问题
若平面 $\pi_{1}$ 的表达式为 $A_{1} x$ $+$ $B_{1} y$ $+$ $C_{1} z$ $+$ $D_{1}$ $=$ $0$, 平面 $\pi_{2}$ 的表达式为 $A_{2} x$ $+$ $B_{2} y$ $+$ $C_{2} z$ $+$ $D_{2}$ $=$ $0$. 此外，平面 $\pi_{1}$ 和 $\pi_{2}$ 的法向量分别为 $\vec{n_{1}}$ $=$ $(A_{1}, B_{1}, C_{1})$ 和 $\vec{n_{2}}$ $=$ $(A_{1}, B_{2}, C_{2})$.那么，若 $\pi_{1}$ 与 $\pi_{2}$ 之间的夹角为 $\theta$ $(0 \leqslant \theta \leqslant \frac{\pi}{2})$, 则 $\cos \theta$ $=$ $?$
选项
[A]. $\cos \theta$ $=$ $\frac{|A_{1}A_{2} + B_{1}B_{2} + C_{1} C_{2}|}{(A_{1}^{2} + B_{1}^{2} + C_{1}^{2}) \times (A_{2}^{2} + B_{2}^{2} + C_{2}^{2})}$[B]. $\cos \theta$ $=$ $\frac{A_{1}A_{2} + B_{1}B_{2} + C_{1} C_{2}}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} \times \sqrt{A_{2}^{2} + B_{2}^{2} + C_{2}^{2}}}$
[C]. $\cos \theta$ $=$ $\frac{|A_{1}A_{2} + B_{1}B_{2} + C_{1} C_{2}|}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} \times \sqrt{A_{2}^{2} + B_{2}^{2} + C_{2}^{2}}}$
[D]. $\cos \theta$ $=$ $\frac{|A_{1}A_{2} – B_{1}B_{2} – C_{1} C_{2}|}{(A_{1}^{2} + B_{1}^{2} + C_{1}^{2}) \times (A_{2}^{2} + B_{2}^{2} + C_{2}^{2})}$
$\textcolor{red}{\cos \theta}$ $=$ $\frac{|\vec{n_{\textcolor{orange}{1}}} \cdot \vec{n_{\textcolor{cyan}{2}}}|}{|\vec{n_{\textcolor{orange}{1}}}| |\vec{n_{\textcolor{cyan}{2}}}|}$ $=$ $\frac{|A_{\textcolor{orange}{1}}A_{\textcolor{cyan}{2}} + B_{\textcolor{orange}{1}}B_{\textcolor{cyan}{2}} + C_{\textcolor{orange}{1}} C_{\textcolor{cyan}{2}}|}{\sqrt{A_{\textcolor{orange}{1}}^{2} + B_{\textcolor{orange}{1}}^{2} + C_{\textcolor{orange}{1}}^{2}} \times \sqrt{A_{\textcolor{cyan}{2}}^{2} + B_{\textcolor{cyan}{2}}^{2} + C_{\textcolor{cyan}{2}}^{2}}}$