# Proving Coplanarity of 3 vectors

Let $a,b,c$ be three vectors such that $|a|=|b|=|c|=\sqrt{2}$ and $a\cdot b = b\cdot c = a\cdot c = -1$ .

How can I prove that they are all coplanar?

I found that the angle between every two of them is $120$ degrees, and tried to use this in order to prove that $( a\times b )\cdot c =0$ but without any luck. All I know is that $|a\times b | = \sqrt{3}$ ...

Will someone help me understand this ?

• how is it possible that you haven't luck? If they are coplanar, there should be some common plane - find it! Normals of planes defined by any pair of vectors will be colinear which is your desired proof...
– V-X
Commented Jan 6, 2014 at 7:38

A nice example of such vectors would be $a=(0,1,-1)$, $b=(-1,0,1)$, $c=(1,-1,0)$. In this case, $a+b+c=0$. Maybe this linear dependency holds for all such triples? Try to compute $$(a+b+c)\cdot (a+b+c).$$

• awesome! Thanks a lot !!!! Commented Jan 6, 2014 at 7:32

Here is one way to do it, in outline.

Given vectors $a$ and $b$ with length $\sqrt{2}$ and $a \cdot b = -1$, they form an angle of $120^{\circ}$ as you said. Now place a vector $c_0$ of length $\sqrt{2}$ in the plane of $a$ and $b$, forming an angle of $120^{\circ}$ with each of them. Then $a$, $b$ and $c_0$ satisfy all the hypotheses.

Now let $c$ be any vector satisfying the hypotheses, except perhaps the one about its length being $\sqrt{2}$. Prove that $c - c_0$ is orthogonal to $a$ and $b$, and therefore to all vectors in the plane generated by $a$ and $b$, including $c_0$.

Since $c$ consists of two orthogonal components $c - c_0$ and $c_0$, you can apply the Pythagorean Theorem to conclude that if $c - c_0 \ne 0$, then $c$ is longer than $c_0$.

This proves that if $c$ is to satisfy all the hypotheses, it must in fact be $c_0$.