※ 引述《spss520 (剩下一個月啦)》之銘言： : Use Lagranges multiplier to prove that the minimum distance from a : point (X_0,Y_0,Z_0) to a plane ax+by+cz+d=0 : is ∣a(X_0)+b(Y_0)+c(Z_0)+d∣ : ___________________________ : _______________ : √a^2+b^2+c^2 : 呵呵～上次打錯了 : 更正囉！！ : 謝謝大家 pf: Let f(x,y,z) = (x - (x_0))^2 + (y - (y_0))^2 + (z - (z_0))^2 subject to ax + by + cz + d = 0 F(x,y,z) = (x - (x_0))^2 + (y - (y_0))^2 + (z - (z_0))^2 + (λ)(ax + by + cz + d) Fx = 2(x - (x_0)) + (aλ) = 0 --------(1) Fy = 2(y - (y_0)) + (bλ) = 0 --------(2) Fz = 2(z - (z_0)) + (cλ) = 0 --------(3) ax + by + cz + d = 0 --------(4) (1)*(a) => 2a(x - (x_0)) + (a^2)(λ) = 0 ------(5) (2)*(b) => 2b(x - (x_0)) + (b^2)(λ) = 0 ------(6) (3)*(c) => 2c(x - (x_0)) + (c^2)(λ) = 0 ------(7) (5) + (6) + (7) (2)(ax + by + cz - ((a)(x_0) + (b)(y_0) + (c)(z_0))) + (a^2 + b^2 + c^2)(λ) = 0 (2)((a)(x_0) + (b)(y_0) + (c)(z_0) + d) => λ = --------------------------------------- a^2 + b^2 + c^2 Suppose we can get (x_1 , y_1 , z_1) minimum value of f(x,y,z) is f(x_1 , y_1 , z_1) = ((x_1) - (x_0))^2 + ((y_1) - (y_0))^2 + ((z_1) - (z_0))^2 1 1 1 = (---)(a^2)(λ^2) + (---)(b^2)(λ^2) + (---)(c^2)(λ^2) 4 4 4 1 = (---)(λ^2)(a^2 + b^2 + c^2) 4 1 (4)(((a)(x_0) + (b)(y_0) + (c)(z_0) + d)^2) = (---)(a^2 + b^2 + c^2)(-------------------------------------------) 4 (a^2 + b^2 + c^2)^2 ((a)(x_0) + (b)(y_0) + (c)(z_0) + d)^2 = --------------------------------------- a^2 + b^2 + c^2 Therefore , Shortest distance is |(a)(x_0) + (b)(y_0) + (c)(z_0) + d| √(f((x_1),(y_1),(z_1))) = --------------------------------------- √(a^2 + b^2 + c^2) --

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