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Math Help - Vectors in a tetrahedron

  1. #1
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    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 <br />
\hat n<br />
    then <br />
\hat n = \frac{{(a - c) \wedge (b - a)}}{{\left| {(a - c) \wedge (b - a)} \right|}}<br />
    distance d is the projection of c onto a, so <br />
d = (c.\hat n)\hat n<br />

    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.
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  2. #2
    Member Henderson's Avatar
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    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}

     <br />
h = \frac{l\sqrt(6)}{3}<br />
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  3. #3
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    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?
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