Following this article on MathWorld define the plane passing through a point $x_0$ perpendicular to a vector $n$ as the set of all points $x$ satisfying $$n \cdot (x - x_0) = 0.$$ Define a normal vector to a plane to be a nonzero vector $N$ that satisfies $$N \cdot (x_1 - x_2) = 0$$ for all points $x_1$ and $x_2$ on the plane.

Question: Let $\alpha$ be a plane. If $N_1$ and $N_2$ are both normal vectors to $\alpha$, must they be parallel to each other (i.e., does there exist a nonzero number $c$ such that $N_1 = cN_2$)?

I've tried to resolve the question using the cross product and some linear algebra as follows: Let $x_1$ and $x_2$ be two distinct points on $\alpha$ and set $v = (x_1 - x_2) \times N_1$. Then $v$ is orthogonal to both $(x_1 - x_2)$ and $N_1$; so the set $\{v, N_1, x_1 - x_2\}$ is an orthogonal basis of $\mathbb{R}^3$. It follows that there exist numbers $c_1$, $c_2$ and $c_3$ such that $$N_2 = c_1v + c_2N_1 + c_3(x_1 - x_2).$$ The dot product $N_2 \cdot (x_1 - x_2)$ is zero since $N_2$ is a normal vector to $\alpha$; hence $$0 = N_2 \cdot (x_1 - x_2) = c_3\lVert x_1 - x_2 \rVert^2,$$ where $\lVert v \rVert = \sqrt{v \cdot v}$ for all vectors $v$. Since $x_1 \neq x_2$, this forces $c_3 = 0$. It remains to prove that $c_1 = 0$. I'm stuck on this point; all ideas are welcome.

  • $\begingroup$ Yes ${}{}{}{}{}{}$ $\endgroup$
    – MPW
    Oct 31, 2015 at 8:04

1 Answer 1


Look at dimensions.

First, let $$ W = \{ x_1-x_2 \mid x_1,x_2\text{ are in the plane}\} $$ It is an easy matter of algebra to see that this is the same as $$ W = \{ x_1-x_0 \mid (x_1-x_0)\cdot n = 0 \} $$ which is a linear subspace of $\mathbb R^3$, namely the solution space of $$ n_x x + n_y y + n_z z = 0 $$ which has dimension $2$ because it is the null space of the $1\times 3$ matrix $[n_x\;n_y\;n_z]$, and by the rank-nullity theorem the dimension must be either 2 or 3 and can be 3 only if $n=0$.

Thus, $W$ has some basis $\{w,u\}$. It should be easy to see that $N$ is a normal vector exactly if $N\cdot w=0$ and $N\cdot u=0$ -- one doesn't need to check for all members of $W$, only the two basis vectors.

This means that the set of all normal vectors is the solutions to $$ w_x x + w_y y + w_z z = 0 \\ u_x x + u_y y + u_z z = 0 $$ which is two linear equations of three unknowns. By the rank-nullity-theorem again this solution space has dimension 1, because the matrix $$\begin{bmatrix} w_x & w_y & w_z \\ u_x & u_y & u_z \end{bmatrix}$$ has rank 2 (if the rank were any lower, $\{w,u\}$ would not be linearly independent, contradicting their being a basis for $W$).

This means the set of normal vectors is an $1$-dimensional linear subspace.

Finally, if you have two normal vectors $N_1$ and $N_2$, they must be linearly dependent, because a linear subspace cannot contain a linearly independent subset with more elements than its dimension. And if you have two linearly dependent vectors, one is always a multiple of the other.

  • $\begingroup$ Thanks a lot for your thorough answer to the question. It had not occurred to me to look at it that way; I was stuck on the idea of using the cross product. $\endgroup$
    – jhck
    Oct 31, 2015 at 9:10

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