1）由 (a+b)/3=1/3*a+1/3*b=1/3*a+1/(3t)*tb ,
则 1/3+1/(3t)=1 ,所以 t=1/2 .
2）令 |a|=|b|=2m ,则 a*b=|a|*|b|*cos60=2m^2 ,
所以,由 (a-tb)^2=a^2-2t*a*b+t^2*b^2=4m^2(1-t+t^2)
=4m^2*[(t-1/2)^2+3/4] ,得
当 t=1/2 时,|a-tb| 值最小 .