Are there any unit vector dot product identites?

Question

Given 3d unit vectors $\hat{r}_1,\hat{r}_2,\hat{r}_3,\hat{r}_4$, (i.e. where $\hat{r}_i\cdot \hat{r}_i=1$ and $\hat{r}_i \in \mathcal{R}^3$) are there any identities for tying together their dot products $\hat{r}_i\cdot \hat{r}_j$?

For example, something like $\hat{r}_1\cdot\hat{r}_2+\hat{r}_2\cdot\hat{r}_3 = \hat{r}_1\cdot\hat{r}_3$? (NOT true)

Context: I'm working on an under determined system of equations involving dot products between 4 or more unknown unit vectors. My gut feeling is I'm missing something in the geometry that would give me the last piece of the puzzle. Edit: I removed most of the context because it just distracted from the question and the answer (given below, based on Daniel's comment).

Answer 1

The determinant of the Gramian matrix is zero if the vectors are linearly dependent. Since any combination of 4 or more vectors in $\mathcal{R}^3$ are guaranteed to be linearly dependent, each such combination can give me an equation to work with.

For example, for unit vectors $\hat{r}_1,\hat{r}_2,\hat{r}_3, \hat{r}_4$ the Gramian matrix is: $$ \left( \begin{array}{cccc} 1 & \hat{r}_1.\hat{r}_2 & \hat{r}_1.\hat{r}_3 & \hat{r}_1.\hat{r}_4 \\ \hat{r}_2.\hat{r}_1 & 1 & \hat{r}_2.\hat{r}_3 & \hat{r}_2.\hat{r}_4 \\ \hat{r}_3.\hat{r}_1 & \hat{r}_3.\hat{r}_2 & 1 & \hat{r}_3.\hat{r}_4 \\ \hat{r}_4.\hat{r}_1 & \hat{r}_4.\hat{r}_2 & \hat{r}_4.\hat{r}_3 & 1 \\ \end{array} \right) $$

And setting the determinant to 0 gives:

$$ 0=-\hat{r}_1.\hat{r}_2 \hat{r}_2.\hat{r}_1+\hat{r}_1.\hat{r}_3 \hat{r}_3.\hat{r}_2 \hat{r}_2.\hat{r}_1+\hat{r}_1.\hat{r}_4 \hat{r}_4.\hat{r}_2 \hat{r}_2.\hat{r}_1-\hat{r}_1.\hat{r}_3 \hat{r}_3.\hat{r}_4 \hat{r}_4.\hat{r}_2 \hat{r}_2.\hat{r}_1-\hat{r}_1.\hat{r}_4 \hat{r}_3.\hat{r}_2 \hat{r}_4.\hat{r}_3 \hat{r}_2.\hat{r}_1+\hat{r}_1.\hat{r}_2 \hat{r}_3.\hat{r}_4 \hat{r}_4.\hat{r}_3 \hat{r}_2.\hat{r}_1-\hat{r}_1.\hat{r}_3 \hat{r}_3.\hat{r}_1+\hat{r}_1.\hat{r}_2 \hat{r}_2.\hat{r}_3 \hat{r}_3.\hat{r}_1-\hat{r}_2.\hat{r}_3 \hat{r}_3.\hat{r}_2-\hat{r}_1.\hat{r}_4 \hat{r}_4.\hat{r}_1+\hat{r}_1.\hat{r}_2 \hat{r}_2.\hat{r}_4 \hat{r}_4.\hat{r}_1+\hat{r}_1.\hat{r}_4 \hat{r}_2.\hat{r}_3 \hat{r}_3.\hat{r}_2 \hat{r}_4.\hat{r}_1-\hat{r}_1.\hat{r}_3 \hat{r}_2.\hat{r}_4 \hat{r}_3.\hat{r}_2 \hat{r}_4.\hat{r}_1+\hat{r}_1.\hat{r}_3 \hat{r}_3.\hat{r}_4 \hat{r}_4.\hat{r}_1-\hat{r}_1.\hat{r}_2 \hat{r}_2.\hat{r}_3 \hat{r}_3.\hat{r}_4 \hat{r}_4.\hat{r}_1-\hat{r}_2.\hat{r}_4 \hat{r}_4.\hat{r}_2-\hat{r}_1.\hat{r}_4 \hat{r}_2.\hat{r}_3 \hat{r}_3.\hat{r}_1 \hat{r}_4.\hat{r}_2+\hat{r}_1.\hat{r}_3 \hat{r}_2.\hat{r}_4 \hat{r}_3.\hat{r}_1 \hat{r}_4.\hat{r}_2+\hat{r}_2.\hat{r}_3 \hat{r}_3.\hat{r}_4 \hat{r}_4.\hat{r}_2+\hat{r}_1.\hat{r}_4 \hat{r}_3.\hat{r}_1 \hat{r}_4.\hat{r}_3-\hat{r}_1.\hat{r}_2 \hat{r}_2.\hat{r}_4 \hat{r}_3.\hat{r}_1 \hat{r}_4.\hat{r}_3+\hat{r}_2.\hat{r}_4 \hat{r}_3.\hat{r}_2 \hat{r}_4.\hat{r}_3-\hat{r}_3.\hat{r}_4 \hat{r}_4.\hat{r}_3+1 $$

Thanks to Daniel for pointing out this property in the comments. I think this property captures exactly what I was intuitively grasping for.

FWIW doing this on Mathematica was fairly easy:

g=Outer[Dot,{r1,r2,r3,r4},{r1,r2,r3,r4}]; g //MatrixForm
reps = {r1.r1->1, r2.r2->1, r3.r3->1,r4.r4->1,r5.r5->1, r1->Subscript[ OverHat[r],1], r2->Subscript[ OverHat[r],2], r3->Subscript[ OverHat[r],3], r4->Subscript[ OverHat[r],4], r5->Subscript[ OverHat[r],5]}
Det[g]/.reps
%//TeXForm
g  /.reps //MatrixForm //TeXForm