∵OA=i+3k,OB=j+3k∴AB=OB-OA=j-i设OA=b,OB=a,AB=C由余弦定理：a²+b²-2abcosC=c²∴cosC=(a²+b²-c²)/2ab∴sinC=√{1-[(a²+b²-c²)/2ab]²}∴S△OAB=1/2absinC=..希...最后答案是多少