If $A,B,C$ are collinear, prove $\vec{A}\times \vec{B} + \vec{B}\times \vec{C} + \vec{C}\times \vec{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$ Prove that if $A, B$ and $C$ are collinear, then
$\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$.
So far, I know that $\vec{AB}, \vec{BC}, \vec{AC}$ are scalar multiples and their cross product is zero. I'm not sure how to apply this to prove $\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A}$ though.
 A: Since $A$, $B$ and $C$ are collinear, you can write $C$ as $A+\lambda(B-A)\require{cancel}$. So\begin{align}A\times B+B\times C+C\times A&=A\times B+B\times(A+\lambda(B-A))+(A+\lambda(B-A))\times A\\&=\cancel{A\times B+B\times A}+\lambda B\times(B-A)+\cancel{A\times A}+\lambda(B-A)\times A\\&=-\lambda B\times A+\lambda B\times A\\&=0.\end{align}
A: Setting up some vector identities:
$$ 
\overrightarrow{AB} = \overrightarrow{B} -\overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{B} = - \overrightarrow{B} \times \overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{A} = \overrightarrow{0}
$$
Proof:
$$
\begin{align}
&\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} \\
&= \overrightarrow{A} \times \overrightarrow{B} - \overrightarrow{A} \times \overrightarrow{C} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{A} \times (\overrightarrow{B} - \overrightarrow{C}) + \overrightarrow{B} \times \overrightarrow{C} \\
&= (\overrightarrow{B} - \overrightarrow{AB}) \times \overrightarrow{CB} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{B} \times \overrightarrow{CB} - \overrightarrow{AB} \times \overrightarrow{CB} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{B} \times (\overrightarrow{CB} + \overrightarrow{C}) - \overrightarrow{0} \\
&= \overrightarrow{B} \times (\overrightarrow{B} - \overrightarrow{C} + \overrightarrow{C}) \\
&= \overrightarrow{0}
\end{align}
$$
A: Geometrically, the cross product of $A$ and $B$ will have length equal to the (signed) area of the parallelogram with sides $OA$ and $OB$. The area is thus twice that of the triangle $OAB$. Then:
$$\textrm{area}(OAB) + \textrm{area}(OBC) = \textrm{area}(OAC),$$
and reversing the last triangle (thereby flipping the sign),
$$\textrm{area}(OAB) + \textrm{area}(OBC) + \textrm{area}(OCA) = 0.$$
A: Suppose WLOG that $B$ is on the segment $AC$, then
$$
B=\lambda A+\mu C,\qquad \lambda+\mu=1
$$
so
\begin{align}
  &A\times B+B\times C+C\times A=\\
  &\quad=A\times(\lambda A+\mu C)+(\lambda A+\mu C)\times C+C\times A=\\
  &\quad=\mu A\times C+\lambda A\times C-A\times C=\\
  &\quad=(\lambda + \mu-1)A\times C=0
\end{align}
