Three points X, Y, Z have position vectors x,y,z. Show that X, Y and Z are collinear iff x ^ y + y ^ z + z ^ x = 0.
Here "^" denotes the vector cross product
If X, Y and Z are collinear then $\displaystyle z=\lambda x + (1-\lambda)y$ for some scalar $\displaystyle \lambda$. You can then verify that $\displaystyle y\times z + z\times x + x\times y=0$ (remembering that the cross product of a vector with itself is always 0).
For the converse, if $\displaystyle y\times z + z\times x + x\times y=0$ then $\displaystyle 0 = x.(y\times z) + x.(z\times x) + x.(x\times y) = x.(y\times z)$ (since the other two terms are 0). So $\displaystyle y\times z$ is orthogonal to x. But it is also orthogonal to y and z. If x, y and z are linearly independent then $\displaystyle y\times z$ is orthogonal to the whole space and is therefore 0. But that would mean that y and z are not linearly independent. That contradiction shows that the three vectors must be linearly dependent.
So one of them, z say, is a linear combination of the others, $\displaystyle z = \lambda x + \mu y$. Substitute that value for z into the equation $\displaystyle y\times z + z\times x + x\times y=0$ and you will find that $\displaystyle \mu = 1-\lambda$, which is the condition for X, Y and Z to be collinear.