cross product

Cross Product

The definition of the cross product is:

(a⃗⨯b⃗)^i:=ε^{ijk}⋅a^j⋅b^k

... where ε^{ijk} is the Levi-Civita Symbol.

Where does the angle come from?

First, there is an identity connecting the cross product to the scalar product:

((a⃗⨯b⃗)∙(c⃗⨯d⃗))^i=ε^{ijk}⋅a^j⋅b^k⋅ε^{ilm}⋅c^l⋅d^m
((a⃗⨯b⃗)∙(c⃗⨯d⃗))^i=ε^{ijk}⋅ε^{ilm}⋅a^j⋅b^k⋅c^l⋅d^m
((a⃗⨯b⃗)∙(c⃗⨯d⃗))^i=(δ^{jl}⋅δ^{km}-δ^{jm}⋅δ^{kl})⋅a^j⋅b^k⋅c^l⋅d^m
((a⃗⨯b⃗)∙(c⃗⨯d⃗))^i=(δ^{jl}⋅δ^{km}⋅a^j⋅b^k⋅c^l⋅d^m)-(δ^{jm}⋅δ^{kl}⋅a^j⋅b^k⋅c^l⋅d^m)
((a⃗⨯b⃗)∙(c⃗⨯d⃗))^i=(a^j⋅b^k⋅c^j⋅d^k)-(a^j⋅b^k⋅c^k⋅d^j)

So the identity is:

(a⃗⨯b⃗)∙(c⃗⨯d⃗)=(a⃗∙c⃗)⋅(b⃗∙d⃗)-(b⃗∙c⃗)⋅(a⃗∙d⃗)

If the vectors c and a are the same and the vectors d and b are the same, it follows:

|a⃗⨯b⃗|^2=|a⃗|^2⋅|b⃗|^2-(a⃗∙b⃗)^2
|a⃗⨯b⃗|^2=|a⃗|^2⋅|b⃗|^2⋅(1-(\cos φ)²)
|a⃗⨯b⃗|^2=|a⃗|^2⋅|b⃗|^2⋅(\sin φ)²
|a⃗⨯b⃗|=|a⃗|⋅|b⃗|⋅\sin φ

Hence the angle in the cross product comes from the angle in the scalar product:

Scalar Product

Given a triangle with sides a and b and c=b-a, the Law of Cosines says:

|c⃗|²=|a⃗|²+|b⃗|²-2⋅|a⃗|⋅|b⃗|⋅\cos φ
|b⃗-a⃗|²=|a⃗|²+|b⃗|²-2⋅|a⃗|⋅|b⃗|⋅\cos φ

On the other hand, using only the scalar product distributivity:

(b⃗-a⃗)⋅(b⃗-a⃗)=b⃗∙b⃗-a⃗∙b⃗-a⃗∙b⃗+a⃗∙a⃗
(b⃗-a⃗)⋅(b⃗-a⃗)=a⃗∙a⃗+b⃗∙b⃗-2⋅a⃗∙b⃗

So since (b⃗-a⃗)⋅(b⃗-a⃗)=|b⃗-a⃗|², it follows that:

2⋅a⃗∙b⃗=2⋅|a⃗|⋅|b⃗|⋅\cos φ
a⃗∙b⃗=|a⃗|⋅|b⃗|⋅\cos φ

Author: Danny (remove the ".nospam" to send)

Last modification on: Sat, 04 May 2024 .