Parametric equation for a plane perpendicular to a vector

Question

The implicit equation for a plane perpendicular to a given vector at the origin is $ax + by + cz = 0$.

I can write this in parametric form as $x = t, y = u, z = -\frac{at + bu}{c}$.

The only problem is that I can't use this equation when $c = 0$, and $a$, $b$ and $c$ are each going to be zero at different times so I can't avoid division by zero by shuffling around the coordinates. Is there a way to formulate this plane parametrically without restricting values of $a$, $b$ and $c$?

Answer 1

A solution is to parametrize the plane by $\mathbb R^3$. This may seem a strange idea since a plane has dimension $2$ and $\mathbb R^3$ has dimension $3$, but, for the plane of equation $ax+by+cz=0$, such a parametrization is $$ x=\gamma b-\beta c,\quad y=\alpha c-\gamma a,\quad z=\beta a-\alpha b,\quad (\alpha,\beta,\gamma)\in\mathbb R^3. $$ Then, as was to be expected, the parameter $(\alpha,\beta,\gamma)$ corresponding to a given point $(x,y,z)$ in the plane is not uniquely defined. The set of parameters corresponding to $(x,y,z)$ is exactly the affine line $L(x,y,z)$ in $\mathbb R^3$ defined as $$ L(x,y,z)=(\alpha,\beta,\gamma)+\mathbb R\cdot(a,b,c)=\left\{(\alpha+\lambda a,\beta+\lambda b,\gamma+\lambda c)\mid\lambda\in\mathbb R\right\}, $$ where $(\alpha,\beta,\gamma)$ denotes any parameter corresponding to $(x,y,z)$. Note that $L(x,y,z)$ depends on $(x,y,z)$ but not on the choice of $(\alpha,\beta,\gamma)$.

Answer 2

let the point set on the plane: P, and the origin: Q

given a normal vector N, then dot(P-Q, N)=0 --------EQU 1

subsitute arbitrary value x,y in to the EQU 1, then you can get a value z. M=(x,y,z) is on that plane.

after that, let U=M - Q, V=N, and W=cross(U,V), then you can get a local frame [U,V,W] of that plane.

any linear combination of U,W can be on the plane. that is, x*U + u*W.

(notice the cross product, cross(U,V), the order is vital. you can change that to cross(V,U) if you want a different handness)

Answer 3

We can find two vectors $u$ and $v$ that are orthogonal to the normal vector of the plane $n = (a, b, c)$. Because not all $a, b, c$ are zero, we can find two components, such at least one of them is not zero. Assume these are $a, b$. The vector $u = (-b, a, 0)$ is orthogonal to $n$ and lies hence in the plane. The vector $v = n \times u = (-ac, -bc, a^2 + b^2)$ is orthogonal to both $n, u$ and lies also in the plane.

Now, the plane can be parametrized as

$$ x = t_1 u + t_2 v$$

where $x$ is an arbitrary vector in the plane given by the scalar parametes $t_1, t_2$.

Answer 4

Since $(a,b,c)\ne0$ at least two of the three vectors $$u:=(b,-a,0),\quad v:=(0,c,-b),\quad w:=(-c,0,a)$$ are nonzero, linearly independent, and orthogonal to $(a,b,c,)$. If, e.g., $b\ne0$ then this is the case for $u$ and $v$, and your plane can be parametrized in the form $$x(t_1,t_2):= t_1 u+t_2 v\ .$$