已知α∈[0,π],向量a=（(√3)cosα,sinα）,向量b=（0,-1）,求向量a减去向量b的模的最大值和最小值,并求出相应的α的值
a-b=((√3)cosα,sinα+1）
故︱a-b︱=√[3cos²α+(sinα+1)²]=√[(3cos²α+sin²α+2sinα+1)
=√[3(1-sin²α)+sin²α+2sinα+1]=√(-2sin²α+2sinα+4)=√[-2(sin²α-sinα)+4]
=√{-2[(sinα-1/2)²-1/4]+4}=√[-2(sinα-1/2)²+9/2]≦√(9/2)=3(√2)/2
∵α∈[0,π],0≦sinα≦1,-1/2≦sinα-1/2≦1/2,0≦(sinα-1/2)²≦1/4,-1/2≦-2(sinα-1/2)²≦0
∴√(-1/2+9/2)=2≦︱a-b︱≦3(√2)/2
故当α=π/2时︱a-b︱获得最小值2；当sinα=1/2,即α=π/6或5π/6时,︱a-b︱获得最大
值3(√2)/2.