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{"id":28252,"date":"2018-04-20T17:50:09","date_gmt":"2018-04-20T10:50:09","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=28252"},"modified":"2018-04-23T09:44:17","modified_gmt":"2018-04-23T02:44:17","slug":"bai-tap-trac-nghiem-hinh-hoc-12-chuong-3-bai-1-he-toa-do-trong-khong-gian-phan-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-hinh-hoc-12-chuong-3-bai-1-he-toa-do-trong-khong-gian-phan-2\/","title":{"rendered":"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 – Ch\u01b0\u01a1ng 3 – B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian (ph\u1ea7n 2)"},"content":{"rendered":"

C\u00e2u 5:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz , cho h\u00ecnh h\u1ed9p ABCD.A’B’C’D’ c\u00f3 A(1;0;0), B(1;2;0), D(2;-1;0), A\u2019(5;2;2). T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m C\u2019 l\u00e0:<\/p>\n

A. (3;1;0)\u00a0\u00a0\u00a0 B. (8;3;2)\u00a0\u00a0\u00a0 C. (2;1;0)\u00a0\u00a0\u00a0 D. (6;3;2)<\/p>\n

C\u00e2u 6:<\/b>\u00a0Cho hai vect\u01a1\u00a0a\u2192<\/i>,\u00a0b\u2192<\/i>\u00a0thay \u0111\u1ed5i nh\u01b0ng lu\u00f4n th\u1ecfa m\u00e3n:<\/p>\n

\n

\"\"<\/p>\n

Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a<\/p>\n

\"\"<\/p>\n

A. 11\u00a0\u00a0\u00a0 B. -1\u00a0\u00a0\u00a0 C. 1\u00a0\u00a0\u00a0 D. 0<\/p>\n

C\u00e2u 7:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t c\u1ea7u (S) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0– 2x – 2y – 4z + 5 = 0<\/p>\n

Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai?<\/p>\n

A. M\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m I(1;1;2) v\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng 2.<\/p>\n

B. Ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u00e0: (x – 1)2<\/sup>\u00a0+ (y – 1)2<\/sup>\u00a0+ (z – 2)2<\/sup>\u00a0= 1<\/p>\n

C. Di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u00e0 \u03c0<\/p>\n

D. Th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i c\u1ea7u (S) l\u00e0 4\u03c0\/3<\/p>\n

C\u00e2u 8:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho t\u1ee9 di\u1ec7n \u0111\u1ec1u ABCD c\u00f3 A(0;1;2). G\u1ecdi H l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a A l\u00ean m\u1eb7t ph\u1eb3ng (BCD). Cho H(4;-3;-2). T\u1ecda \u0111\u1ed9 t\u00e2m I v\u00e0 b\u00e1n k\u00ednh R c\u1ee7a m\u1eb7t c\u1ea7u (S) ngo\u1ea1i ti\u1ebfp t\u1ee9 di\u1ec7n ABCD l\u00e0:<\/p>\n

A. I(2; -1; 0); R = 2\u221a3 \u00a0\u00a0\u00a0 C. I(3; -2; -1); R = 3\u221a3<\/p>\n

B. I(4; -3; -2); R = 4\u221a3 \u00a0\u00a0\u00a0 D. I(3; -2; -1); R = 9<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
5-D<\/td>\n6-C<\/td>\n7-C<\/td>\n8-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 5:<\/b><\/p>\n

V\u00ec ACC\u2019A\u2019, ABCD l\u00e0 nh\u1eefng h\u00ecnh b\u00ecnh h\u00e0nh n\u00ean \u00e1p d\u1ee5ng quy t\u1eafc h\u00ecnh b\u00ecnh h\u00e0nh ta c\u00f3:<\/p>\n

\"\"<\/p>\n

T\u1eeb \u0111\u00f3 suy ra:<\/p>\n

\"\"<\/p>\n

V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/p>\n

L\u01b0u \u00fd. \u0110\u00e1p \u00e1n A sai do cho r\u1eb1ng t\u1ecda \u0111\u1ed9 c\u1ee7a C\u2019 l\u00e0 t\u1ed5ng t\u1ecda \u0111\u1ed9 c\u1ee7a hai \u0111i\u1ec3m B v\u00e0 D.<\/p>\n

