Equation of plane

点法式方程

A (x - x_{0}) + B (y - y_{0}) + C (z - z_{0}) = 0

推导过程

称垂直于平面的非零向量为平面法向量，记为 $n = (A, B, C)$
在平面上找出不重合的两点 $M_{0} (x_{0}, y_{0}, z_{0}) and M (x, y, x)$ 连接 $M_{0} M = (x - x_{0}, y - y_{0}, z - z_{0})$
$由 M_{0} M ⊥ n \Rightarrow M_{0} M \cdot n = 0$

A (x - x_{0}) + B (y - y_{0}) + C (z - z_{0}) = 0

打开括号得到：

A x - A x_{0} + B y - B y_{0} + C z - C z_{0} = 0 (Let: D = - A x_{0} - B y_{0} - C z_{0}) \Rightarrow A x + B y + C z + D = 0

上面是平面的一般方程。

用法向量来判别。
设有两个平面：

Π_{1} : A_{1} x + B_{1} y + C_{1} z + D_{1} = 0 Π_{2} : A_{2} x + B_{2} y + C_{2} z + D_{2} = 0 则 n_{1} = (A_{1}, B_{1}, C_{1}); n_{2} = (A_{2}, B_{2}, C_{2})

平行：

Π_{1} // Π_{2} \Leftrightarrow n_{1} // n_{2} \Leftrightarrow \frac{A _{1}}{A _{2}} = \frac{B _{1}}{B _{2}} = \frac{C _{1}}{C _{2}}

垂直：

Π_{1} ⊥ Π_{2} \Leftrightarrow n_{1} ⊥ n_{2} \Leftrightarrow n_{1} \cdot n_{2} = 0 \Leftrightarrow A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2} = 0

相交：

\frac{A _{1}}{A _{2}} = \frac{B _{1}}{B _{2}} = \frac{C _{1}}{C _{2}} 不成立

重合（所有参数完全一样）：

\frac{A _{1}}{A _{2}} = \frac{B _{1}}{B _{2}} = \frac{C _{1}}{C _{2}} = \frac{D _{1}}{D _{2}}

就两平面的法向量的夹角（取值只有 $[0, \frac{π}{2}]$ ）

cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣∣ n _{2} ∣}

点 $M (x_{0}, y_{0}, z_{0})$ 到平面 $A x + B y + C z + D = 0$ 的距离公式

d = \frac{A x _{0} + B y _{0} + C z _{0}}{A ^{2} + B ^{2} + C ^{2}}

π_{1} : π_{2} : d = \frac{∣ D _{1} - D _{2} ∣}{A ^{2} + B ^{2} + C ^{2}} A x + B y + C z + D_{1} = 0 A x + B y + C z + D_{2} = 0

两个平面的交线

{A_{1} x + B_{1} y + C_{1} z + D_{1} = 0 A_{2} x + B_{2} y + C_{2} z + D_{2} = 0

{dot: (x_{0}, y_{0}, z_{0}) direction vector: s = (m, n, p) \frac{x - x _{0}}{m} = \frac{y - y _{0}}{n} = \frac{z - z _{0}}{p}

⎩ ⎨ ⎧ x = m t + x_{0} y = n t + y_{0} z = pt + z_{0} {dot: (x_{0}, y_{0}, z_{0}) direction vector: s = (m, n, p)

l_{1} : \frac{x - x _{1}}{m _{1}} = \frac{y - y _{1}}{n _{1}} = \frac{z - z _{1}}{p _{1}} l_{1} : {M (x_{1}, y_{1}, z_{1}) s_{1} = (m_{1}, n_{1}, p_{1}) l_{1} : \frac{x - x _{2}}{m _{2}} = \frac{y - y _{1}}{n _{2}} = \frac{z - z _{2}}{p _{2}} l_{2} : {M (x_{2}, y_{2}, z_{2}) s_{2} = (m_{2}, n_{2}, p_{2}) l_{1} // l_{2} \Leftrightarrow s_{1} // s_{2} \Leftrightarrow \frac{m _{1}}{m _{2}} = \frac{n _{1}}{n _{2}} = \frac{p _{1}}{p _{2}} l_{1} ⊥ l_{2} \Leftrightarrow s_{1} ⊥ s_{2} \Leftrightarrow s_{1} \cdot s_{2} = 0 cos < s_{1}, s_{2} >= \frac{s _{1} \cdot s _{2}}{∣ s _{1} ∣ \cdot ∣ s _{2} ∣}

求直线 $L : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z - 3}{1}$ 在平面 $Π : x - 3 y + 2 z - 5 = 0$ 上的投影直线

直线方向向量 s = (2, - 1, 1) 平面法向量 n = (1, - 3, 2) 点 P 在直线上 P (1, 0, 3) 令 t = \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z - 3}{1} 带入平面解得 t = - \frac{2}{7} 直线与平面的交点 Q (\frac{3}{7}, \frac{2}{7}, \frac{19}{7}) 设 R 在平面上 l_{p r} : \frac{x - 1}{1} = \frac{y}{- 3} = \frac{z - 3}{2} = t R (\frac{6}{7}, \frac{3}{7}, \frac{19}{7})