2<\/sup>\u00a0= 9 v\u00e0 m\u1eb7t ph\u1eb3ng (P): 2x – y – 2z + 2 = 0. L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh c\u00e1c m\u1eb7t ph\u1eb3ng (Q) song song v\u1edbi m\u1eb7t ph\u1eb3ng (P) v\u00e0 ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t c\u1ea7u (S)<\/p>\nA. 2x – y – 2z + 16 = 0 \u00a0\u00a0\u00a0C. 2x – y – 2z – 34 = 0<\/p>\n
B. 2x – y – 2z + 20 = 0\u00a0\u00a0\u00a0D. 2x – y – 2z – 16 = 0<\/p>\n
C\u00e2u 52:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t ph\u1eb3ng (P) thay \u0111\u1ed5i nh\u01b0ng lu\u00f4n \u0111i qua \u0111i\u1ec3m M(2;1;3) v\u00e0 c\u1eaft c\u00e1c tia Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t t\u1ea1i c\u00e1c \u0111i\u1ec3m A, B, C (kh\u00e1c O). Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a th\u1ec3 t\u00edch t\u1ee9 di\u1ec7n OABC l\u00e0:<\/p>\n
A. 54\u00a0\u00a0\u00a0B. 6\u00a0\u00a0\u00a0C. 27\u00a0\u00a0\u00a0D. 81<\/p>\n
C\u00e2u 53:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t ph\u1eb3ng (P) thay \u0111\u1ed5i nh\u01b0ng lu\u00f4n \u0111i qua \u0111i\u1ec3m M(1;2;-3). Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (P) sao cho kho\u1ea3ng c\u00e1ch t\u1eeb O \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P) l\u1edbn nh\u1ea5t<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10.png\")
<\/p>\n
C\u00e2u 55:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t ph\u1eb3ng (P): 3x – 4y + 12 = 0 . L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m I(1;0;3) v\u00e0 (S) giao (P) theo m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c\u00f3 b\u00e1n k\u00ednh r=4<\/p>\n
A. (x – 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z – 3)2<\/sup>\u00a0= 25\u00a0\u00a0\u00a0C. (x – 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z – 3)2<\/sup>\u00a0= 5<\/p>\nB. (x + 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 3)2<\/sup>\u00a0= 25\u00a0\u00a0\u00a0D. (x + 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 3)2<\/sup>\u00a0= 5<\/p>\nC\u00e2u 56:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t c\u1ea7u (S): x2<\/sup>\u00a0+ (y – 1)2<\/sup>\u00a0+ (z + 2)2<\/sup>\u00a0= 25 v\u00e0 m\u1eb7t ph\u1eb3ng (P): 2x – 2y + z + m = 0 . T\u00ecm m sao cho (P) giao (S) theo m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c\u00f3 b\u00e1n k\u00ednh r=3 l\u00e0:<\/p>\nA. m=16\u00a0\u00a0\u00a0C. m=40<\/p>\n
B. m=16 ho\u1eb7c m=-8\u00a0\u00a0\u00a0D. m=40 ho\u1eb7c m=32<\/p>\n
H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n| 50-D<\/td>\n | 51-D<\/td>\n | 52-C<\/td>\n | 53-C<\/td>\n | 54-A<\/td>\n | 55-A<\/td>\n | 56-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 50:<\/b><\/p>\n G\u1ecdi M(0;m;0) v\u1edbi m > 0 ta c\u00f3:<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10-2.png\") <\/p>\n
K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n m > 0 ta \u0111\u01b0\u1ee3c m=4. V\u1eady M(0;4;0)<\/p>\n C\u00e2u 51:<\/b><\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a mp(Q) c\u00f3 d\u1ea1ng 2x – y – 2z + m = 0 v\u1edbi m kh\u00e1c 2<\/p>\n M\u1eb7t c\u1ea7u (S) t\u00e2m I(1;-1;-2) v\u00e0 c\u00f3 b\u00e1n k\u00ednh R=3. M\u1eb7t ph\u1eb3ng (Q) ti\u1ebfp x\u00fac v\u1edbi (S) khi v\u00e0 ch\u1ec9 khi:<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10-3.png\") <\/p>\n
K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n m kh\u00e1c 2 ta \u0111\u01b0\u1ee3c m=-16. V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (Q) l\u00e0: 2x – y – 2z – 16 = 0<\/p>\n \n \n C\u00e2u 52:<\/b><\/p>\n G\u1ecdi A(a; 0; 0), B(0; b; 0), C(0; 0; c) trong \u0111\u00f3 a, b, c > 0. Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (P) l\u00e0:<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bt-12-1.jpg\") <\/p>\n
