Cross product shortcuts

Question

Given two vectors $a=(1,t, t^{3})$ $b=(0,1,3t^{2})$

I have seen that $|a\times b|^{2} = (1+t^{2}+t^{6})(1+9t^{4})-(t+3t^{5})^{2}$

It seems that its has been done in some shorthand way maybe other than calculating the cross product then the absolute value then squaring.

Answer 1

Yes, it's the identity $(a\times b)\cdot (a\times b)=(a\cdot a)(b\cdot b)-(a\cdot b)^2$, which is essentially the cartesian coordinates version of $\sin^2\theta=1-\cos^2\theta$.

Answer 2

Look at $|a\cdot b|^2 = |a|^2 |b|^2 \cos^2(\theta_{a,b}) = |a|^2|b|^2-|a\times b|^2$ so $|a\times b|^2 = a^2 b^2 - (\vec{a}\cdot \vec{b})^2$

I like this method since it doesn't require knowledge of the cross-product identities or levi-civita symbol depending on your flavor preference.