问题
已知一条空间直线由两平面相交形成,且这两个平面的法向量分别为 $\vec{n_{1}}$ $=$ $(A_{1}, B_{1}, C_{1})$ 和 $\vec{n_{2}}$ $=$ $(A_{2}, B_{2}, C_{2})$, 则该直线的方向向量 $\vec{s}$ $=$ $?$选项
[A]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{1}{A_{1}} & \frac{1}{B_{1}} & \frac{1}{C_{1}} \\ \frac{1}{A_{2}} & \frac{1}{B_{2}} & \frac{1}{C_{2}} \end{vmatrix}$[B]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_{1} & B_{1} & C_{1} \end{vmatrix}$
[C]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2} \end{vmatrix}$
[D]. $\vec{s}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ C_{1} & B_{1} & A_{1} \\ C_{2} & B_{2} & A_{2} \end{vmatrix}$
$\textcolor{yellow}{\vec{s}}$ $=$ $\begin{vmatrix} \mathbf{\textcolor{orange}{i}} & \mathbf{\textcolor{orange}{j}} & \mathbf{\textcolor{orange}{k}} \\ A_{\textcolor{red}{1}} & B_{\textcolor{red}{1}} & C_{\textcolor{red}{1}} \\ A_{\textcolor{cyan}{2}} & B_{\textcolor{cyan}{2}} & C_{\textcolor{cyan}{2}} \end{vmatrix}$
其中,向量 $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ 分别为 $x$ 轴、$y$ 轴和 $z$ 轴上的单位向量.