Proving $AB$, $AC$ perpendicular to each other when vectors are median

Question

I want to prove this claim:

Triangle $ABC$ with $A(2,4,6),B(6,2,2),C(0,0,0)$ median, $AC$ and $BC$ perpendicular to each other.

What I did is to the $AB$, $BC$ make a dot product and thought it will be zero, but no result.
there is something I miss here.
Any suggestions?
Thanks!

Cannot understand what the question is. ACand BC cannot be orthogonal, given that both have a vertex in the origin and the other inside the first quarter. — Vincenzo Tibullo, Jul 13 '13 at 09:05
I need to prove that the median in this triangle are orthogonal — Ofir Attia, Jul 13 '13 at 09:11

score 1 · Answer 1 · answered Dec 16 '13 at 14:21

1

The median to $AC$ is $$\frac{A+C}{2}-B=(-5,0,-1), $$ and the median to $BC$ is $$\frac{B+C}{2}-A=(1,-3,-5) .$$ Their dot product is indeed zero.

answered Dec 16 '13 at 14:21

user1337

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Proving $AB$, $AC$ perpendicular to each other when vectors are median

1 Answers1