Problem related to non-coplanar unit vectors,equally inclined to one another at an angle $\theta$ I've lately been facing lot of trouble in solving vector equations.Like the one below :

Let $a,b,c$ be non-coplanar unit vectors,equally inclined to one
  another at an angle $\theta$.If $a×b+b×c=pa+qb+rc$,find the scalars
  p,q and r in terms of $\theta$.

What would be the shortest method to solve this problem?
In my book they took nearly 2 pages!But I guess there might be a shorter method to solve such type of problems.What say ?
 A: First of all, observe that we can safely assume $0 < \theta < \frac{2}{3} \pi$. Indeed, it makes sense to consider $\theta$ to be the smallest (positive) angle between, say, $a$ and $b$. Furthermore, clearly for $\theta = 0$ we have $a = b = c$, while for $\theta = \frac{2}{3} \pi$ we have $a,b,c$ coplanar.
Now, recall that
$$
a \cdot (b \times c) = c \cdot (a\times b)
$$
coincides with the volume of the parallelepiped with sides defined by $a,b,c$, i.e.
$$
V = \sqrt{1 + 2 \cos^3\theta - 3 \cos^2\theta}.
$$
Since we're after a representation of a vector in the $\Bbb{R}^3$-basis $\{a,b,c\}$, let's consider the system of linear equations obtained by taking the dot product of $a \times b + b \times c = pa + qb + rc$ by $a,b$, and $c$, respectively:
$$
\begin{cases}
a \cdot (a \times b) + a \cdot (b \times c) = p(a \cdot a) + q(a \cdot b) + r(a \cdot c) \\
b \cdot (a \times b) + b \cdot (b \times c) = p(b \cdot a) + q(b \cdot b) + r(b \cdot c) \\
c \cdot (a \times b) + c \cdot (b \times c) = p(c \cdot a) + q(c \cdot b) + r(c \cdot c).
\end{cases}
$$
Recalling that $u \cdot (u \times v) = 0$ for every $u,v \in \Bbb{R}^3$, that $u \cdot u = 1$ for every unit vector $u$, and that $a \cdot b = b \cdot c = a \cdot c = \cos\theta$, this becomes
$$
\begin{cases}
p + q \cos\theta + r \cos\theta = V \\
p \cos\theta + q + r \cos\theta = 0 \\
p \cos\theta + q \cos\theta + r = V
\end{cases}
\tag{1} \label{eq:1}
$$
or, in matrix form
$$
A \mathbf{p} =
\begin{pmatrix}
1 & \cos\theta & \cos\theta \\
\cos\theta & 1 & \cos\theta \\
\cos\theta & \cos\theta & 1
\end{pmatrix}
\begin{pmatrix}
p \\
q \\
r
\end{pmatrix}
=
\begin{pmatrix}
V \\
0 \\
V
\end{pmatrix}.
$$
Interestingly, for $0 < \theta < \frac{2}{3} \pi$ we have that $\det A = V^2 > 0$ (try to see why), so $\eqref{eq:1}$ has exactly one solution, which is†
$$
\begin{pmatrix}
\dfrac{1-\cos\theta}{V},\;
\dfrac{2 (\cos\theta-1) \cos\theta}{V},\;
\dfrac{1-\cos\theta}{V}
\end{pmatrix}^T.
$$

† Since doing this by hand is a bit tedious, here's how to compute it in Mathematica
m = {{1, Cos[t], Cos[t]}, {Cos[t], 1, Cos[t]}, {Cos[t], Cos[t], 1}};
v = Sqrt[Det[m]];
p = Assuming[0 <= t <= 2/3 Pi, LinearSolve[m, {v, 0, v}]] // Simplify

