|a|=√2,|b|=3,=π/4
即：a·b=|a|*|b|*cos(π/4)
=3
(λa+b)⊥(a+λb),即：
(λa+b)·(a+λb)=λ|a|^2+λ|b|^2+(λ^2+1)a·b
=2λ+9λ+3(λ^2+1)
=3λ^2+11λ+3=0
即：λ=(-11±√85)/6
題目數(shù)據(jù)有問題吧,結(jié)果這么怪