\u0110\u00e1p \u00e1n B sai do cho r\u1eb1ng t\u1ecda \u0111\u1ed9 c\u1ee7a C\u2019 l\u00e0 t\u1ed5ng t\u1ecda \u0111\u1ed9 c\u1ee7a ba \u0111i\u1ec3m B, D v\u00e0 A\u2019<\/p>\n

\u0110\u00e1p \u00e1n C xu\u1ea5t ph\u00e1t t\u1eeb sai l\u1ea7m r\u1eb1ng<\/p>\n

\"\"<\/p>\n

C\u00e2u 6:<\/b><\/p>\n

\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c vect\u01a1<\/p>\n

\"\"<\/p>\n

D\u1ea5u b\u1eb1ng x\u1ea3y ra khi 2 vect\u01a1<\/p>\n

\"\"<\/p>\n

c\u00f9ng h\u01b0\u1edbng. V\u1eady \u0111\u1ed9 d\u00e0i c\u1ee7a vect\u01a1 |a\u2192<\/i>\u00a0– 2b\u2192<\/i>| \u2265 0 nh\u1ecf nh\u1ea5t b\u1eb1ng 1.<\/p>\n

Suy ra \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/p>\n

L\u01b0u \u00fd. \u0110\u00e1p \u00e1n A l\u00e0 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a<\/p>\n

\"\"<\/p>\n

\u0110\u00e1p \u00e1n B xu\u1ea5t ph\u00e1t t\u1eeb b\u1ea5t \u0111\u1eb3ng th\u1ee9c<\/p>\n

\"\"<\/p>\n

tuy nhi\u00ean \u0111\u00e1p \u00e1n B sai do d\u1ed9 d\u00e0i c\u1ee7a m\u1ed9t vect\u01a1 kh\u00f4ng \u00e2m<\/p>\n

\u0110\u00e1p \u00e1n D xu\u1ea5t ph\u00e1t t\u1eeb nh\u1eadn x\u00e9t<\/p>\n

\"\"<\/p>\n

tuy nhi\u00ean trong tr\u01b0\u1eddng h\u1ee3p n\u00e0y d\u1ea5u b\u1eb1ng kh\u00f4ng x\u1ea3y ra<\/p>\n

C\u00e2u 7:<\/b><\/p>\n

Ta vi\u1ebft l\u1ea1i ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a (S) d\u01b0\u1edbi d\u1ea1ng ch\u00ednh t\u1eafc nh\u01b0 sau:<\/p>\n

x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0– 2x – 2y – 4z + 5 = 0<\/p>\n

<=> (x2<\/sup>\u00a0– 2x + 1) +(y2<\/sub>\u00a0– 2y + 1) + (z2<\/sup>\u00a0– 4z + 4) = 1 + 1 + 4 – 5<\/p>\n

<=> (x – 1)2<\/sup>\u00a0+ (y – 1)2<\/sup>\u00a0+ (z – 2)2<\/sup>\u00a0= 1<\/p>\n

V\u1eady kh\u1eb3ng \u0111\u1ecbnh B \u0111\u00fang.<\/p>\n

M\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m I(1;1;2) v\u00e0 c\u00f3 b\u00e1n k\u00ednh R=1, do \u0111\u00f3 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a (S) l\u00e0 2R=2.<\/p>\n

V\u1eady kh\u1eb3ng \u0111\u1ecbnh A \u0111\u00fang.<\/p>\n

Th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i c\u1ea7u (S) l\u00e0<\/p>\n

\"\"<\/p>\n

V\u1eady kh\u1eb3ng \u0111\u1ecbnh D \u0111\u00fang<\/p>\n

Kh\u1eb3ng \u0111\u1ecbnh C l\u00e0 sai do nh\u1ea7m gi\u1eefa c\u00f4ng th\u1ee9c di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t c\u1ea7u v\u1edbi di\u1ec7n t\u00edch c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n. Di\u1ec7n t\u00edch m\u1eb7t c\u1ea7u (S) l\u00e0: 4\u03c0R2<\/sup>\u00a0= 4\u03c0<\/p>\n

