设向量a的方向角a,β,y满足条件a=β,2a=y,求a,β,y.
2024-08-03高等数学(工本)(00023)
设向量a的方向角a,β,y满足条件a=β,2a=y,求a,β,y.
【正确答案】:由方向余弦的性质,有cos2a+cos2β+cos2y=2cos2a +cos22a =1,因此(2cos2a-1)+cos22a=0,即cos2a+cos22a=cos2a(1+cos2a)=0, 所以cos2a=0或cos2a=-1,即a=π/4或a=π/2,所以a=π/4,β=π/4,y=π/2,或a=π/2,β=π/2,y=π
【正确答案】:由方向余弦的性质,有cos2a+cos2β+cos2y=2cos2a +cos22a =1,因此(2cos2a-1)+cos22a=0,即cos2a+cos22a=cos2a(1+cos2a)=0, 所以cos2a=0或cos2a=-1,即a=π/4或a=π/2,所以a=π/4,β=π/4,y=π/2,或a=π/2,β=π/2,y=π
