Vectors in a plane I have a question where i need to prove that three vectors are all on one plane.
I proved that the angle between ALL the three vectors is 120. 
Is that enough to prove that it is on one plane? because together they are 360.
Basiclly, is there any other way to build three vectors with 120 angle between them, without being all on one plane
Thanks!
 A: Hint:
You only need two vectors $\{\textbf{v}_1,\textbf{v}_2\}$ to span a plane. Show that the third vector $\textbf{v}_3$ is a linear combination of $\textbf{v}_1$ and $\textbf{v}_2$.
A: To answer the question you asked: if you have 3 nonzero vectors in 3-space, and the pairwise angles are all 120 degrees, then they lie in a plane, yes. Let's begin by making all three unit vectors, which won't change any angles. 
Call the first two $u$ and $v$; these clearly span a plane. We might as well pick coordinates on 3-space so that this is the $xy$-plane, and $\mathbf u$ is the vector $(\cos 60^\circ, \sin 60^\circ)$, while $v$ is the vector $(-\cos(60^\circ), \sin(60^\circ))$. 
The set of all vectors in 3-space that are 120 degrees from $\mathbf u$ is a cone with a vertex at the origin and a vertex angle of 120 degrees. You can, alternatively, think of this as a cone around the axis $-\mathbf u$, with vertex angle 60-degrees. Similarly for $\mathbf v$. It's not too hard to see that these two cones touch along the ray $x = 0, y \le 0, z = 0$., which is therefore where your third vector $\mathbf w$ must be located. 
A: use the scalar (NB : see comments below) triple product a dot ( b cross c )
$$\vec{a}\cdot(\vec{b}\times\vec{c})$$
This will be zero for three co-planar vectors
A: Assume WLOG that all the vectors are normalized. Then
$$u\cdot v=v\cdot w=u\cdot w=-\frac12$$
Now, consider the equation
$$au+bv+cu\times v=w$$
and dot-product with $u$:
$$a-\frac b2=-\frac12$$
now with $v$:
$$-\frac a2+b=-\frac12$$
and we get $a=b=-1$. Then
$$-u-v+cu\times v=w$$
and dot-product with $w$:
$$\frac12+\frac12+c[u,v,w]=1$$
so $c=0$ or $[u,v,w]=0$, and both facts tell us the same: $u$, $v$ and $w$ are coplanar.
A: Using spherical trigonometry, there is one triangle with sides of $\frac{2\pi}3$. Using the Law of Cosines for Spherical Triangles, we can compute each angle of the triangle to be
$$
\begin{align}
\cos(\theta)
&=\frac{\cos(\frac{2\pi}3)-\cos(\frac{2\pi}3)\cos(\frac{2\pi}3)}{\sin(\frac{2\pi}3)\sin(\frac{2\pi}3)}\\
&=\frac{-\frac12-(-\frac12)(-\frac12)}{\frac{\sqrt3}2\cdot\frac{\sqrt3}2}\\
&=-1
\end{align}
$$
That is, each angle of the triangle is $\pi$. Thus, the triangle forms a great circle, which is the intersection of a plane through the center of the sphere with the sphere.
This means that the three vectors are coplanar.
