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Let $\mathbf{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$. 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 $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}.$

Since the cross product of two vectors gives an area, and for two vectors to give an area of $0$ they need to be on the same line (or they can be a point, but I'm assuming both are not $\mathbf 0$). However, in this problem, each of the cross products need not necessarily be $0$ since it's their sum that is $0$, and now I'm not sure what to do.

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    $\begingroup$ The points are collinear iff the vectors $b-a$ and $c-a$ point in the same direction. $\endgroup$ – Angina Seng Feb 17 '18 at 13:56
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Let $A$, $B$ and $C$ be the corresponding points. Do you know that they are colinear iff the vectors $b-a$ and $c-a$ are colinear? And as you said, $b-a$ and $c-a$ are colinear iff their cross product is $0$.

Edit

Let $O=(0,0,0)$ be the origin of $\mathbb{R}^3$. Then we have $\vec{OA}=a$, $\vec{OB}=b$ and $\vec{OC}=c$.

$A,B$ and $C$ are aligned iff $\vec{AB}$ and $\vec{AC}$ are parallel, and you can write $\vec{AC}=\lambda\vec{AB}$ for some $\lambda\in\mathbb{R}$.

Here's an example where $\lambda>0$

enter image description here

Here's an example where $\lambda<0$

enter image description here

If you agree that $A,B$ and $C$ are aligned iff $\vec{AB}$ and $\vec{AC}$ are parallel, notice that:

\begin{align*} \vec{AB}&=\vec{AO}+\vec{OB}\\ &=\vec{OB}-\vec{OA}\\ &=b-a. \end{align*}

Similarly, $\vec{AC}=c-a$.

Hence $A$, $B$ and $C$ are aligned iff $b-a$ and $c-a$ are parallel.

The knowledge that these two vectors are parallel is enough. Actually, $c-b=(c-a)-(b-a)$. Since $\vec{AC}=\lambda\vec{AB}$, in other words, $c-a=\lambda(b-a)$, we have $c-b=(\lambda-1)(b-a)$ and thus $c-a$ is indeed parallel to both $b-a$ and $c-a$. Mentioning $c-b$ as well is not wrong, but redundant.

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  • $\begingroup$ I think I might have seen that somewhere, but I'm not familiar with it... Could you explain a little more in depth about the $b-a$ and $c-a$ being collinear? What about $c-b$? $\endgroup$ – space Feb 17 '18 at 14:06
  • $\begingroup$ Sure. I included the explanations. $\endgroup$ – Scientifica Feb 17 '18 at 14:30
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Using that $\,\color{blue}{\mathbf{b} \times \mathbf{c} = -\,\mathbf{c} \times \mathbf{b}}\,$ and $\,\color{red}{\mathbf{a} \times \mathbf{a} = 0}\,$:

$$ \begin{align} \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0} \;\;&\iff\;\; \mathbf{a} \times \mathbf{b} \color{blue}{- \mathbf{c} \times \mathbf{b}} + \mathbf{c} \times \mathbf{a} \color{red}{- \mathbf{a} \times \mathbf{a}} = \mathbf{0} \\ & \iff\;\; (\mathbf{a} - \mathbf{c}) \times \mathbf{b} + (\mathbf{c} - \mathbf{a}) \times \mathbf{a} = 0 \\ & \iff\;\; (\mathbf{a} - \mathbf{c}) \times \mathbf{b} - (\mathbf{a} - \mathbf{c}) \times \mathbf{a} = 0 \\ & \iff\;\; (\mathbf{a} - \mathbf{c}) \times (\mathbf{b} - \mathbf{a}) = 0 \end{align} $$

The latter equality is equivalent to $\,\mathbf{a} - \mathbf{c}\,$ and $\, \mathbf{b} - \mathbf{a}\,$ being collinear.

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    $\begingroup$ Thank you for this clarification — I was kind of missing that link! $\endgroup$ – space Feb 18 '18 at 3:00
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If the vectors are collinear, they are of the form $$ \mathbf{a}=\mathbf{u}+\alpha\mathbf{v}, \quad \mathbf{b}=\mathbf{u}+\beta\mathbf{v}, \quad \mathbf{c}=\mathbf{u}+\gamma\mathbf{v} $$ Then $$ \mathbf{a}\times\mathbf{b}= \mathbf{u}\times\mathbf{u}+\alpha\mathbf{v}\times\mathbf{u}+ \beta\mathbf{u}\times\mathbf{v}+\alpha\gamma\mathbf{v}\times\mathbf{v}= (\beta-\alpha)\mathbf{u}\times\mathbf{v} $$ Similarly $\mathbf{b}\times\mathbf{c}=(\gamma-\beta)\mathbf{u}\times\mathbf{v}$ and $\mathbf{c}\times\mathbf{a}=(\alpha-\gamma)\mathbf{u}\times\mathbf{v}$, so easily $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$.

Conversely, suppose $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$ and set $\mathbf{b}-\mathbf{a}=\mathbf{x}$. Without loss of generality, we can assume $\mathbf{a}\ne\mathbf{b}$ (or there would be just one or two vectors, which are obviously collinear). Then \begin{align} \mathbf{0} &=\mathbf{a}\times(\mathbf{a}+\mathbf{x}) +(\mathbf{a}+\mathbf{x})\times\mathbf{c} +\mathbf{c}\times\mathbf{a} \\[4px] &=\mathbf{a}\times\mathbf{x} +\mathbf{a}\times\mathbf{c} +\mathbf{x}\times\mathbf{c} +\mathbf{c}\times\mathbf{a} \\[4px] &=\mathbf{x}\times(\mathbf{c}-\mathbf{a}) \end{align} which implies $\mathbf{c}-\mathbf{a}=\delta\mathbf{x}$. Thus $$ \mathbf{a}=\mathbf{a}+0\mathbf{x}, \quad \mathbf{b}=\mathbf{a}+1\mathbf{x}, \quad \mathbf{c}=\mathbf{a}+\delta\mathbf{x} $$

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  • $\begingroup$ There is a slight problem if $\mathbf{b}=\mathbf{a}$ and $\mathbf{c}$ is not equal to the other two vectors. $\endgroup$ – Somos Feb 17 '18 at 14:32
  • $\begingroup$ @Somos That case is trivial, isn't it? Anyway, I'll add it. $\endgroup$ – egreg Feb 17 '18 at 14:36

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