# Determining the vectors of a tetrahedron

Suppose that a regular tetrahedron with vertices $A$, $B$, $C$, and $D$ has its centroid at the origin $O$, as in the below schematic. Vectors $OA$, $OB$, $OC$, and $OD$ each have length $\ell$ ($|OA| = |OB| = |OC| = |OD| = \ell$). I wish to determine the four Cartesian vectors $OA$, $OB$, $OC$, and $OD$, given the axes shown in the figure. How can I do this?

Vector $OA$ is clearly $OA = (0, 0, \ell)$.

Vector $OB$ has no $x$ component. From chemistry (see, for example, this Wikipedia article), I know that the vectors make angles of $\cos^{-1}(-1/3) \approx 109.471^{\circ}$ with respect to one another. How should I determine $OB$, $OC$, and $OD$? Thanks for your time.

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You've made a good start. I'll assume that the diagram was meant to say "$x$ out of page".

To get $\angle AOB=\arccos(-1/3)$, we need

$$B=\ell\left(0,-\sin\arccos\left(-\frac13\right),\cos\arccos\left(-\frac13\right)\right)=\ell\left(0,-\sqrt{\frac89},-\frac13\right)\;.$$

Now since $B$, $C$ and $D$ have the same $z$ coordinate, we want $C=(x,y,-\ell/3)$, $D=(-x,y,-\ell/3)$ such that $\angle DOC=\arccos\left(-\frac13\right)$ and $x^2+y^2+\frac19\ell^2=\ell^2$. From $\angle DOC$ we get $(-x^2+y^2+\frac19\ell^2)/(x^2+y^2+\frac19\ell^2)=-1/3$, and thus $-x^2+2y^2+\frac29\ell^2=0$. Adding that to $x^2+y^2+\frac19\ell^2=\ell^2$ yields $3y^2+\frac13\ell^2=\ell^2$ and thus $y=\frac{\sqrt2}3\ell$, and then $x=\sqrt\frac23\ell$. So the points you want are

\begin{align} A&=\ell(0,0,1)\;,\\ B&=\ell\left(0,-\sqrt{\frac89},-\frac13\right)\;,\\ C&=\ell\left(\sqrt\frac23,\frac{\sqrt2}3,-\frac13\right)\;,\\ D&=\ell\left(-\sqrt\frac23,\frac{\sqrt2}3,-\frac13\right)\;. \end{align}

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Thanks! I've fixed my figure, with $x$ indeed out of the page. –  Andrew Nov 7 '12 at 23:37
@Andrew: You may verify that $A,B,C,D$ are of length $\ell$, and the dot product between any two of them is $-1/3$ when $\ell=1$. –  user1551 Nov 7 '12 at 23:39