How to manually calculate the cross product of vectors?

General Discussions
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1

How can I calculate the cross product of two vectors manually without using any programming libraries?

For example, given vectors a = (1, 2, 3) and b = (4, 5, 6), how can I compute their cross product in Python?

2

Hey, I’ve got some experience working with vector math, so let me share the basics first. To manually calculate the cross product, you can use the formula:

𝑐

π‘Ž Γ— 𝑏

( π‘Ž 2 𝑏 3 βˆ’ π‘Ž 3 𝑏 2 , π‘Ž 3 𝑏 1 βˆ’ π‘Ž 1 𝑏 3 , π‘Ž 1 𝑏 2 βˆ’ π‘Ž 2 𝑏 1 ) c=aΓ—b=(a 2 ​ b 3 ​ βˆ’a 3 ​ b 2 ​ ,a 3 ​ b 1 ​ βˆ’a 1 ​ b 3 ​ ,a 1 ​ b 2 ​ βˆ’a 2 ​ b 1 ​ ) Here’s a straightforward implementation in Python:

Manual computation of the cross product

def cross_product_python(a, b): c1 = a[1] * b[2] - a[2] * b[1] c2 = a[2] * b[0] - a[0] * b[2] c3 = a[0] * b[1] - a[1] * b[0] return (c1, c2, c3)

a = (1, 2, 3) b = (4, 5, 6) result = cross_product_python(a, b) print(β€œCross product:”, result) This gives you the cross product of two 3D vectors step by step."

3

Great explanation, @shashank_watak! Building on that, I’ve worked on simplifying such calculations for better readability. If you want to make it more compact, you can compute the cross product in a single line of code using list comprehension. This is especially useful for those who like cleaner code:

# Single-line computation using list comprehension
def cross_product_python(a, b):
    return (
        a[1] * b[2] - a[2] * b[1],
        a[2] * b[0] - a[0] * b[2],
        a[0] * b[1] - a[1] * b[0]
    )

a = (1, 2, 3)
b = (4, 5, 6)
result = cross_product_python(a, b)
print("Cross product:", result)

This concise version produces the same result while keeping the logic straightforward and focused.

4

Both methods are excellent! But here’s a cool technique I’ve picked up from my experience with linear algebra. You can approach the cross product using a determinant-based method, mimicking how a 3x3 matrix determinant is calculated. It adds a nice mathematical touch:

# Custom determinant-based computation
def cross_product_python(a, b):
    c1 = (a[1] * b[2] - a[2] * b[1])
    c2 = -(a[0] * b[2] - a[2] * b[0])
    c3 = (a[0] * b[1] - a[1] * b[0])
    return (c1, c2, c3)

a = (1, 2, 3)
b = (4, 5, 6)
result = cross_product_python(a, b)
print("Cross product:", result)

This method highlights the matrix-like structure of the computation, making it ideal for those who appreciate a more visual or structured approach to understanding the cross product python implementation.