Given vectors PQ=ai^+bj^​ and PS=ai^−bj^​ are adjacent sides of a parallelogram PQRS, so area of parallelogram PQRS=∣PQ×PS∣∴​=2ab=8ab=4​ (given) … (i) ​According to the question,∣u∣=∣ projection vector of w=i^+j^​ along PQ∣=∣∣​a2+b2​(i^+j^​)⋅(ai^+bj^​)​∣∣​=a2+b2​∣a+b∣​and, similarly, ∣v∣=∣ projection vector of w=i^+j^​ along PS ∣=a2+b2​∣(i^+j^​)⋅(ai^+bj^​)∣​=a2+b2​∣a−b∣​∵⇒⇒​∣u∣+∣v∣=∣w∣a2+b2​∣a+b∣​+a2+b2​∣a−b∣​=2​∣a+b∣+∣a−b∣=2​a2+b2​​If a≥b>0, then 2a=2​a2+b2​⇒⇒⇒​4a2​=2a2+2b2a2a​=b2=b​From Eqs. (i) and (ii), we geta∴​=2=ba+b​=4​and the length of diagonal PR=∣ai^+bj^​+ai^−bj^​∣=∣2ai^∣=2a=4​And, the angle bisector of vector PQ and PS is along the vector±a2+b2​λ​((ai^+bj^​)±(ai^−bj^​))