a tetrahedron has the points (0,0,0), (3,0,0), (2,4,0), (1,3,5). For each face of the tetrahedron, find a vector whose length is equal to the area of the face and is perpendicular to the face pointing outward.
Hello multimasterThis is all about finding the vector product of the pairs of vectors representing the four edges of the tetrahedron, taken two at a time.
For instance if O is the point $\displaystyle (0,0,0)$, A is $\displaystyle (3, 0, 0)$ and B is $\displaystyle (2, 4, 0)$ then the edges OA and OB are represented by the vectors $\displaystyle \vec{a}=3\vec{i}$ and $\displaystyle \vec{b} = 2\vec{i} + 4\vec{j}$ respectively. The area of the face OAB is then $\displaystyle \tfrac{1}{2}\vec{a}\times\vec{b}$, and this vector is perpendicular to the plane OAB, in a direction given by the right-hand rule. But you need this vector to be pointing outwards from the tetrahedron, and, if you sketch a diagram, you'll see that it points inwards (in the positive direction of $\displaystyle \vec{k}$). So, in fact, the vector you need is $\displaystyle \tfrac{1}{2}\vec{b}\times\vec{a} $, which is given by
$\displaystyle \frac{1}{2}\begin{vmatrix}\vec{i} & \vec{j} & \vec{k}\\2 & 4 & 0\\3 & 0 & 0\end{vmatrix} = -6\vec{k}$
If you call C the point $\displaystyle (1, 3, 5)$, and $\displaystyle \vec{c} = \vec{i} + 3\vec{j}+ 5\vec{k}$, then you can find the vectors for the areas of the faces OAC and OBC in the same way.
To find the area of the face ABC, you'll first need vectors representing two of its edges. For instance $\displaystyle \vec{AB} = \vec{b} - \vec{a} = -\vec{i} + 4\vec{j}$, and $\displaystyle \vec{AC} = \vec{c} - \vec{a} = -2\vec{i} + 3\vec{j} + 5\vec{k}$.
For more on how this works see Cross product - Wikipedia, the free encyclopedia
Can you complete it OK now?
Grandad