Let OA→=a→, OB→=12a→+4b→ and OC→=b, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then area of the quadrilateral OABCarea of S is equal to __________. [2024]
(1)
We have,
Area of parallelogram, S=|OA→×OC→|=|a→×b→| ... (i)
Area of quad. OABC = Area of △OAB + Area of △OBC
=12|OA→×OB→|+12|OB→×OC→|
=12|a→×(12a→+4b→)|+12|(12a→+4b→)×b→|
=12×(4|a→×b→|+12|a→×b→|)=8|a→×b→|
∴ Area of quad. OABCArea of parallelogram=8|a→×b→||a→×b→|=8.