If O is the origin and Px1,y1,Qx2,y2 are two points OP×OQsin∠POQ=
x1x2+y1y2
x1y2+x2y1
x1y2−x2y1
none of these
We, have,
Area of ΔOPQ=Absolute value of 120 0 1x1 y1 1x2 y2 1
⇒ Area of ΔOPQ=12x1y2−x2y1
Also, Area of ΔOPQ=12OP×OQ×sin∠POQ
∴ 12OP×OQsin∠POQ=12x1y2−x2y1⇒ OP×OQsin∠POQ=x1y2−x2y1