C\u00e2u 53:<\/b><\/p>\n G\u1ecdi H l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a O tr\u00ean m\u1eb7t ph\u1eb3ng (P). Ta c\u00f3<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10-8.png\") <\/p>\n
D\u1ea5u b\u1eb1ng x\u1ea3y ra khi v\u00e0 ch\u1ec9 khi H tr\u00f9ng M.<\/p>\n Ta ch\u1ecdn\u00a0np<\/sub>\u2192<\/i>\u00a0=\u00a0a\u2192<\/i>\u00a0= (1; 2; -3). T\u1eeb \u0111\u00f3 suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (P) l\u00e0 :<\/p>\n1(x – 1) + 2(y – 2) – 3(z + 3) = 0 <=> x + 2y – 3z – 14 = 0<\/p>\n C\u00e2u 54:<\/b><\/p>\n M\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m I(1 ;2 ;-2) v\u00e0 c\u00f3 b\u00e1n k\u00ednh R=1. Kho\u1ea3ng c\u00e1ch t\u1eeb I \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P) l\u00e0<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10-9.png\") <\/p>\n
Do \u0111\u00f3 m\u1eb7t ph\u1eb3ng (P) kh\u00f4ng c\u1eaft m\u1eb7t c\u1ea7u (S).<\/p>\n G\u1ecdi K v\u00e0 H l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a A v\u00e0 I tr\u00ean m\u1eb7t ph\u1eb3ng (P). Ta c\u00f3:<\/p>\n d(A; (P)) = AK \u2264 AH \u2264 IA + IH = R + h = 10<\/p>\n D\u1ea5u b\u1eb1ng x\u1ea3y ra khi v\u00e0 ch\u1ec9 khi A l\u00e0 giao \u0111i\u1ec3m c\u1ee7a tia \u0111\u1ed1i c\u1ee7a tia IH v\u1edbi m\u1eb7t c\u1ea7u (S). V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/p>\n C\u00e2u 55:<\/b><\/p>\n Kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m I \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P) l\u00e0 :<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10-10.png\") <\/p>\n
B\u00e1n k\u00ednh c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u00e0 :<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10-11.png\") <\/p>\n
V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u00e0 : (x -1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z – 3)2<\/sup>\u00a0= 25<\/p>\nC\u00e2u 56:<\/b><\/p>\n Kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m I c\u1ee7a m\u1eb7t c\u1ea7u (S) \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P) l\u00e0 :<\/p>\n ![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-phuong-trinh-mat-phang-10-12.png\") <\/p>\n<\/div>\n<\/div>\n
<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 50:\u00a0Trong kh\u00f4ng gian Oxyz, t\u00ecm nh\u1eefng \u0111i\u1ec3m M tr\u00ean tia Oy sao cho kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m M \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P): x + 2y – 2z + 1 = 0 b\u1eb1ng 3 A. M(0;13;0)\u00a0\u00a0\u00a0C. M(0;4;0) ho\u1eb7c M(0;-5;0) B. M(0;-5;0)\u00a0\u00a0\u00a0D. M(0;4;0) C\u00e2u 51:\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t c\u1ea7u (S): (x – 1)2\u00a0+ (y […]<\/p>\n","protected":false},"author":3,"featured_media":27802,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1450,1448],"yoast_head":"\n B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 \u2013 B\u00e0i 2: Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (ph\u1ea7n 10)<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-hinh-hoc-12-bai-2-phuong-trinh-mat-phang-phan-10\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 \u2013 B\u00e0i 2: Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (ph\u1ea7n 10)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 50:\u00a0Trong kh\u00f4ng gian Oxyz, t\u00ecm nh\u1eefng \u0111i\u1ec3m M tr\u00ean tia Oy sao cho kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m M \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P): x + 2y – 2z + 1 = 0 b\u1eb1ng 3 A. 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M(0;13;0)\u00a0\u00a0\u00a0C. M(0;4;0) ho\u1eb7c M(0;-5;0) B. M(0;-5;0)\u00a0\u00a0\u00a0D. 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