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?
Best Answer
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.