本文介绍了如何在Python中从一组线性方程式绘制平面?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个包含三个方程的线性系统:
I have a linear system with three equations:
解集为(29,16,3),这是这些平面的交点.
The solution set is (29, 16, 3), which is a point at the intersection of these planes.
希望有人可以使用Matplotlib在3D空间中绘制这些平面,以便清楚地可视化问题.
Hoping if anyone can plot these planes in a 3D-space using Matplotlib to visualize the problem clearly.
推荐答案
您的第三个等式说:
-4x + 5y + 9z-9 = 0
-4x + 5y + 9z - 9 = 0
或者一般来说,您的方程式是
or in general an equation of yours is
a x + b y + c z + d = 0
a x + b y + c z + d = 0
正常是(a,b,c)
- 如果a不为0,则平面上的点为(-d/a,0,0)
- 如果b不为0,则平面上的点为(0,-d/b,0)
- 如果c不为0,则平面上的点为(0,0,-d/c)
您可以将其插入到绘制库中,该库中包含法向矢量和平面上的一个点,并执行3次(每个平面一个).
You plug this into the plotting library that takes a normal vector and a point on the plane, and do this 3 times (one for each plane).
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
point1 = np.array([0,0,0])
normal1 = np.array([1,-2,1])
point2 = np.array([0,-4,0])
normal2 = np.array([0,2,-8])
point3 = np.array([0,0,1])
normal3 = np.array([-4,5,9])
# a plane is a*x+b*y+c*z+d=0
# [a,b,c] is the normal. Thus, we have to calculate
# d and we're set
d1 = -np.sum(point1*normal1)# dot product
d2 = -np.sum(point2*normal2)# dot product
d3 = -np.sum(point3*normal3)# dot product
# create x,y
xx, yy = np.meshgrid(range(30), range(30))
# calculate corresponding z
z1 = (-normal1[0]*xx - normal1[1]*yy - d1)*1./normal1[2]
z2 = (-normal2[0]*xx - normal2[1]*yy - d2)*1./normal2[2]
z3 = (-normal3[0]*xx - normal3[1]*yy - d3)*1./normal3[2]
# plot the surface
plt3d = plt.figure().gca(projection='3d')
plt3d.plot_surface(xx,yy,z1, color='blue')
plt3d.plot_surface(xx,yy,z2, color='yellow')
plt3d.plot_surface(xx,yy,z3, color='cyan')
plt.show()
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