Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to show the following:

If $v, w \in \mathbb{R}^3 \setminus \{(0,0,0)\}$ such that the set of vectors orthogonal to both of them is a plane through the origin, then each is a scalar multiple of the other.

I've proved that if $u^\perp = \{x \in \mathbb{R}^3 : x \perp u\}=\{x \in \mathbb{R}^3 : \langle x , u \rangle=0\}$, the set of all vectors which are ortogonal to $u$, then $v^\perp=w^\perp$ [since $v^\perp \cap w^\perp$ is a plane through origin (given) then dim$(v^\perp \cap w^\perp)$=dim$(v^\perp)$=dim$(w^\perp)$=2 and $v^\perp \cap w^\perp \subseteq v^\perp$,$v^\perp \cap w^\perp \subseteq w^\perp$]. Will this be of any help? I can't go any further. Many thanks.

share|cite|improve this question
The answer to this question is evident from the uniqueness of dot product. Since the dot product of orthogonal vectors is zero, and we have the two vectors having dot product zero with SAME PLANE. Hence, they ought to be scalar multiples of each other. – NikBels Dec 15 '11 at 17:54
up vote 2 down vote accepted

Take an orthogonal basis of the plane $\{{\bf v}_1,{\bf v}_2\}$. The set $\{{\bf v},{\bf v}_1,{\bf v}_2\}$ must be independent.

Indeed if $c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2=\bf 0$, then $${\bf v}_1\cdot ( c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2)=0 \quad\Rightarrow \quad c_2=0 $$ and $${\bf v}_2\cdot ( c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2)=0 \quad\Rightarrow \quad c_3=0. $$

we then must have $c_1=0$.

So, $\{{\bf v},{\bf v}_1,{\bf v}_2\}$ is an independent set and thus a basis of $\Bbb R^3$.

Now write $\bf u$ in terms of this basis. We have for some constants $c_1$, $c_2$, and $c_3$

$${\bf u}=c_1{\bf v}+c_2 {\bf v}_1+c_3 {\bf v}_2.$$


$$0={\bf v}_1\cdot {\bf u}= c_1 {\bf v}_1\cdot{\bf v} +c_2 {\bf v}_1\cdot {\bf v}_1+c_3 {\bf v}_1\cdot {\bf v}_2=c_2 ||{\bf v}_1||^2;$$

thus, $c_2=0$. Similarly, taking dot products with ${\bf v}_2$, it follows that $c_3=0$. Thus ${\bf u}=c_1{\bf v} $.

share|cite|improve this answer

Hint: For finite dimensional vector spaces, if you have subspaces $A,B$ such that $\text{dim}(A) = \text{dim}(B)$ and $A \subseteq B$, then $A = B$.

share|cite|improve this answer
I don't really understand your hint. Are you pointing towards proving $\mathbb{R}v=\mathbb{R}w$? In that case how shall I show either $\mathbb{R}v \subseteq \mathbb{R}w$ or $\mathbb{R}w \subseteq \mathbb{R}v$? I can't explicitly find any non-zero point common in them. – Sayantan Dec 15 '11 at 17:36
Using what you already have above, my hint gets you that $v^\perp = w^\perp$. Which implies that $\text{span}(v)=\text{span}(w)$. – Brandon Carter Dec 15 '11 at 17:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.