试试写个具体计算表达式出来,方便以后用。
$\newcommand{\relph}[1]{\mathrel{\phantom{#1}}}$
接着3#,有
\[\text{二面角}~P\text{-}AB\text{-}Q=\bigl\langle\vv m,\vv n\bigr\rangle=\arccos\frac{\vv m\cdot\vv n}{\bigl|\vv m\bigr|\cdot\bigl|\vv n\bigr|},\]
设 $A(x_A,y_A,z_A)$, $B(x_B,y_B,z_B)$, $P(x_P,y_P,z_P)$, $Q(x_Q,y_Q,z_Q)$,则有
\begin{align*}
\vv m\cdot\vv n&=\bigl(\vv{PA}\times\vv{PB}\bigr)\cdot\bigl(\vv{QA}\times\vv{QB}\bigr)\\
&=\begin{vmatrix}
\vv{PA}\cdot\vv{QA} & \vv{PA}\cdot\vv{QB}\\
\vv{PB}\cdot\vv{QA} & \vv{PB}\cdot\vv{QB}\\
\end{vmatrix}\\
&=\sum_{x,y,z}(x_P-x_A)(x_Q-x_A)\sum_{x,y,z}(x_P-x_B)(x_Q-x_B)\\
&\relph{=}{}-\sum_{x,y,z}(x_P-x_A)(x_Q-x_B)\sum_{x,y,z}(x_P-x_B)(x_Q-x_A),
\end{align*}
以及
\begin{align*}
\bigl|\vv m\bigr|&=\sqrt{\begin{vmatrix}
y_P-y_A & z_P-z_A\\
y_P-y_B & z_P-z_B
\end{vmatrix}^2+\begin{vmatrix}
z_P-z_A & x_P-x_A\\
z_P-z_B & x_P-x_B
\end{vmatrix}^2+\begin{vmatrix}
x_P-x_A & y_P-y_A\\
x_P-x_B & y_P-y_B
\end{vmatrix}^2},\\
\bigl|\vv n\bigr|&=\sqrt{\begin{vmatrix}
y_Q-y_A & z_Q-z_A\\
y_Q-y_B & z_Q-z_B
\end{vmatrix}^2+\begin{vmatrix}
z_Q-z_A & x_Q-x_A\\
z_Q-z_B & x_Q-x_B
\end{vmatrix}^2+\begin{vmatrix}
x_Q-x_A & y_Q-y_A\\
x_Q-x_B & y_Q-y_B
\end{vmatrix}^2},
\end{align*}
代入…… 还是算了吧……
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