## Monday, September 22, 2008

### A X (B X C)

Any intro to physics course will teach you the identity

$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})$

It's easy to prove by explicitly writing out each vector in component form, cranking through both sides of the equation, finally getting the same thing on each. But there's some insight to be gained in looking at the identity from a purely geometric point of view. Warning: the following is a bit dense, but also short. Go one sentence at a time and you should be fine.

Together, $\mathbf{B}$ and $\mathbf{C}$ are a basis for a plane. $\mathbf{B} \times \mathbf{C}$ is perpendicular to the plane. Take that perpendicular vector product, cross it with $\mathbf{A}$, and you get a vector perpendicular to the perpendicular to the plane. That's back in the plane again. So we know that the double cross product is in the plane somewhere. But any vector in the plane can be written as a sum of $\mathbf{B}$ and $\mathbf{C}$ (that's what it means to be a basis), so we only need to find the right coefficients. Those coefficients come from requiring that we find a combination of $\mathbf{B}$ and $\mathbf{C}$ perpendicular to $\mathbf{A}$. Just dot the expression $\mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})$
with $\mathbf{A}$, and you'll get zero.

That proves both sides of the equation

$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})$

point in the same direction, but doesn't prove they have the same length. It's not crazy difficult to prove the they have the same length without using a coordinate system, but I don't have a neat insightful trick for it, either.

I started by noting that you could break $\mathbf{A}$ down into perpendicular and parallel components to the plane defined by the other two vectors, and that as far as the identity is concerned the perpendicular part of $\mathbf{A}$ doesn't matter (plug it in and you'll see). Then, because both sides of the equation are proportional to the lengths of all three vectors, you might as well divide by their lengths to make them unit vectors. All that's left is to find the lengths of both sides of the equation in terms of trig functions of the angles between the vectors. You have tree angles between the coplanar vectors. Write one as a sum of the other two, expand out all the trig functions, and fiddle around with it a bit until you see that it's actually a trig identity. That gets the final part of the theorem in there.