# Parametric equation for a plane perpendicular to a vector

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 $c$?

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None of the three answers given so far is an improvement over what you'd already got (and I don't think there can be any). In each answer, there's a case distinction according to which of $a,b,c$ is non-zero; you can do the same in the solution you already had. (craftsman.don's answer hides the case distinction in "given a normal vector $N$", which also requires a case distinction because no normal vector will work for all cases.) – joriki Oct 4 '11 at 17:58
@joriki: I think the critique in your comment does not apply to my answer (which was written much later). – Did Jan 9 '12 at 16:19
@Didier: You're right, it doesn't :-) That's a nice approach. So I should more specifically say "I don't think there can be any with two parameters". – joriki Jan 9 '12 at 16:27
@joriki: Thanks. It would be nice to make more apparent the connection with (some form of) projective space but, until now, I have been too lazy to do so. – Did Jan 9 '12 at 16:42

## 4 Answers

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 $(\alpha,\beta,\gamma)$.

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Can you please brief how you obtained the parametrization? Thanks. – Kumara Mar 28 '13 at 8:24
Looking for a parametrization with symmetries, since all this is reminiscent of a projective geometry setting. – Did Mar 28 '13 at 8:44
I couldn't get it. Could you please be little elaborate? – Kumara Mar 28 '13 at 9:28
One is looking for vectors in the plane $P$ of equation $ax+by+cz=0$. Obviously $(0,c,-b)$ is in $P$ since $a\cdot0+b\cdot c+c\cdot(-b)=0$. By symmetry, $(-c,0,a)$ and $(b,-a,0)$ are in $P$ as well. And every linear combination of them is in $P$. This is the parametrization in the post. – Did Mar 28 '13 at 10:26

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 $t_1, t_2$.

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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 $$x(t_1,t_2):= t_1 u+t_2 v\ .$$

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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)

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I'm afraid I don't understand. Don't I already have $U,V,W$ as $a,b,c$? What is the parametric equation where $(x, y, z)$ is given by $(t, u)$? – jnm2 Oct 4 '11 at 15:28
U,V,W are vectors, not scalar value. You first find out the basis U,V,W, and then parametric equation is Ut + Wu. I make a mistake in the above answer, I am sorry. I have edit it. – craftsman.don Oct 4 '11 at 15:47