Câu 8.a.  = +x t1 2 = = −d y t:2z t∈ ⇒ + −I d I t t t(1 2 ; ; 2 )Ta có: IA2=IB2=R2⇔(1-2t)2+(1-t)2+(2t)2=(-3-2t)2+(3-t)2+(2+2t)2⇔t = -1 ⇒I(1; -1; 2) ⇒R2 = 17 Vậy (S): (x−1)2+(y+1)2+(z−2)2 =17

A.  = +X T1 2 = = −D Y T

DAP AN KHOI B MON TOAN 2012

Câu 8.a.  = +x t1 2 = = −d y t:2z t∈ ⇒ + −I d I t t t(1 2 ; ; 2 )Ta có: IA

=IB

⇔(1-2t)

+(1-t)

+(2t)

=(-3-2t)

+(3-t)

+(2+2t)

⇔t = -1 ⇒I(1; -1; 2) ⇒R

= 17 Vậy (S): (x−1)

+(y+1)

+(z−2)

=17