1) Phuong trinh mat cau (S): (x -1)^ + (y - 2)^ + (z - 3)^ = 5 2) Goi I la tam mat caụ Vi I e Ox l(x;0;0)
Ta CO l A ^ = IB^ o (x - 1 ) 2 + 2^ +1^ = (x - 3)2 +1^ + (-2)^ <:> x = 2 Suy ra tam 1(2; 0;0) va ban kinh R^ = IB^ = 6
Vay phuong trinh mat cau (S): (x - 2)^ + y^ + z^ = 6 . 3) V i mat cau (S) eo tam 1(3;-2;4) va tiép xiic voi mp(P)
Suy ra R = d(I,(P)) = 12.3 + 2 + 2.4 + 41 20
Vay phuong trinh mat cau (S):(x -3)^ + (y + if +{z-^f
4) Goi C p C j X a Ian lugt la hinh chieu ciia C len cac true toa dp Ox,Oy,Oz
ISuy ra Ci (2; 0; 0), C2 (0; -4; 0), C3 (0; 0; 3).
G i a s u (S):x2 +y2 + z2 - 2 a x - 2 b y - 2 c z + d - 0
Do (S) d i qua C C p C j / ^ a nen ta c6 h$ phuong trinh \ + 4 29 a = 4 a = l u d + 16 b = -2 b = 8 « • 3 d + 9 c = 6 d = 0 - 2 d = 0 - -4a + d = - 4 8b + d = -16 -6c + d = -9
Vay phuong trinh m | t cau (S): x^ + y^ + - 2x + 4y - 3z = 0 . V i tam I cua mat cau nam tren mp(Oxy) nen l(x;y;0)
Phumig phdp giai Todn Hinh hoc theo chuyen di- Nguyen Phu Khanh, Nguyen T&'t Thu Ta c6: I M2 =I N ^ r-6x + 2y = l 2 x - 4 y = -3 ^ 10 4 y = 5 UO 5 . Va R2 = I M ^ = 109 20
Vay phuong trinh mat cau (S): | ~ ~ 1 f r 4 f4 2 109
''-5; + z =• 20 Bai 3.4.2. Lap phuang trinh mat cau S(I,R) biet Bai 3.4.2. Lap phuang trinh mat cau S(I,R) biet
1) Mat cau c6 tam thuoc duong thang A : ~ ^ = = va tiep xiic v6i mat phang (â): 3x + 2y + z - 6 = 0 va mat phang ( a j ) : 2x + 3y + z = 0
2) Mat cau c6 tam 1(1;3;5) va cat Á: ^ ^ = ^ ^ = -^ tai hai diem A,B sao c h o A B = 12
3) Mat cau c6 tam thuQc duong thang d : ~ Y~ ^ ^ ~ ~ y ~ ' M(l;l;4) < ^. x + 2 y + 2 z - 4
va tiep xuc voi d : = = .
^ 1 1 - 4
Jiitang đn gidi
1) V i t a m l e A nen I(-2 +1; 1 + 2 t ; - 1 - 2 t ) .
Mat cau tiep xiic voi hai mat phang (â) va (a2) nen : d ( l ( a i ) ) = d(I,(a2)) = R |3(-2 + t) + 2(l + 2 t ) - l - t - 6 2(-2 + t) + 3(l + 2 t ) - l - t Suy ra 2^+12 V22+3^+12 6 t - l l | = |7t-2 • 6 t - l l = 7 t - 2 6 t - l l = 2 - 7 t t = -9 t = l
• Neu t = 1 thi I ( - l ; 3; - 3),R = -y= nen phucmg trinh m^t cau
(x + l ) 2 + ( y _ 3 ) 2 + ( z + 3)2^25 14
65
• Neu t = -9 thi I ( - l l ; -17; 17),R = nen phuang trinh m|it cau
(x +11)^ + (y +17)2+(z-17)2 = 2 4225
14
2) Duang thSng Á qua diem M(2; -3; 0) va c6 vec ta chi phuang la u ^ , ( - l ; 1; 1)
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Ta CO IM(1; 6; 5) nen [ I M , u ^ J = (1; - 4; 5), do do
" A - ^ ( - 1)2+I2+I2
Vi mat cau cat Á tai hai diem A,B nen ban kfnh mat cau dugc xac dinh AB^'
theo cong thuc : R = d (I, Á) + = 14 + 36 = 50
Vay mat cau can tim c6 phuong trinh la: (x -1)2 + (y - 3)2 + (z -5)2 = 50. 3) Duong thang d' qua diem N(-2;-2;4) va c6 u^. = ( l ; l ; - 4 ) la VTCP
Vi tam mat cau I € d nen 1(2 +1; - 1 ; 3 + 2t)
Taco M I = (t + l ; - t - l ; 2 t - l ) , N I = (4 + t ; 2 - t ; - l + 2t) nen
' N I , U d - ] = ( 2 t - 7 ; 6 t + 15;2t + 2).
