How to prove this condition if three vectors are colinear?

Question

So I was given this problem:

Let $\vec{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\vec{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$. Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if $$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \mathbf{0}.$$

I tried plugging it in and it looks like it is true:

Let $\vec{a} = \begin{pmatrix}2\\4\\6\end{pmatrix}$, $\vec{b} = \begin{pmatrix}4\\8\\12\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}8\\16\\24\end{pmatrix}$. $$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} =\begin{pmatrix}0\\0\\0\end{pmatrix} +\begin{pmatrix}0\\0\\0\end{pmatrix} +\begin{pmatrix}0\\0\\0\end{pmatrix}\\ = \begin{pmatrix}0\\0\\0\end{pmatrix}\\ $$

How could I prove this to be true? I am not entirely sure where to begin.

Answer 1

We have $|v \times w| = |v||w|\sin \theta$, where $\theta$ is the angle between $u$ and $v$. Therefore three points $a$, $b$ and $c$ are collinear if and only $$(b - a) \times (c - a) = 0$$ Expanding $$b \times c - b\times a - a \times c + a \times a = 0$$ Since $a \times a = 0$ and anticommutivity $$a\times b + b \times c + c \times a = 0$$

Answer 2

Vectors $\vec{A}$ and $\vec{B}$ are colinear if

$$\vec{A}=k\vec{B}$$

Then the vector product naturally comes out to be $0$.

Answer 3

Hint: Collinear vectors have the same direction vector.