# Prove that three vectors are coplanar

Three vectors are given: $$u,v,w$$. It is given that: $$|u|=|v|=|w|= \sqrt{2}$$; $$u\cdot v=u\cdot w=v\cdot w=-1$$. Prove that vectors $$u,v,w$$ are coplanar (on the same plane).

I have a few ideas, but I don't know if they are helpful in this case:

I know that three vectors are co-planar if $$u\cdot(v x w)=0$$. In addition, I assume that you can prove it with linear dependence, but I don't know how to use it here.

In addition, I thought that maybe the angle between the vectors can be of help- $$120$$ degrees between every $$2$$ vectors- but does that necessarily mean that they are on the same plain- co-planar?

The sum of three angles formed by three non-coplanar vectors is always less than 360 degrees. However instead of proving this general statement one can arrive at the required result immediately, observing that the vectors in the problem form an equilateral triangle. An easy check shows:

$$(u+v+w)\cdot(u+v+w)=|u|^2+|v|^2+|w|^2+2u\cdot v+2v\cdot w+2w\cdot u=0\\ \implies u+v+w=0.$$

As the three vectors are linearly dependent, they are coplanar.

• Why did you think of adding the the vectors to one another in the first place? Why does "u+v+w=0" mean that they are co-planar? Did you write that they are linearly dependent because 1∙u+1∙v+1∙w=0, the coefficients are not all 0? Commented Mar 28, 2019 at 21:15
• @noamAzulay It is not hard to notice that the vectors form equilateral triangle. The non-coplanar vectors are linearly independent and therefore cannot add to 0. And you are right $1\ne0$ quite certainly.
• @noamAzulay As I wrote in another comment, the equality $u+v+w=0$ is evident, as soon as one realizes that the angles between the vectors are 120 degrees. The same method can be applied in fact to any three vectors such that the angles between them satisfy $\alpha+\beta+\gamma=2\pi$ (or $\alpha+\beta-\gamma=0$). The only complication will be in finding the correct coefficients. As soon as they are found one proceeds with the same demonstration $(c_1u+c_2v+c_3w)^2=0$.