# Describe the Tetrahedron that use those 4 points

I want to describe the tetrahedron that use those 4 points $$A(1,2,3) , B(-1,2,0) , C(1,0,1) , D(0,1,1)$$

So I figured by drawing it down that I have to make vectors between each points

$$\vec V1=\vec {OA} - \vec {OB} , \vec V2=\vec {OA} - \vec {OC}, \vec V3=\vec {OB} - \vec {OC}$$

$$\vec V4=\vec {OB} - \vec {OD} , \vec V5=\vec {OC} - \vec {OD}$$

$$\vec V6=\vec {OA} - \vec {OC} - \vec {OD}$$ $$\vec V7=\vec {OB} - \vec {OA} - \vec {OD}$$ $$\vec V8=\vec {OA} - \vec {OC} - \vec {OB}$$

not quit sure about the last ones.

Then we have $$\vec V1 = [0, 0, -3]$$ $$\vec V2 = [0, -2, -2]$$

I am wondering if

1. I'm doing the right thing
2. Why do we have to use - between points instead of + to form a vector ?
-
What do you mean by describe? Do you mean you want to find the vectors that make up the sides of the tetrahedron? – Jeremy Oct 10 '12 at 3:39

A tetrahedron with vertices $\bf{a,b,c,d}$ has volume abs$(\frac{1}6|\bf{(a-b)(b-c))(c-d)}|)$. In this case $V=\left| \begin{array}{ccc} 2&-2&1\\ 0&2&-1\\ 3 &-1&0 \end{array} \right|=\frac{1}{3}$

Centroid is at $\bf{\frac{a+b+c+d}4}$

Surface area $=\frac{1}2\bf{|(a-b)\times(b-c)|+|(b-c)\times(c-d)|+|(c-d)\times(d-a)|+|(d-a)\times(a-b)|}=\frac{2\sqrt{11}+2\sqrt{3}+\sqrt{5}+\sqrt{14}}2$

Using - gives the difference between the vertices, whereas + gives combined distances from the origin.

-

I don't understand what you are trying to do, but maybe I can help with your second question by suggesting you look at something simpler. Consider (in fact: draw) the points $A=(1,0)$ and $B=(0,1)$. If you add them, you get $(1,1)$; if you subtract, $(1,-1)$. Can you see why one of these has something to do with the line segment joining $B$ to $A$, and the other one doesn't?

-