If ABCD is a parallelogram, then its vector area in terms of the diagonals AC and BD is ( AC × BD )
Proof:
(AC × BD) | = | (AB + BC) × (BA + AD) |
= | [AB × BA + AB ×AD + BC × BA + BC × AD] | |
= | [AB × AD + (–CB) × BA] | |
= | [AB × AD + (–CB) × CD] (∵ BA = CD) | |
= | (AB × AD) + (CD × CB) | |
= | vector area of Δ ABD + vector area of Δ CDB | |
= | vector area of ABCD. |