Plot vector subtraction in matplotlib

Solution 1:

According to this and the matplotlib documentation here, you need to use angles='xy', scale_units='xy', scale=1 to get the right units.

See code below:

import numpy as np
import matplotlib.pyplot as plt


b = np.array([4, 1])
d = np.array([-3, 3])
m = d-b
print(np.linalg.norm(d))
o = np.array([
    [0, 0, b[0]],
    [0, 0, b[1]]
])
v = np.array([
    [b[0], b[1]],
    [d[0], d[1]],
    [m[0], m[1]]
])

fig, ax = plt.subplots()
ax.quiver(*o, v[:, 0], v[:, 1],angles='xy', scale_units='xy', scale=1)
ax.set_xlim([-4,4])
ax.set_ylim([-4,4])
plt.show()

And the output:

Solution 2:

It's more convenient to store coordinates of vectors like so:

S = np.stack([*o, v[:, 0], v[:, 1]])
>>> S
array([[ 0,  0,  4],
       [ 0,  0,  1],
       [ 4, -3, -7],
       [ 1,  3,  2]])

Now you can access coords of starting points and directions in this way:

>>> S[:2], S[2:]
(array([[0, 0, 4],
        [0, 0, 1]]),
 array([[ 4, -3, -7],
        [ 1,  3,  2]]))

As well as coords of ending points:

>>> S[:2] + S[2:]
array([[ 4, -3, -3],
       [ 1,  3,  3]])

So you can get all the points on a plane:

>>> pts = np.hstack([S[:2], S[:2] + S[2:]])
array([[ 0,  0,  4,  4, -3, -3],
       [ 0,  0,  1,  1,  3,  3]])

For later part, look into an answer of jylls and use:

plt.quiver(*S, angles='xy', scale_units='xy', scale=1)

Finally, as you've got a nice plot, you might be interested how to specify boundaries for your plot. You can do it like so:

x_min, y_min = np.min(pts, axis=1)
x_max, y_max = np.max(pts, axis=1)
plt.xlim([x_min, x_max])
plt.ylim([y_min, y_max])
plt.show()