Lagrange's Identity: |a x b|^{2} = |a|^{2}|b|^{2} - (a.b)^{2}. This connects dot and cross products and is extremely useful in JEE. It follows directly from |a x b| = |a||b|sin(theta) and a.b = |a||b|cos(theta), since s$in^{2}$(theta) + c$os^{2}$(theta) = 1. Application: Given |a|, |b|, and a.b, you can find |a x b| without computing the cross product component-wise. Conversely, given |a x b| and magnitudes, you can find the dot product.