∆ QUA M (-2; 2; -3), VTCP A (2;3; 2)R=; AM ( 2; 2; 1)UUUUR= − −;...
2. ∆ qua M (-2; 2; -3), VTCP a (2;3; 2)r=; AM ( 2; 2; 1)uuuur= − −r uuuur∧ = + + =⇒ a AM ( 7; 2;10)r uuuur∧ = − − ⇒ d( A, ∆) = a AM 49 4 100 153r =3+ +4 9 4 17aVẽ BH vuông góc với ∆2 =4. ∆AHB ⇒ R
2
= 153 42516+ 17 = 17 =25Ta có : BH = BCPhương trình (S) : x2
+y2
+ +(z 2)2
=25π π = −− = − + − ÷Câu VII.b: (1 3i)3
z 1 i3 3− . (1 3i) 2 cos( ) i sin( )⇒ (1− 3i)3
=8 cos(