Trong không gian Oxyz cho A(a;0;0),B(0;a;0),C(a;a;0),D(0;0;d) với >0,d>0. Gọi A,B là hình chiếu vuông góc của O xuống hai đường thẳng DA,DB
a) Viết phương trình mặt phẳng chứa đường thẳng OA,OB. Chứng minh rằng mặt phẳng đó vuông góc với CD
b) Tính d theo a để ^AOB=450

a) Ta có:  AD=(a;0;d),AC=(0;a;0),BD=(0;a;d),ACmp(Oxz)
ACOA(vìOA(Oxz))
OAAD nên OA cùng phương với [AC,AD]=(ad;0;a2)=a(d;0;a)
Tương tự: OAmp(Oyz)OBOAOBDB
Vậy OB//[OA,DB]=(0;ad;a2)=a(0;d;a)
Do đó mặt phẳng (OAB) có cặp vecto chỉ phương là: (d;0;a)(0;d;a)
Vecto pháp tuyến là: (ad;ad;d2)=d(a;a;d)
Phương trình mặt phẳng (OAB):ax+aydz=0
Vecto pháp tuyến của mặt phẳng (OAB) là: n=(a;a;d)=DC do đó CDmp(OAB)

b)  cos(AOB)=|cos(n1,n2)|=|n1.n2||n1|.|n2|=cos450=22. Do đó:
|a2|d2+a2.d2+a2=222a2=2(a2+d2)2a2=a2+d2d2=(21)a2d=(21)a
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