问题
若直线 $L_{1}$ 的表达式为 $\frac{x-x_{1}}{l_{1}}$ $=$ $\frac{y-y_{1}}{m_{1}}$ $=$ $\frac{z-z_{1}}{n_{1}}$, 直线 $L_{2}$ 的表达式为 $\frac{x-x_{2}}{l_{2}}$ $=$ $\frac{y-y_{2}}{m_{2}}$ $=$ $\frac{z-z_{2}}{n_{2}}$. 此外，直线 $L_{1}$ 和 $L_{2}$ 的方向向量分别为 $\vec{s_{1}}$ $=$ $(l_{1}, m_{1}, n_{1})$ 和 $\vec{s_{2}}$ $=$ $(l_{2}, m_{2}, n_{2})$.那么，若 $L_{1}$ 与 $L_{2}$ 之间的夹角为 $\theta$, 且 $(0 \leqslant \theta \leqslant \frac{\pi}{2})$, 则 $\cos \theta$ $=$ $?$
选项
[A]. $\cos \theta$ $=$ $\frac{|l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}|}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \times \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$[B]. $\cos \theta$ $=$ $\frac{|l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}|}{\sqrt{l_{1} + m_{1} + n_{1}} \times \sqrt{l_{2} + m_{2} + n_{2}}}$
[C]. $\cos \theta$ $=$ $\frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) \times (l_{2}^{2} + m_{2}^{2} + n_{2}^{2})}$
[D]. $\cos \theta$ $=$ $\frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \times \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$
$\textcolor{red}{\cos \theta}$ $=$ $\frac{\textcolor{yellow}{|}\vec{s_{\textcolor{orange}{1}}} \cdot \vec{s_{\textcolor{cyan}{2}}}\textcolor{yellow}{|}}{\textcolor{yellow}{|}\vec{s_{\textcolor{orange}{1}}}\textcolor{yellow}{|} \textcolor{yellow}{|}\vec{s_{\textcolor{cyan}{2}}}\textcolor{yellow}{|}}$ $=$ $\frac{\textcolor{yellow}{|}l_{\textcolor{orange}{1}} l_{\textcolor{cyan}{2}} + m_{\textcolor{orange}{1}} m_{\textcolor{cyan}{2}} + n_{\textcolor{orange}{1}} n_{\textcolor{cyan}{2}}\textcolor{yellow}{|}}{\sqrt{l_{\textcolor{orange}{1}}^{2} + m_{\textcolor{orange}{1}}^{2} + n_{\textcolor{orange}{1}}^{2}} \times \sqrt{l_{\textcolor{cyan}{2}}^{2} + m_{\textcolor{cyan}{2}}^{2} + n_{\textcolor{cyan}{2}}^{2}}}$