设a={2,-3,1},b={1,-2,3},c={2,1,2},向量r满足r⊥a,r⊥b,Prjc r=21,求r.
问题描述:
设a={2,-3,1},b={1,-2,3},c={2,1,2},向量r满足r⊥a,r⊥b,Prjc r=21,求r.
向量r在向量c上的投影为21.{21,15,3},
答
设r=(x,y,z)
因为r⊥a,r⊥b,
则2x-3y+z=0.(1)
x-2y+3z=0.(2)
又Prjcr=21
设向量r与向量c的夹角为θ,cosθ=(2x+y+2z)/(3√x²+y²+z²)
则Prjcr=√x²+y²+z²×cosθ=√x²+y²+z²×(2x+y+2z)/(3√x²+y²+z²)
=(2x+y+2z)/3=21,即2x+y+2z=63.(3)
由(1)(2)(3)式可解得x=21,y=15,z=3