+cz does not equal d, then it does not lie in the plane, if it does equal d, then it does lie in the plane. – Chill2Macht Apr 23 '16 at 20:59
• You can convert the plane to parameter-form $\vec{x} = \vec{a} + \lambda \vec{u} + \mu \vec{v} \quad \text{with } \lambda, \mu \in \mathbb{R}$ and pick arbitrary values for $\lambda$ and $\mu$, then compute the resulting $\vec{x}$ which is in the plane. – Maximilian Gerhardt Apr 23 '16 at 21:01
• no, i mean how can I calculate a set of unknown points from ax+by+cz=d. the x, y, z intercepts are one way to approach it but I wanted to know if there was another way (a=b=0, solve for z, etc.) – math_Artist88 Apr 23 '16 at 21:02
• how could I convert the plane eqtn(ax+by+cz=d) to parametric form. I am not familiar with this concept – math_Artist88 Apr 23 '16 at 21:05
• Think of $\vec{a}$ as the base-point, then you have two vectors $\vec{u}, \vec{v}$ which are two "direction vectors" from this point which describe the whole plane. You can compute these from the three original points $A,B,C$ quite easily. Pick one as the base-point (e.g. $A$), then $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{AC}$. Else, there are tons of questions here and material on the internet about this. – Maximilian Gerhardt Apr 23 '16 at 21:12

From your comment I finally understood what you are looking for:

If you have a plane defined by $$a x + b y + c z = d$$ then you also have the following properties:

• Plane normal direction: $$\hat{n} = \begin{pmatrix} \frac{a}{\sqrt{a^2+b^2+c^2}} \\ \frac{b}{\sqrt{a^2+b^2+c^2}} \\ \frac{c}{\sqrt{a^2+b^2+c^2}} \end{pmatrix}$$

• Point on plane closest to the origin (position of plane) $$\vec{r} = \begin{pmatrix} \frac{a d}{a^2+b^2+c^2} \\ \frac{b d}{a^2+b^2+c^2} \\ \frac{c d}{a^2+b^2+c^2} \end{pmatrix}$$

• Distance of plane from the origin $$r = \frac{d}{\sqrt{a^2+b^2+c^2}}$$

• Directions along the plane (not unit vectors), and perpendicular to $$\hat{n}$$. \begin{align} \hat{e}_1 & = \begin{pmatrix} c-b \\ a-c \\ b-a \end{pmatrix} & \hat{e}_2 & = \begin{pmatrix} a (b+c)-b^2-c^2 \\ b (a+c) -a^2-c^2 \\ c (a+b)-a^2 - b^2 \end{pmatrix} \end{align}

You can verify that $$\hat{e}_1 \cdot \hat{n} =0$$, $$\hat{e}_2 \cdot \hat{n}=0$$ and $$\hat{e}_1 \cdot \hat{e}_2 =0$$, where $$\cdot$$ is the dot (inner) product.

Confirmation via GeoGebra

NOTES: Please edit the equation to make it clear you are looking for the plane properties when given a plane in equation form.

• This post would be a good approach except I do not know how to calculate the center point of the rectangle stackoverflow.com/questions/22769430/… – math_Artist88 Apr 24 '16 at 2:29
• a normal vector and a point will give you a plane equation. im trying to go backwards from the plane equation to find a point at the center of the plane – math_Artist88 Apr 24 '16 at 2:37
• perhaps, i should rephrase the question. how could I generate a rectangle (corner1, corner2, corner3, corner4) from the plane equation – math_Artist88 Apr 24 '16 at 19:40
• From the information I in this answer use the two direction vectors to form a rectangle. – ja72 Apr 24 '16 at 22:22
• There were typos in the equations that I finally fixed. – ja72 May 22 at 22:48

You pick arbitrarely two values $a,b$ for $x$ and $y$, you plug them in the equation, you solve it for $z$ and obtain a value $c$. The point $(a,b,c)$ is a point of your plane.

• but how do I know the two arbitrary values for x and y are on the plane? – math_Artist88 Apr 23 '16 at 21:11
• if your equation has an equation of the form $ax+bx+cx=d$ and $a \neq 0$ and $b \neq 0$, you can pick $x$ and $y$ as you wish they will be one the plane. – Jennifer Apr 23 '16 at 21:16