已知：x²+y²=k,z²+w²=k,xz+yw=0,则xy+zw的值为——

Question

已知：x²+y²=k,z²+w²=k,xz+yw=0,则xy+zw的值为——

Answer 1

∵x²+y²=k,z²+w²=k,
∴x=√kcosα,y=√ksinα
z=√kcosβ,w=√ksinβ
∵xz+yw=0
∴k(cosαcosβ+sinαsinβ)=0
∴cos(α-β)=0
∴α-β=mπ+π/2,m∈Z
∴α=β+mπ+π/2,m∈Z
∴2α=2β+2mπ+π
∴xy+zw
=k(cosαsinα+cosβsinβ)
=k/2(2cosαsinα+2cosβsinβ)
=k/2(sin2α+sin2β)
=k/2[sin(2β+2mπ+π)+sin2β]
=k/2[-sin2β+sin2β]
=0