What is the breakdown of a curl squared? i.e..

Question

anonymous · Answer

Vector Cross product!

anonymous · Answer

The identity I've found states : $$|\vec{A} \times \vec{B}|^2 = |\vec{A}||\vec{B}| - (\vec{A}\cdot\vec{B})^2$$but my original statement wasn't the magnitude of them sqaured, does that matter?

anonymous · Answer

what about this http://upload.wikimedia.org/math/4/f/8/4f8ae548c9ac2c8336d2ac2eb939e750.png

anonymous · Answer

Original equation should be...$$(\vec{B} \times \vec{A})^2$$

irishboy123 · Answer

|B⃗ x A⃗|^2 = |B⃗|^2 |A⃗|^2 sin^2 theta =  |B⃗|^2 |A⃗|^2 (1- cos^2 theta ) =  |B⃗ |^2|A⃗ |^2− |B⃗⋅A⃗ |^2

i don't think you can "square" a vector.  if you apply the nearest thing, ie the dot product, you end up with the same as above, (B⃗ x A⃗)•(B⃗ x A⃗) =  |B⃗|^2 |A⃗|^2 sin^2 theta.......

irishboy123 · Answer

refresh screen

anonymous · Answer

Ah I did, thanks.