Cho A(3;2;9),B(1;2;3)
a) Viết phương trình mặt phẳng trung trực AB
b) Tính cosin của góc hợp bởi mặt phẳng trung trực và các mặt phẳng tọa độ

a) Gọi (α) là mặt phẳng trung trực của AB. Khi đó (α)AB tại trung điểm I của AB. Ta có:
I(x;y;z){x=3+12=1y=222=0z=9+32=6I(1;0;6)
(α) có vecto pháp tuyến:
n=AB=(4;4;6)(α):4x4y6z+D=0
(α) qua I(1;0;6)D=40
Vậy phương trình mặt phẳng trung trực của AB là:
(α):2x2y3z+20=0

b) Ta có: nα=(4;4;6)
cos[^(α),Oxy)]=|cos(n,k)|=|n.k|n|.|k||=|616+16+36|=|668|=317
cos[^(α),Oyz)]=|cos(n,i)|=|n.i|n|.|i||=|4217|=217
cos[^(α),Oxz)]=|cos(n,j)|=|n.j|n|.|j||=|4217|=217
 
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