Vi mat cau qua diem M va tiep xiic voi d ' nen M I - d(I,d') = R.
Do do M I - [ N I , U d ' J i -J 2 J(2t - 7)2 + (6t +15)2 + (2t + 2)2 Hay V(2t -1)2 + 2(1 +1)2 = ^ . / ' ^ -4 o 18(6t2 + 3) = 44t2 + 160t + 278 « 1^+1^+4^ t = - l .=1
Voi t = - l thi 1(1; 1;1), R = 3 nen phuong trinh mat cau
3N/34
( x - l) 2+ ( y - l) 2+ ( z - l) 2= 9
Voi t = - thi I
2 i l2 2 , ; - ^ ; 1 0 nen phuong trinh mat cau 'x 2 { 7^ + y + - V 2) ^ Ij y + - V 2) + (z-10)^ = 2 153 14
Bai 3.4.3. Trong khong giian Oxyz, cho mat cau (S) c6 phuang trinh x2 + y2 + z2 + 2x - 4y + 6z + m = 0.
Tim m sao cho:
1) Mat cau tiep xiic voi mat ph4ng (P): x - 2y + 2z - 1 = 0.
2) Mat cau cat mat phSng (Q) :2x - y - 2z +1 = 0 theo giao tuyen la mpt duong
tron CO dien tich bang 4n.
Phuotig phlip giiii Toiiit Hiiih hoc theo chtiyen ile- Nguyen Phii Khdiih, Ngiii/eii Till Tim
3) M a t cau cat d u o n g thang A : — — = = ' ^ tai hai diem phan birt A , B
sao cho tarn giac TAB vuong ( I la tarn mat cau).
Jlaufrig đn gidi
Dieu kien ton tai mat cau 14 - m > 0.
Mat cau (S) c6 tarn I ( - 1 ; 2; - 3 ) va ban kinh R = V l 4 - m . 1) Khoang each t u tarn mat cau den mat phang la
^ 2 + ( _ 2 ) 2 + 2 2
Vi mat cau tiep xiic voi mat phang nen R = 4 <=> m = -2. Vay gia t i i can tim cua m la m - - 2 ,
2) T a c 6 d ( I ; ( Q ) ) = -i|;14tlL = l .
Vl^+(-2)2+22
Vi d u o n g tron giao tuyen c6 di$n tich la An nen c6 ban kinh r = 2 , do do ban kinh cua mat cau la: R^ = + d^{l, (Q)) o R 2= 5 o l 4 - m = 5 < : : > m - 9 .
Vay gia trj can tim la: m = 9.
3) Goi H ( - l - t; 2t; 2 - t) la hinh chieu cua 2; - 3) tren A .
Ta CO i H ( - t ; 2t - 2; 5 - t) va u ^ ( - l ; 2; - 2) nen
I H l A < ^ I H . u ^ ^ - 0 o t + 4 t - 4 - 1 0 + 2 t - 0 o t = 2
Vay i H ( - 2 ; 2; 3) I H = 4v7.
Do lam giac l A B can tai I nen vuong can tai I , do do tarn giac I H A cunj^ vuong can tai H , v i the R = l A = V2.IH = ^ 3 4 ^ V l 4 - m = ^34 => m = -20.
Vay gia t r i can t i m ciia m la m = -20.
Bdi 3.4.4. Trong khong gian Oxyz, cho mat cau (S) c6 phuong trinh ( S ) : ( x - l) 2+ ( y - l) 2+ ( z - l) 2= 2 5
va mat phang ( a ) : 2x + 2y + z + 7 = 0.
1) Chung m i n h rang mat phang (a) cat mat cau (S) theo mot d u o n g tron. Xac djnh lam va t i m ban kinh ciia d u o n g tron dọ
2) Lap phuong trinh mat phang (P) d i qua hai diem A ( l ; - 1 ; 2 ) , B(3;5;-2) va (P)
cat mat cau (S) theo mpt d u o n g Iron c6 ban kinh nho nhat.
Jiicang dan gidi
M a t cau (S) c6 tam 1(1; 1; 1), ban kinh R = 5 .
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