ka-b与a+2b垂直,则(ka-b)*(a+2b)=0,即k|a|^2+(2k-1)(a*b)-2|b|^2=0. |a|=5,|b|=4,a*b=|a|*|b|*cos60°=10. 所以k|a|^2+(2k-1)(a*b)-2|b|^2=25k+10(2k-1)-32=0,k=14/15.采纳哦