向量a=(2,3),向量b=(3,-2),
则a^2=4+9=13,b^2=4+9=13,a•b=6-6=0.
设向量ka+b与a+kb夹角为θ,
则cosθ=(ka+b)•(a+kb)/ [|ka+b||a+kb|]
(ka+b)•(a+kb)= ka^2+(k^2+1) a•b+ kb^2=26k.
|ka+b|^2= k^2 a^2+ b^2+2k a•b=13 k^2+13,
|a+kb|^2= a^2+ k^2 b^2+2k a•b=13 k^2+13,
∴cosθ=26k/(13 k^2+13)
因为向量ka+b与a+kb夹角为60°,所以cosθ=1/2.
即26k/(13 k^2+13) =1/2.
解得k=2±√3.