-1
$\begingroup$

This question already has an answer here:

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


I was thinking of proving that the area of the triangle formed by the three points is 0. I thought the box product, $|(a \times b)\bullet c|$, would be helpful but I don't know how to relate that to the equation. All help is greatly appreciated.

$\endgroup$

marked as duplicate by Yuna Kun, Community Mar 31 '17 at 1:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Why delete and repost the same question? You should edit the first post if you need to add or make changes. $\endgroup$ – dxiv Mar 31 '17 at 1:05
0
$\begingroup$

You don't need to complicate this problem. We know that $x\times y=0$ if and only if $x=my$ where $m$ is a real number. Given $a$, $b$, $c$, we want to look at $a-b$ and $b-c$ $$(a-b)\times(b-c)=a\times b - a\times c -b\times b +b\times c=a\times b +b\times c+c\times a$$ If $a$,$b$, $c$ are collinear, then the left hand side is 0, therefore the right hand side is 0. Or the other way around, if RHS=0 then LHS=0, therefore $(a-b)=m(b-c)$.

$\endgroup$
  • $\begingroup$ How does $(a-b)\times(b-c) = 0 \iff (a-b) = m(b-c)$ imply that $a\parallel b, \;a\parallel c, $ and $b\parallel c$ exactly? $\endgroup$ – Vlad Mar 31 '17 at 1:34
  • $\begingroup$ $a$ is not parallel to $b$ or to $c$. The difference between two pairs have to be parallel. If you think in a plane (1,0),(1,1), and (1,-1) are not parallel, but they represent point on a vertical line through $x=1$ $\endgroup$ – Andrei Mar 31 '17 at 1:45
  • 1
    $\begingroup$ I agree with you now, after I saw geometrical justification here $\endgroup$ – Vlad Mar 31 '17 at 2:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.