Find the unknown angle Suppose I have $4$ unit vectors in $3$D and I know all the $^4C_2=6$ angles between them. These angles provide the complete description of this group of vectors. Now, I want to add anther unit vector to the mix. How many additional angles do I need to uniquely identify this new vector?
I'm  guessing that I need just $3$ more vectors since its in $3$D space. But what about the remaining angle? How do I represent that one?
 A: You need to pick two vectors $V_1$ and $V_2$ from the four vectors that you have, and specify angles of the added vector (the fifth vector) from these two vectors.  There can be two vectors making $\theta_1$ with $V_1$ and $\theta_2$ with $V_2$.  So you just need one more angle with a third vector $V_3$ to uniquely specify your added vector, but keep in mind that this third angle can take only two specific values.
As an explicit numerical example, suppose
$V_1 = (1, 0, 0) $
$V_2 = (0, 1, 1)/\sqrt{2} $
$V_3 = (2, 0, -1)/\sqrt{5} $
$V_4 = (1, 1, 1)/\sqrt{3} $
Now suppose I have a fifth unit vector $V_5 = (1, 2, 3)/\sqrt{14} $
That I want to uniquely identify using angles from the given vectors.
We note that
$V_1 \cdot V_5 = \dfrac{1}{\sqrt{14}} = \cos \theta_1 $
and
$V_2 \cdot V_5 = \dfrac{5}{\sqrt{28}} = \cos \theta_2$
and
$V_3 \cdot V_5 = \dfrac{-1}{\sqrt{70}} =\cos \theta_3$
And now we want to go from the $\theta_1, \theta_2, \theta_3$ to the vector $V_5$.
So we'll let $V_5 = a V_1 + b V_2 + c V_3 $ then using the dot product between $V_5$ and $V_1, V_2, V_3$ we'll have the linear system in the cosines of the angles between $V_5$ and $V_1, V_2, V_3$, as follows
$ \begin{bmatrix} V_1 \cdot V_5 \\ V_2 \cdot V_5 \\ V_3 \cdot V_5 \end{bmatrix} = \begin{bmatrix} 1 && V_1 \cdot V_2 && V_1 \cdot V_3 \\ V_2 \cdot V_1 && 1 && V_2 \cdot V_3 \\ V_3 \cdot V_1 && V_3 \cdot V_2 && 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} $
Substituting the values in the above matrix equation we arrive at
$\displaystyle \begin{bmatrix} \dfrac{1}{\sqrt{14}} \\ \dfrac{5}{\sqrt{28}} \\ 
 \dfrac{-1}{\sqrt{70}} \end{bmatrix} = \begin{bmatrix} 1 && 0 && \dfrac{2}{\sqrt{5}} \\ 0 && 1 && -\dfrac{1}{\sqrt{10}} \\ \dfrac{2}{\sqrt{5}} && -\dfrac{1}{\sqrt{10}} && 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} $
Solving this system numerically, gives us
$ a = 0.801784, b = 0.755929, c = -0.59761 $
Now using
$ V_5 = a V_1 + b V_2 + c V_3 $
gives us
$\begin{equation} \begin{split}
V_5 &= (0.267261, 0.534522, 0.801784) \\
&= (0.267261) ( 1, 2, 3 ) \\
&= (1,2,3) / \sqrt{14} \end{split}\end{equation} $
