# The many ways in which to express a plane

There are many ways to express a plane of $$R^3$$. I am focusing on two of them.

The first is the cartesian equation $$Ax + By + Cz + D = 0$$.

The second is to give two direction vectors $$u$$ and $$v$$ and a point $$P$$ of the plane.

My question is: how can I obtain two ortogonal direction vectors $$u$$ and $$v$$ and a point $$P$$ from the cartesian equation $$Ax + By + Cz + D = 0$$? How can I obtain the cartesian equation from the direction vectors and a point of the plane?

$$(a,b,c)=\vec n$$ is a vector normal to the plane.

From the knowledge of the Cartesian equation, choose the vector among $$(0,c,-b), (-c,0,a)$$ and $$(b,-a,0)$$ with the largest norm (do this to avoid degeneracies). This gives you a first vector perpendicular to $$\vec n$$, let $$\vec u$$. Then set $$\vec v=\vec n\times\vec u$$, and you have your second vector.

For the point, you can project the origin orthogonally onto the plane, i.e. find $$\lambda$$ such that $$\lambda(a,b,c)$$ fulfills the plane equation. This yields

$$\lambda(a^2+b^2+c^2)+d=0.$$

The converse is easier.

Compute

$$(a,b,c)=\vec u\times\vec v$$ and expand

$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0.$$

From $$Ax+By+Cz+D=0$$

you get first the normal vector to the plane $$n=(A,B,C)$$.

then you can take

$$u=(0,C,-B)$$

and $$v$$ as the vectorial product of $$n$$ by $$u$$.

To get the cartesian equation from two vectors $$u,v$$ and a point $$P$$,

$$det(PM,u,v)=0$$

with $$M=(x,y,z)$$.

• Really thanks for the help, but I need two specifications. Firstly, I understand where $n$ come from and the use of the vectorial product to gain $v$, but why $u$ is egual to $(0,C,-B)$? Second: in the formula $det(PM, u, v)=0$ the product $PM$ is a vectorial product? Nov 19, 2018 at 19:41
• For planes of the form $Ax + D = 0$ this procedure gives $u = v = (0, 0, 0)$. Nov 19, 2018 at 20:17
• @Travis In this case $n=(A 0,0),u=(0,1,0),v=(0,0,1)$. Nov 19, 2018 at 20:19

As other answers point out, you know the normal vector $$n=[A,B,C]$$. Suppose you have one vector $$u\perp n$$, $$|u|\neq 0$$, then clearly you can find $$v\perp u$$, $$v\perp n$$ using $$v=u\times n$$.

So the problem reduces to finding a single nonzero vector perpendicular to $$n$$. In 3D, there is no single (non-branching) formula that will do this: see this question.

We can write $$Ax+By+Cz+D=0$$ as: $$(A,B,C)\cdot (x,y,z) = -D$$

That is, all points $$(x,y,z)$$ that have the same dot product with $$(A,B,C)$$, which is $$-D$$. The shortest vector for which this is the case, is the one that is a parallel to $$(A,B,C)$$. It means that this vector must be perpendicular to the plane to achieve that. Therefore the vector $$\mathbf n$$: $$\mathbf n = \frac{(A,B,C)}{\sqrt{A^2+B^2+C^2}}$$ is the normal vector of the plane with length $$1$$.

Consequently, we can write the equation as: $$\mathbf n \cdot (x,y,z) = -\frac{D}{\sqrt{A^2+B^2+C^2}} = d$$ To find a vector $$\mathbf P$$ in the plane, any vector with $$\mathbf n \cdot \mathbf P = d$$ will do. That's because the plane consists of all points with the same dot product with $$\mathbf n$$, which is $$d$$. We can pick the one that represents the shortest distance: $$\mathbf P = d\mathbf n = -\frac{D}{A^2+B^2+C^2}(A,B,C)$$

And since $$\mathbf n$$ is a unit vector, the distance of the origin to the plane must be $$|d|=\frac{|D|}{\sqrt{A^2+B^2+C^2}}$$.

To find 2 vectors in the plane, we need 2 independent vectors that are perpendicular to the normal vector $$(A,B,C)$$. It suffices if the dot product is $$0$$. Without loss of generality, let's assume that $$A\ne 0$$. Then we can pick: $$\mathbf u =(B,-A,0)\quad\text{and}\quad \mathbf v = (C,0,-A)$$

• Thanks @Travis. Fixed it by assuming without loss of generality that $A\ne 0$. Nov 19, 2018 at 20:21

Going one direction:

Given $$Ax + By + Cz + D = 0$$

$$(B,-A, 0)$$ is a vector in the plane

$$(A,B,\frac {A^2 + B^2}{C})$$ is a vector in the plane orthogonal to the first.

$$(0,0,-\frac {D}{C})$$ is a point in the plane

Of course, this is just one way to find orthogonal vectors and a point in the plane.

Going the other direction....

given, vectors in the plane $$u,v$$ and point in the plane $$P$$

$$u\times v = N$$ is normal vector in the plane.

Suppose the components are $$N = (N_x,N_y,N_z)$$

then $$N_x x + N_y y+ N_z z - N\cdot P = 0$$ will be an equation for the plane.