Question

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.

Answer 1

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