设a⊥b,=π/3,=π/6,|a|=1,|b|=2,|c|=3,求|a+b+c|长度

(a+b+c)^2
=a^2+b^2+c^2+2ab++2ac+2bc
=1*1+2*2+3+0+2*1*√3cosπ/3+2*2*√3cosπ/6
=14+√3
所以|a+b+c|=√(14+√3)