# 线面交点

1. 没有交点；
2. 有且只有一个交点；
3. 有无数个交点。

## 代数形式

${\displaystyle (\mathbf {p} -\mathbf {p_{0}} )\cdot \mathbf {n} =0}$

${\displaystyle \mathbf {p} =d\mathbf {l} +\mathbf {l_{0}} \quad d\in \mathbb {R} }$

${\displaystyle (d\mathbf {l} +\mathbf {l_{0}} -\mathbf {p_{0}} )\cdot \mathbf {n} =0}$

${\displaystyle d\mathbf {l} \cdot \mathbf {n} +(\mathbf {l_{0}} -\mathbf {p_{0}} )\cdot \mathbf {n} =0}$

${\displaystyle d={(\mathbf {p_{0}} -\mathbf {l_{0}} )\cdot \mathbf {n} \over \mathbf {l} \cdot \mathbf {n} }.}$

${\displaystyle \mathbf {l} \cdot \mathbf {n} =0}$，则直线与平面平行。此时，如果(${\displaystyle \mathbf {p_{0}} -\mathbf {l_{0}} )\cdot \mathbf {n} =0}$，则该直线在平面内，即直线上所有的点都是交点。否则，直线与平面没有交点。

${\displaystyle \mathbf {l} \cdot \mathbf {n} \neq 0}$，则直线与平面有且只有一个交点。解得 ${\displaystyle d}$，则交点的坐标为

${\displaystyle d\mathbf {l} +\mathbf {l_{0}} }$.

## 参数形式

${\displaystyle \mathbf {l} _{a}+(\mathbf {l} _{b}-\mathbf {l} _{a})t,\quad t\in \mathbb {R} }$

${\displaystyle \mathbf {p} _{0}+(\mathbf {p} _{1}-\mathbf {p} _{0})u+(\mathbf {p} _{2}-\mathbf {p} _{0})v,\quad u,v\in \mathbb {R} }$

${\displaystyle \mathbf {l} _{a}+(\mathbf {l} _{b}-\mathbf {l} _{a})t=\mathbf {p} _{0}+(\mathbf {p} _{1}-\mathbf {p} _{0})u+(\mathbf {p} _{2}-\mathbf {p} _{0})v}$

${\displaystyle \mathbf {l} _{a}-\mathbf {p} _{0}=(\mathbf {l} _{a}-\mathbf {l} _{b})t+(\mathbf {p} _{1}-\mathbf {p} _{0})u+(\mathbf {p} _{2}-\mathbf {p} _{0})v,}$

${\displaystyle {\begin{bmatrix}x_{a}-x_{0}\\y_{a}-y_{0}\\z_{a}-z_{0}\end{bmatrix}}={\begin{bmatrix}x_{a}-x_{b}&x_{1}-x_{0}&x_{2}-x_{0}\\y_{a}-y_{b}&y_{1}-y_{0}&y_{2}-y_{0}\\z_{a}-z_{b}&z_{1}-z_{0}&z_{2}-z_{0}\end{bmatrix}}{\begin{bmatrix}t\\u\\v\end{bmatrix}}}$

${\displaystyle \mathbf {l} _{a}+(\mathbf {l} _{b}-\mathbf {l} _{a})t}$

${\displaystyle u,v\in [0,1],\;\;\;(u+v)\leq 1,}$

${\displaystyle {\begin{bmatrix}t\\u\\v\end{bmatrix}}={\begin{bmatrix}x_{a}-x_{b}&x_{1}-x_{0}&x_{2}-x_{0}\\y_{a}-y_{b}&y_{1}-y_{0}&y_{2}-y_{0}\\z_{a}-z_{b}&z_{1}-z_{0}&z_{2}-z_{0}\end{bmatrix}}^{-1}{\begin{bmatrix}x_{a}-x_{0}\\y_{a}-y_{0}\\z_{a}-z_{0}\end{bmatrix}}.}$