作业帮求2X— y+Z=6 x+y+2Z=3平面间的夹角.
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求2X— y+Z=6 x+y+2Z=3平面间的夹角.
求2X— y+Z=6 x+y+2Z=3平面间的夹角.
求2X— y+Z=6 x+y+2Z=3平面间的夹角.
设两定平面的方程为：
A1X+B1Y+C1Z+D1=0.(1)
A2X+B2Y+C2Z+D2=0.(2)
它们的法线矢量分别为{A1,B1,C1}和{A2,B2,C2},令这
两法线矢量的夹角为φ,那么这两平面的夹角就是φ,于是
cosφ=(A1A2+B1B2+C1C2)/[√(A1²+B1²+C1²)√(A2²+B2²+C2²)]
=（2*1-1*1+1*2)/[√(4+1+1)√(1+1+4)]
=1/2
φ=60度
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