Proving linear independence about unit outward normals of a pyramid In proving that every pyramid (4 faces) admits a unique inscribed ball, I encountered the following problem:

Given any pyramid whose unit outward normal vectors are $n_{1,2,3,4}$, prove that the set of vectors $\{n_1-n_{2,3,4}\}$ are linearly independent. 

This turns out surprisingly complicated. I feel we will have to do some inner product calculation, since if their lengths were flexible, the claim would be clearly false. So I suppose
$$\sum_{j=2}^4 \alpha_j (n_1 - n_j) = 0$$ and end up with 
$$n_1 =\frac{\sum_j \alpha_j n_j}{\sum_j\alpha_j}\tag{*}$$
But after performing inner product (with themselves) on both sides, I get little information. 
One fact I'm aware of: any three of $\{n_{1,2,3,4}\}$ are linearly independent, since three faces intersect at a single point iff their outward normals are independent. 
Could anybody give any insight? Thanks!

I attempted another approach to simplifying this problem. Since $n_{2,3,4}$ form a basis in $\Bbb R^3$, there exist unique $\lambda_{2,3,4}$ such that 
$$n_1 = \sum_{j=2}^4\lambda_j n_j.$$
But from $(*)$ we also must have that $\sum_{j=2}^4\lambda_j = 1$. Therefore, it suffices to show the following system of linear equations has no solution
$$\begin{bmatrix}
n_2 & n_3 & n_4 \\
1 & 1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
\lambda_2 \\ \lambda_3 \\ \lambda_4
\end{bmatrix} = 
\begin{bmatrix}
n_1 \\ 1
\end{bmatrix}$$
which again amounts to showing that
$$A:=\begin{bmatrix}
n_1 & n_2 & n_3 & n_4\\
1 & 1 & 1 & 1
\end{bmatrix}$$
has full rank, i.e. rank 4 which is greater than the coefficient matrix's rank 3. Givien that each $n_{1,2,3,4}$ is unit length, and any three of them are independent, is there any easy way to show $A$ is full rank? Thanks!
 A: I will write out my take on this problem as a series of hints. Currently is it incomplete. This proof is admittedly a bit long, but it is does not take any hand-wavy approaches to steps. There may be a much better way to approach this problem. You can stop reading at any time if you feel you know how to proceed.

Hint 1: Use the Inner Product
As you stated, we wish to show that
$$a_1(n_1-n_2)+a_2(n_1-n_3)+a_3(n_1-n_4)=0$$
has no nontrivial solution. We can rearrange this to
$$(a_1+a_2+a_3)n_1=a_1n_2+a_2n_3+a_3n_4$$
Assume we have a nontrivial solution to this. That is, not all $a_1,a_2,a_3=0$. As you suggested, the inner product may be able to help us here. We will assume these normal vectors have unit length for simplicity. Note that
$$n_i\cdot n_j=||n_i||\cdot||n_j||\cos\theta=\cos\theta\leq1$$
How could we use this to simplify the previous equation?

Hint 2: Simplifying the Equation
The inner products tell us
$$(a_1+a_2+a_3)(n_1\cdot n_1)=a_1(n_2\cdot n_1)+a_2(n_3\cdot n_1)+a_3(n_4\cdot n_1)$$
$$a_1+a_2+a_3=a_1\cos\theta_2+a_2\cos\theta_3+a_3\cos\theta_4$$
$$a_1(1-\cos\theta_2)+a_2(1-\cos\theta_3)+a_3(1-\cos\theta_4)=0$$
Why does this imply linear independence? At this point, you could wave your hands and say this equation can't be true in general. That would result in a much shorter proof. However, I will prove that below.

Hint 3: Restriction on the Inner Product
Say we orient the pyramid so it has a "bottom" flush with the $x-y$ plane, and let that face have normal vector $n_1$. Logically, it will be pointing straight down, and at least one other $n_{2,3,4}$ will be pointing somewhat up.  Call it $n_2$. Thus, we then have
$$n_1\cdot n_2=\cos\theta_2<0$$
Since $\theta_2>\pi/2$. We must now split into cases:
Case A: One $n_{3,4}$ is pointing up, and the other down.
Case B: All $n_{2,3,4}$ are pointing up.
In case A, we have, for simplicity, $\theta_3>\pi/2$ and $\theta_4<\pi/2$
In case B, we have all $\theta\leq\pi/2$.
What does this tell us about our simplified equation?

Hint 4: Multiple Cases
Note that, if $\cos\theta_j\leq0$,
$$|a_i(1-\cos\theta_j)|\geq|a_i|$$
If $\cos\theta_j>0$
$$|a_i(1-\cos\theta_j)|<|a_i|$$
Note that the equation from hint 2 is equivalent to
$$|a_1(1-\cos\theta_2)+a_2(1-\cos\theta_3)+a_2(1-\cos\theta_4)|=0$$
There are two possible cases here:


*

*One or two of these $a_i<0$.

*All or none of these $a_i<0$.


You should see why I grouped these conditions like this. Can either of these cases be true?

Hint 5: Neither Case is Correct
For case B2, we break up the above equation to see
$$0=|a_1(1-\cos\theta_2)|+|a_2(1-\cos\theta_3)|+|a_3(1-\cos\theta_4)|\geq|a_1|+|a_2|+|a_3|$$
The right side is a positive number, but we have it less than or equal to $0$. So case B2 can't hold. I do not currently have time to handle the other three cases, hopefully you can work them out.

Wrapping Up
We've now exhausted all possible routes. Everything we can do has ended in a contradiction. Thus, the only possibility is that our initial non-trivial assumption was false. That is, we must have $a_1=a_2=a_3=0$, showing that $\{n_1-n_{2,3,4}\}$ are LI.
A: Finally, I figured out this relatively easy approach. From $(*)$ we know that if $\{n_1 - n_{2,3,4}\}$ are dependent then we must have $\lambda_{j}\in\Bbb R,\,j=2,3,4$ such that $\sum_j\lambda_j=1$ and $n_1 = \sum_j\lambda_j n_j$.
We put the starting points of all $n_i$s to the origin $O$. Then the endpoints of $n_{2,3,4}$ must form a plane located away from the origin, since they form a basis. Now since $\sum_j\lambda_j=1$, the endpoint of $n_1$ lies on this plane too. Let $n$ be the unit normal of this plane that points away from the origin, then if we denote the distance of this plane to the origin by $d$, we have
$$0<d = n\cdot n_i,\quad i=1,2,3,4.$$
However, as a result of Gauss-Ostrogradski theorem (aka divergence theorem), we must have 
$$\sum_{i=1}^4 S_i n_i = 0 \tag{**}$$
where $S_i>0$ denote areas of the faces corresponding to $n_i$ respectively. Now dot multiply $n$ on both sides of $(**)$ to get a contradiction.
