m=a+tb=(1,2)+t(cosa,sina)=(1+tcosa,2+tsina)當(dāng)a=π/4時(shí),
則m=(1+(t根號(hào)2)/2,2+(t根號(hào)2)/2),
于是|m|^2=[1+(t根號(hào)2)/2]^2+[2+(t根號(hào)2)/2]^2=t^2+(3倍根號(hào)2)t+5=[t+(3倍根號(hào)2)/2]^2+1/2
顯然當(dāng)t=-(3倍根號(hào)2)/2時(shí),|m|取得最小值；
2、若a⊥b,則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)的夾角為π/4,
故(a-b)*m=(1-cosa,2-sina)(1+tcosa,2+tsina)=5-t+(t-1)(cosa+2sina)=5-t,
|a-b|=根號(hào)[(1-cosa)^2+(2-sina)^2]=根號(hào)[6-2(cosa+2sina)]=根號(hào)6,
|m|=根號(hào)[(1+tcosa)^2+(2+tsina)^2]=根號(hào)[5+t^2+2t(cosa+2sina)]=根號(hào)(5+t^2),
于是cosπ/4=[(a-b)*m]/[|a-b|*|m|]=(5-t)/[(根號(hào)6)(根號(hào)(5+t^2))]
即(5-t)/[(根號(hào)6)(根號(hào)(5+t^2))]=(根號(hào)2)/2
整理得：t^2+5t-5=0解得t=(-5-3倍根號(hào)5)/2或t=(-5+3倍根號(hào)5)/2
綜上當(dāng)t=(-5-3倍根號(hào)5)/2或t=(-5+3倍根號(hào)5)/2時(shí),向量a-b和向量m的夾角為π/4.