V22+2^+12 = 4 < R , suy ra (a) cat mat cau (S) theo
duong tron tam H ban kinh r = ^R^ - d 2(I,(a)) = 3
H la h i n h chieu cua I len mat phang ( a ) , suy ra p h u o n g trinh cua H I la: x = l + 2t
' y = l + 2t z = l + t
Toa do diem I la nghiem ciia h ^
x = l + 2t y ^ l + 2t z = l + t -1 2 x + 2 y + z + 7 = 0 x = y = - 1 z 3 Vay tam H 5 _ 5 _ \ " 3 ' 3 ' 3
2) Ta CO A B = (2; 6;-4) nen phuong trinh d u o n g thang A B :
x = l + t y = - l + 3t y = 2 - 2 t
V i l A < R nen mat phang ( P ) d i qua A B luon cat mat cau (S) theo d u o n g
tron CO ban kinh r - 725- d 2(I, ( P ) ) .
Do d o r nho nhat <=> d(I, ( P ) ) i o n nhát.
Gpi K, H Ian hxgl la h i n h chieu cua I len A B va ( P ), ta luon c6 I H < I K nen suy ra d(I, ( P ) ) Ion nhat o H = K
Do H e AB => H ( l +1; - 1 + 3t;2 - 2t) ^ I H = (t; 3t - 2; 1 - 2t)
V i I H l A B r r > i H A B = 0 o t + 3 ( 3 t - 2 ) - 2 ( l - 2 t ) = : 0 o t = ^ = ^ i H = [ ^ ^ ; - ^ ; - ^ Vay p h u o n g trinh ( a ) : 4x - 2y - z - 4 = 0 .
Bdi 3.4.5. Trong khong gian Oxyz cho d u o n g thang d la giao tuyen cua hai mat phang ( a ) : 2 x - 2 y - z + l = 0, (P):x + 2 y - 2 z - 4 = 0 va mat cau (S) c6 phuong trinh x^ + y^ + + 4x - 6y + m = 0 . T i m m de d u o n g thang d cat mat cau (S) tell hai diem phan bi^t A , B sao cho AB = 8.
Jiuang dan gidi
Taco i i ; ' = ( 2 ; - 2 ; - l ) , IT^ = (1;2;-2) Ian lu(?t la V T P T cua (a) va (P) S u y r a u = i f n ^ n j ] = (2;1;2) la VTCP cua d u o n g thang d
3
Phucmg phdpgidi Toan Hinh hoc theo chuyen dc- Nguyen Phu Khanh, Nguyen Tat Thii
Hon nua diem Ă6; 4; 5) la diem chung ciia hai mat phSng (a) va ((3) nen A e d
Mat cau (S) c6 tam I(-2;3;0), ban kinh R = 7 l 3 - m vai m < 13 . IA = (8;1;5): I A , u = (-3;-6;6)=:>d(I,d) = 3
AB
G<?i H la trung diem AB A H = — = 4 va IH = 3
Trong tam giac vuong IHA ta c6: l = IH^ + AH^ » = 9 +16 <:> 13 - m = 25 <=> m = -12 . Vay m = -12 la gia trj can tim.
Bdi 3.4.6. Trong khong gian voi he true toa dp Oxyz cho mat phang
(P):2x + 2y + z - m ^ - 3 m = 0 va mat cau (S): (x - i f + ( y +1)^ + (z-1)^ = 9 . Tim m de mat phang (P) tiep xiic voi mat cau (S). Voi m vua tim duac hay xac dinh toa dp tiep diem.
J-Iu6ng đn gidi
Mat cau (S) c6 tam 1(1;-1;1), ban kinh R = 3. Goi A la duong thang di qua I , vuong goc voi (?)
x - 1 _ y + 1 _ z - 1 Suy ra phuong trinh ^ • 2 2 1 Mat phang (P) tiep xiic voi mat cau (S)
o d ( I , ( P ) ) = R o + 3 m - l - 3 o m'^ + 3m-10 = 0
m^ + 3m + 8 = 0 V N <=> m = -5, m = 2 Khi do (P): 2x + 2y + z -10 = 0. Tpa dp tiep diem A la nghi#m Gtia h?:
x - 1 y + 1 z - 1 ,
2 2 1 /giaihf nay tadupc x = 3,y = l,z = 2=^ Ă3;l;2). 2x + 2y + z-10 = 0
Bai 3.4.7. Trong khong gian vdi h? tpa dp De cac vuong goc Oxyz cho I(l;2;-2) va mat phang ( p ) : 2x + 2y + z + 5 = 0
1) Lap phuang trinh mat cau (S) tam I sao cho giao ciia (S) vai mp(P) la
duong tron (C) c6 chu vi bang 8n
2) Chung minh rang mat cau (S) noi trong phan 1 tiep xiic voi duong thang A : 2 x - 2 = y + 3 = z
3) Lap phuang trinh m^t phang (Q) chua duong thang A va tiep xiic vai (S).
Cty TNHH MTV DWH Khang Viet
JJu&ng đn gidi
^) Goi R, r Ian lupt la ban kinh ciia mat cau (S) va duong tron (C). Ta c6: 27ir = STT r = 4 va d(I,(P)) = 3 nen R = ^r^ +d^{l,{P)) = 5 . Vly phuang trinh mat cau (S): (x -1)^ + (y - if + (z + if = 25.
2) Duong thang A c6 u7 = (l;2;2) la VTCP va di qua Ăl;-3;0).
Suy ra A I = (0; 5; -2) ^ [u ^, AI] = (-14; 2;5) => d(I, A) =
Vay duong thang A tiep xiic vai mat cau (S).
Cdch2.
x = l + t Phuang trinh tham so ciia A: -
[u A " 'A l l
" A - 5 - 5
y = -3 + 2t, thay vao phuang trinh mat cau (S) z = 2t
tadupc: t 2+ ( 2 t - 5) 2+ ( 2 t + 2) 2= 2 5 » ( 3 t - 2) 2= 0 o t = - 5 5 4 Suy ra mat cau (S) va A giao nhau tai mot diem ^ ( 3 ' " 3 ' 3 ) • Vay duong thang A tiep xiic vai mat cau (S) tai M .
3) Vi mp(Q) chiia A va tiep xiic voi mat cau (S) nen M la tiep diem ciia mp(Q) va mat cau (S)
•fl 11 10^ Do do (Q) la mat phang di qua M va nhan ^M^^ ^, 3 ' 3 j Vay phuong trinh mat phing (Q): 2x - l l y + lOz - 35 = 0.
lam VTPT.