Given $ v = (-2a\sin(2t),2a\cos(2t), 2a\cos(t) )$
and $k= (-4a\cos(2t),-4a\sin(2t),-2a\sin t)$
Finding $|v \times k|$. Now I am finding it hard to compute determinant. There must be an easy way because this question was of 3 points only in test I gave yesterday. Thanks for help
EDIT : This question is originally part of following question : Find the curvature and torsion at any point t of the curve $x=a\cos(2t) , y=a\sin(2t), z=2a\sin(t)$
Since $v\cdot k=-4a^2\sin t\cos t$, $|v|^2=4a^2(1+\cos^2 t)=4a^2(2-\sin^2 t)$ and $|k|^2=4a^2(4+\sin^2 t)$, we can use an identity
$$|v\times k|^2=|v|^2|k|^2-(v\cdot k)^2\\=16a^4(8-2\sin^2 t-\sin^4 t-\sin^2t(1-\sin^2 t))\\=16a^4(8-3\sin^2 t).$$