思路：用已知的三维点坐标拟合一个平面
- 三维平面方程
cos α ⋅ X i + cos β ⋅ Y i + cos γ ⋅ Z i + d = 0 \cos \alpha \cdot X_i+\cos \beta \cdot Y_i+\cos \gamma \cdot Z_i +d=0 cosα⋅Xi+cosβ⋅Yi+cosγ⋅Zi+d=0 - 点到直线距离
d i s t a n c e = ∣ A x 0 + B y 0 + C z 0 + D ∣ A 2 + B 2 + C 2 distance=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}} distance=A2+B2+C2∣Ax0+By0+Cz0+D∣ - 优化的目标函数
d i = ∣ cos α ⋅ X i + cos β ⋅ Y i + cos γ ⋅ Z i − d ∣ d_i = |\cos\alpha\cdot X_i+\cos\beta\cdot Y_i+\cos \gamma \cdot Z_i-d| di=∣cosα⋅Xi+cosβ⋅Yi+cosγ⋅Zi−d∣
约束条件： cos 2 α + cos 2 β + cos 2 γ = 1 \cos^2\alpha+\cos^2 \beta+\cos^2\gamma=1 cos2α+cos2β+cos2γ=1
拉格朗日对偶问题：
f = ∑ n = 0 N d i 2 − λ ⋅ ( cos 2 α + cos 2 β + cos 2 γ − 1 ) f=\sum_{n=0}^N d_i^2-\lambda\cdot(\cos^2\alpha+\cos^2\beta+\cos^2\gamma-1) f=∑n=0Ndi2−λ⋅(cos2α+cos2β+cos2γ−1) - 求目标函数f的极小值
∂ f ∂ d i = 2 ⋅ ∑ n = 0 N d i = 2 ∑ n = 0 N ∣ cos α ⋅ X i + cos β ⋅ Y