Lagrange's Identity states: |a x b|^{2} = |a|^{2}|b|^{2} - (a.b)^{2}. This follows directly from s$in^{2}$(theta) + c$os^{2}$(theta) = 1 applied to the dot and cross product formulas. This identity is extremely powerful in JEE because it allows you to find |a x b| when you know magnitudes and dot product, or vice versa, without computing the full cross product. For example, if |a| = 3, |b| = 5, a.b = 9, then |a x b| = sqrt(225 - 81) = 12. It can also be written as (a.b)^{2} + |a x b|^{2} = |a|^{2}|b|^{2}, showing that dot and cross products completely determine the geometric relationship.