Areas in the plane: The idea behind cross products is a
generalization of the formula for the area of a parallelogram in the
plane.
So suppose we have two vectors a=⟨a1,a2⟩=a1i+a2j and b=⟨b1,b2⟩=b1i+b2j in the plane. We want to
compute the area A(a,b) of the parallelogram with
vertices at the origin, a, b and a+b. This area function should have a few simple properties:
It should scale correctly: For any constant λ,
A(λa,b)=A(a,λb)=λA(a,b). This forces us to allow A(a,b) to be negative sometimes.
If b is counter-clockwise of a, then A(a,b) is
positive. If b is clockwise of a, then A(a,b) is minus the area of the parallelogram.
For any vectors a, b and c, A(a+c,b)=A(a,b)+A(c,b)
and A(a,b+c)=A(a,b)+A(a,c)
.
The fancy mathematical term for properties (1) and (2) is that A(a,b) is a bilinear function of a and b.
For any vector a, A(a,a)=0, since the "parallelogram"
is squashed flat.
Notice that 0=A(a+b,a+b),=A(a,a)+A(a,b)+A(b,a)+A(b,b),=A(a,b)+A(b,a),soA(b,a)=−A(a,b).
Unlike ordinary multiplication, the area function cares about
which entry comes first.
This means that A(i,i)=A(j,j)=0, that
A(i,j)=1 (since i and j span the unit square),
and that A(j,i)=−1.
Putting this together, we get the formula for area:
A(a,b)=A(a1i+a2j,b1i+b2j)=a1b1A(i,i)+a1b2A(i,j)+a2b1A(j,i)+a2b2A(j,j)=a1b2−a2b1.
You may have seen this formula before as the
determinant
of the 2×2 matrix [a1a2b1b2].
The following video describes determinants of
2×2 and 3×3 matrices, and explains how they are
related to areas and volumes.