Question

设向量a,b满足|a|＝|b|＝|a＋b|＝1,则|a－tb|(t∈R)的最小值为( )

Answer 1

a与b夹角为120°,ab=-1/2

|a-tb|^2=(a-tb)^2=a^2+b^2*t^2-2abt=t^2+t+1=(t+1/2)^2+3/4

|a-tb|最小值为√3/2