Dot product, cosine, and distance

Drag the tips of a and b. Three ways to compare two vectors update as you move them.

Scale both to length 1

Dot product a · b6.50

Cosine similarity0.82

Angle35°

Distance ‖a − b‖1.80

Challenges

0 of 3 complete

Get a dot product above 8 while the cosine similarity stays below 0.5. (not yet)
Make the vectors point almost the same way (cosine above 0.95) but more than 3 apart. (not yet)
Turn on Scale both to length 1 and set the cosine similarity to 0.50. Read the distance. (not yet)

a = (3.00, 1.00), b = (1.50, 2.00). a · b = 3.00 × 1.50 + 1.00 × 2.00 = 6.50. Lengths ‖a‖ = 3.16, ‖b‖ = 2.50, so cos θ = 6.50 / (3.16 × 2.50) = 0.82.

Try this

Make the angle 90 degrees. What happens to the dot product and the cosine?
Choose Same direction, then drag b farther out. Which measures change?
Turn on Scale both to length 1 and check that a smaller distance always means a larger cosine.

The three measures in NumPyPython with NumPy

import numpy as np

# The Similar example from the demo.
a = np.array([3.0, 1.0])
b = np.array([1.5, 2.0])

def compare(a, b):
    dot = a @ b                                          # multiply matching entries, add
    cos = dot / (np.linalg.norm(a) * np.linalg.norm(b))  # divide out both lengths
    angle = np.degrees(np.arccos(np.clip(cos, -1, 1)))
    dist = np.linalg.norm(a - b)                         # straight-line distance
    print(f"dot {dot:.2f}, cosine {cos:.2f}, angle {angle:.0f}°, distance {dist:.2f}")

compare(a, b)
compare(a, 3 * b)  # b three times as long: same cosine, other numbers change

# Scaled to length 1, the dot product is the cosine and distance follows from it.
ua, ub = a / np.linalg.norm(a), b / np.linalg.norm(b)
compare(ua, ub)
print(f"2 - 2 cos = {2 - 2 * (ua @ ub):.4f}")
print(f"distance squared = {np.sum((ua - ub) ** 2):.4f}")

dot 6.50, cosine 0.82, angle 35°, distance 1.80
dot 19.50, cosine 0.82, angle 35°, distance 5.22
dot 0.82, cosine 0.82, angle 35°, distance 0.60
2 - 2 cos = 0.3556
distance squared = 0.3556