Question

已知向量a=(1,2),b=(cosa,sina),设m=a+tb(t为是实数)

Answer 1

m=a+tb=(1,2)+t(cosa,sina)=(1+tcosa,2+tsina)当a=π/4时,
则m=(1+(t根号2)/2,2+(t根号2)/2),
于是|m|^2=[1+(t根号2)/2]^2+[2+(t根号2)/2]^2=t^2+(3倍根号2)t+5=[t+(3倍根号2)/2]^2+1/2
显然当t=-(3倍根号2)/2时,|m|取得最小值；
2、若a⊥b,则ab=0,即(1,2)(cosa,sina)=0,
cosa+2sina=0向量a-b(1-cosa,2-sina)和向量m(1+tcosa,2+tsina)的夹角为π/4,
故(a-b)*m=(1-cosa,2-sina)(1+tcosa,2+tsina)=5-t+(t-1)(cosa+2sina)=5-t,
|a-b|=根号[(1-cosa)^2+(2-sina)^2]=根号[6-2(cosa+2sina)]=根号6,
|m|=根号[(1+tcosa)^2+(2+tsina)^2]=根号[5+t^2+2t(cosa+2sina)]=根号(5+t^2),
于是cosπ/4=[(a-b)*m]/[|a-b|*|m|]=(5-t)/[(根号6)(根号(5+t^2))]
即(5-t)/[(根号6)(根号(5+t^2))]=(根号2)/2
整理得：t^2+5t-5=0解得t=(-5-3倍根号5)/2或t=(-5+3倍根号5)/2
综上当t=(-5-3倍根号5)/2或t=(-5+3倍根号5)/2时,向量a-b和向量m的夹角为π/4.