# Thread: Vectors in a tetrahedron

1. ## Vectors in a tetrahedron

I'm stuck on this question:

Q: The vectors a, b, and c are of equal length l, and define the positions of the points A, B, and C relative to O. If OABC is a regular tetrahedron, find the distance of one vertex from the opposite face.

My attempt at the question was as follows:
Let the vector normal to the plane ABC be $
\hat n
$

then $
\hat n = \frac{{(a - c) \wedge (b - a)}}{{\left| {(a - c) \wedge (b - a)} \right|}}
$

distance d is the projection of c onto a, so $
d = (c.\hat n)\hat n
$

but if I substitute equation 1 into 2 it gives rise to a very complicated equation and I know I need to use the fact that a, b, and c are of length l.

Any help and pointers would be much appreciated.

2. Without using vectors, I placed one of the faces (an equilateral triangle) as the base. The top vertex is right above the center of that face. The center of the face is $\frac{l\sqrt(3)}{3}$ from the center, and the top vertex is $l$ from the center. Using Pyth. Theorem,

$(\frac{l\sqrt(3)}{3})^2 + h^2 = l^2$

$\frac{l^2}{3} + h^2 = l^2$

$h^2 = \frac{2l^2}{3}$

$
h = \frac{l\sqrt(6)}{3}
$

3. Thanks for your help, I guess that's a lo easier than what I was trying to do. But unfortunately I have to do the question using vectors (point of the whole exercise...)

I know I need to do the dot product of the position vector of the centre of face with something else, but I don't know what else, so any pointers please?