以c→=xa→+yb→形式引入,考查向量相关知识,很多同学感到很困难,它常与几何图形相结合,通过几个例子说明常见转化方法.一、变形,发现几何关系求解例1已知平面内不共线的四点O,A,B,C满足OB→=1/3OA→+2/3OC→,则|AB→|∶|BC→|=.解∵OB→=1/3OA→+2/3OC→,∴OB→-OC→=1/3OA→-1/3OC→,得CB→=1/3CA→, The introduction of c → = xa → + yb → form, to examine the vector-related knowledge, many students feel very difficult, it is often combined with geometry, through a few examples of common transformation methods. It is known that the four points O, A, B and C which are not collinear in the plane satisfy OB → = 1 / 3OA → + 2 / 3OC →, then | AB → |: | BC → | = 3OA → + 2 / 3OC →, ∴OB → -OC → = 1 / 3OA → -1 / 3OC →, CB → = 1/3 CA →,