C\u00e2u 8:<\/b><\/p>\n

Do ABCD l\u00e0 t\u1ee9 di\u1ec7n \u0111\u1ec1u n\u00ean H l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c BCD v\u00e0 I tr\u00f9ng v\u1edbi tr\u1ecdng t\u00e2m G c\u1ee7a t\u1ee9 di\u1ec7n ABCD. Ta c\u00f3:<\/p>\n

\"\"<\/p>\n

T\u1eeb \u0111\u00f3 ta c\u00f3:<\/p>\n

\"\"<\/p>\n

V\u1eady \u0111\u00e1p \u00e1n C \u0111\u00fang<\/p>\n

L\u01b0u \u00fd. \u0110\u00e1p \u00e1n A sai do nh\u1eadn \u0111\u1ecbnh I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AH<\/p>\n

\u0110\u00e1p \u00e1n B sai do cho r\u1eb1ng I tr\u00f9ng H<\/p>\n

\u0110\u00e1p \u00e1n D sai do t\u00ednh to\u00e1n nh\u1ea7m b\u00e1n k\u00ednh R<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 5:\u00a0Trong kh\u00f4ng gian Oxyz , cho h\u00ecnh h\u1ed9p ABCD.A’B’C’D’ c\u00f3 A(1;0;0), B(1;2;0), D(2;-1;0), A\u2019(5;2;2). T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m C\u2019 l\u00e0: A. (3;1;0)\u00a0\u00a0\u00a0 B. (8;3;2)\u00a0\u00a0\u00a0 C. (2;1;0)\u00a0\u00a0\u00a0 D. (6;3;2) C\u00e2u 6:\u00a0Cho hai vect\u01a1\u00a0a\u2192,\u00a0b\u2192\u00a0thay \u0111\u1ed5i nh\u01b0ng lu\u00f4n th\u1ecfa m\u00e3n: Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a A. 11\u00a0\u00a0\u00a0 B. -1\u00a0\u00a0\u00a0 C. 1\u00a0\u00a0\u00a0 D. 0 C\u00e2u 7:\u00a0Trong kh\u00f4ng gian Oxyz, […]<\/p>\n","protected":false},"author":3,"featured_media":28256,"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":"\nB\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 - Ch\u01b0\u01a1ng 3 - B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian (ph\u1ea7n 2)<\/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-chuong-3-bai-1-he-toa-do-trong-khong-gian-phan-2\/\" \/>\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 - Ch\u01b0\u01a1ng 3 - B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian (ph\u1ea7n 2)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 5:\u00a0Trong kh\u00f4ng gian Oxyz , cho h\u00ecnh h\u1ed9p ABCD.A’B’C’D’ c\u00f3 A(1;0;0), B(1;2;0), D(2;-1;0), A\u2019(5;2;2). T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m C\u2019 l\u00e0: A. (3;1;0)\u00a0\u00a0\u00a0 B. (8;3;2)\u00a0\u00a0\u00a0 C. (2;1;0)\u00a0\u00a0\u00a0 D. (6;3;2) C\u00e2u 6:\u00a0Cho hai vect\u01a1\u00a0a\u2192,\u00a0b\u2192\u00a0thay \u0111\u1ed5i nh\u01b0ng lu\u00f4n th\u1ecfa m\u00e3n: Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a A. 11\u00a0\u00a0\u00a0 B. -1\u00a0\u00a0\u00a0 C. 1\u00a0\u00a0\u00a0 D. 0 C\u00e2u 7:\u00a0Trong kh\u00f4ng gian Oxyz, […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-hinh-hoc-12-chuong-3-bai-1-he-toa-do-trong-khong-gian-phan-2\/\" \/>\n<meta property=\"og:site_name\" content=\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam\" \/>\n<meta property=\"article:published_time\" content=\"2018-04-20T10:50:09+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2018-04-23T02:44:17+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-tap-trac-nghiem-he-toa-do-trong-khong-gian-2-3.png\" \/>\n\t<meta property=\"og:image:width\" content=\"108\" \/>\n\t<meta property=\"og:image:height\" content=\"40\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"H\u00e0 Trang\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"H\u00e0 Trang\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"4 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-hinh-hoc-12-chuong-3-bai-1-he-toa-do-trong-khong-gian-phan-2\/\",\"url\":\"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-hinh-hoc-12-chuong-3-bai-1-he-toa-do-trong-khong-gian-phan-2\/\",\"name\":\"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 - 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