Find the suitable vector Consider $\lbrace \vec{B}_{1},\vec{B}_{2},\ldots,\vec{B}_{n}\rbrace$ be a set of $n>2$ three-dimensional non-unit vectors. My question is can I find a unit vector $\hat{A}$ such that,
$$\hat{A}.\vec{B}_{1}=\hat{A}.\vec{B}_{2}=\hat{A}.\vec{B}_{3}=\hat{A}.\vec{B}_{4}=\ldots=\hat{A}.\vec{B}_{n}$$
For $n=3$, I can easily obtain
$$\hat{A}=\frac{\vec{B}_{1}\times\vec{B}_{2}+\vec{B}_{2}\times\vec{B}_{3}+\vec{B}_{3}\times\vec{B}_{1}}{||\vec{B}_{1}\times\vec{B}_{2}+\vec{B}_{2}\times\vec{B}_{3}+\vec{B}_{3}\times\vec{B}_{1}||}$$
However, what I want is a general construction of $\hat{A}$ for arbitrary $n$.
 A: The requirement that $A$ has length 1 can be replaced by the requirement that $A$ is nonzero, because then if you have found a solution you can always divide by the norm, and it will remain a solution.
Define an $(n-1)\times n$ matrix $M$ with elements $M_{ij}=(B_{i+1}-B_i)_j$, $i=1,2,\ldots n-1$, $j=1,2,\ldots n$. Remove the $j_0$'th column to convert $M$ into a square matrix ${M}^{(j_0)}$ of size $(n-1)\times(n-1)$. Which column you remove does not matter, provided that the determinant of ${M}^{(j_0)}$ is nonzero. The elements of the removed column form a vector $v$ of length $n-1$. Solve the set of $n-1$ linear equations for the $n-1$ unknowns $a_j$,
$$\sum_{j=1}^{n-1}M_{ij}^{(j_0)}a_j=v_i,\;\;i=1,2,\ldots n-1.$$
The solution is given by Cramer's rule,
$$a_i=\frac{\det X^{(i)}}{\det M^{(j_0)}},$$
where the matrix $X^{(i)}$ is obtained by replacing the $i$-th column of $M^{(j_0)}$ by the column vector $v$.
Now the desired vector $A$ has elements $a_1,a_2,\ldots a_{j_0-1},-1,a_{j_0},\ldots a_{n-1}$. Divide by the norm to obtain a unit vector, and you're done.
