Calc iii question/analytic geometry

[jaejoon89]

Given vertices of regular tetrahedron OABC, prove that vector OA + OB + OC is perpendicular to plane ABC...

I've been racking my brain on this one, can't figure it out... would appreciate some help

I thought I'd center it on the origin:
(0, 0, 0)
(1, 0, 0)
(0, 1, 0)
(0, 0, 1)

A + B + C = (1, 0, 0) + (0, 1, 0) +(0, 0, 1) = (1, 1, 1)
(not sure if that's right - seems like it would be outside the tetrahedron)

No idea where to go from there.

 

[quantumdude]

The problem here is that your tetrahedron isn't regular. Consider triangle OAB. Two of the sides have length 1, but the hypoenuse has length $\sqrt{2}$. You'll have to choose different vectors.

 

[jaejoon89]

Is there an easier way to do it? Showing that this is true for one regular tetrahedron doesn't necessarily/formally mean it holds for all of them, does it?

Also, even when I switch coordinates, it doesn't work. For example, a regular tetrahedron (... at least wikipedia says this one is regular)

O (1, 1, 1)
A (-1, -1, 1)
B (-1, 1, -1)
C (1, -1, -1)

vector OA = A - O = (-2, -2, 0)
Similarly, vector OB = (-2, 0, -2)
OC = (0, -2, -2)

AB = (0, 2, -2)
AC = (2, 0, -2)

AB x AC = (-4, -4, -4)
OA + OB + OC = (-4, -4, -4)

And when you take the dot product of those it should be 0 (if they are perpendicular, as they are supposed to be)... but it isn't 0

What's wrong? Also, again is showing it holds for one regular tetrahedron sufficient?

 

[quantumdude]

Is there an easier way to do it?

If there is, I don't see it.

Showing that this is true for one regular tetrahedron doesn't necessarily/formally mean it holds for all of them, does it?

No, but you're trying to do hear is easily generalized. Instead of trying to come up with a tetrahedron whose sides are of unit length just let the length be some variable, say $a$. Then *presto* you're working with any possible regular tetrahedron.