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{"id":22135,"date":"2018-03-16T16:49:36","date_gmt":"2018-03-16T09:49:36","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=22135"},"modified":"2018-03-16T16:49:36","modified_gmt":"2018-03-16T09:49:36","slug":"hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/","title":{"rendered":"H\u00ecnh h\u1ecdc – Ch\u01b0\u01a1ng 3 – B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian"},"content":{"rendered":"

B\u00e0i 1 (trang 80 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Trong h\u1ec7 t\u1ecda \u0111\u1ed9 (0,\u00a0i\u2192<\/i>,j\u2192<\/i>,k\u2192<\/i>) cho c\u00e1c vect\u01a1<\/p>\n

u\u2192<\/i>=i\u2192<\/i>-2j\u2192<\/i>;v\u2192<\/i>=3i\u2192<\/i>+5(j\u2192<\/i>–k\u2192<\/i>\u00a0);w\u2192<\/i>=2i\u2192<\/i>–k\u2192<\/i>+3j\u2192<\/i><\/p>\n

a) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1 \u0111\u00f3;<\/p>\n

b) T\u00ecm c\u00f4sin c\u1ee7a c\u00e1c g\u00f3c (v\u2192<\/i>,i\u2192<\/i>),(v\u2192<\/i>,j\u2192<\/i>\u00a0)v\u00e0 (v\u2192<\/i>,k\u2192<\/i>)<\/p>\n

c) T\u00ednh c\u00e1c t\u00edch v\u00f4 h\u01b0\u1edbng\u00a0u\u2192<\/i>,v\u2192<\/i>,u\u2192<\/i>.w\u2192<\/i>,v\u2192<\/i>.w\u2192<\/i><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\u00a0\"\"<\/div>\n
\n

B\u00e0i 2 (trang 80 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Trong h\u1ec7 t\u1ecda \u0111\u1ed9 (0,i\u2192<\/i>,j\u2192<\/i>,k\u2192<\/i>) cho v\u00e9ct\u01a1 t\u00f9y \u00fd kh\u00e1c\u00a00\u2192<\/i>. Ch\u1ee9ng minh r\u1eb1ng: cos2<\/sup>(u\u2192<\/i>\u00a0,i\u2192<\/i>)+cos2<\/sup>(u\u2192<\/i>,j\u2192<\/i>\u00a0)=1 .<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

<\/div>\n

Gi\u1ea3 s\u1eed\u00a0u\u2192<\/i>=(a,b,c), ta c\u00f3:<\/p>\n

\"\"<\/p>\n

B\u00e0i 3 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

T\u00ecm g\u00f3c gi\u1eefa hai vect\u01a1\u00a0u\u2192<\/i>\u00a0v\u00e0\u00a0v\u2192<\/i>\u00a0trong m\u1ed7i tr\u01b0\u1eddng h\u1ee3p<\/p>\n

a) u\u2192<\/i>=(1,1,1);v\u2192<\/i>=(2,1,-1)<\/p>\n

b) u\u2192<\/i>=3.i\u2192<\/i>+4j\u2192<\/i>;v\u2192<\/i>=-2j\u2192<\/i>+3k\u2192<\/i><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\"\"<\/p>\n

B\u00e0i 4 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Bi\u1ebft |u\u2192<\/i>|=2;|v\u2192<\/i>\u00a0|=5, g\u00f3c gi\u1eefa\u00a0u\u2192<\/i>\u00a0v\u00e0\u00a0v\u2192<\/i>\u00a0b\u1eb1ng 2\u03c0\/3. T\u00ecm k \u0111\u1ec3 vect\u01a1\u00a0p\u2192<\/i>=k\u00a0u\u2192<\/i>+17v\u2192<\/i>\u00a0vu\u00f4ng v\u1edbi vect\u01a1\u00a0q\u2192<\/i>=3u\u2192<\/i>–v\u2192<\/i>.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

<\/div>\n

\u0110\u1ec3\u00a0p\u2192<\/i>\u00a0vu\u00f4ng g\u00f3c v\u1edbi\u00a0q\u2192<\/i>\u00a0th\u00ec\u00a0p\u2192<\/i>.q\u2192<\/i>=0<\/p>\n

<=> (k\u00a0u\u2192<\/i>+17v\u2192<\/i>\u00a0)(3u\u2192<\/i>–v\u2192<\/i>\u00a0)=0<\/p>\n

<=> 3k.u\u2192<\/i>2<\/sup>-k.u\u2192<\/i>.v\u2192<\/i>+51\u00a0v\u2192<\/i>.u\u2192<\/i>-17v\u2192<\/i>2<\/sup>=0<\/p>\n

<=> 12k+5k-255-17.25=0 (v\u00ec\u00a0u\u2192<\/i>.v\u2192<\/i>=|u\u2192<\/i>||v\u2192<\/i>\u00a0| cos\u2061(u\u2192<\/i>,v\u2192<\/i>)=-5<\/p>\n

<=> 17k=680 <=> k = 40. V\u1eady K = 40 l\u00e0 gi\u00e1 tr\u1ecb c\u1ea7n t\u00ecm.<\/p>\n

B\u00e0i 5 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Trong kh\u00f4ng gian t\u1ecda \u0111\u1ed9 Oxyz, cho \u0111i\u1ec3m M (a, b, c).<\/p>\n

a) T\u00ecm t\u1ecda \u0111\u1ed9 h\u00ecnh chi\u1ebfu (vu\u00f4ng g\u00f3c) c\u1ee7a M tr\u00ean c\u00e1c m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 v\u00e0 c\u00e1c tr\u1ee5c t\u1ecda \u0111\u1ed9.<\/p>\n

b) T\u00ecm kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m M \u0111\u1ebfn c\u00e1c m\u1eb7t t\u1ecda \u0111\u1ed9, \u0111\u1ebfn c\u00e1c tr\u1ee5c t\u1ecda \u0111\u1ed9.<\/p>\n

c) T\u00ecm t\u1ecda \u0111\u1ed9 c\u00e1c \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi M qua c\u00e1c m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\"Gi\u1ea3i<\/p>\n

a) H\u00ecnh chi\u1ebfu c\u1ee7a M l\u00ean mp(Oxy) t\u1ecda \u0111\u1ed9 l\u00e0: (a, b, 0)<\/p>\n

T\u01b0\u01a1ng t\u1ef1, h\u00ecnh chi\u1ebfu c\u1ee7a M l\u00ean mp(Oxz) v\u00e0 mp(Oyz) l\u1ea7n l\u01b0\u1ee3t c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0: (a, 0, c) v\u00e0 (0, b, c).<\/p>\n

H\u00ecnh chi\u1ebfu c\u1ee7a M l\u00ean c\u00e1c tr\u1ee5c Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t c\u00f3 t\u1ecda \u0111\u1ed1 l\u00e0: (a, 0, 0), (0, b, 0), (0, 0, c)<\/p>\n

b) Ta c\u00f3: d(M,(Oxy)=|c|,d(M,(Oxz)=|b|,d(M,9Oyz))=|a|<\/p>\n

\"Gi\u1ea3i<\/p>\n

c) \u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a M = (a, b, c) qua c\u00e1c m\u1eb7t ph\u1eb3ng (Oxy), (Oxz) v\u00e0 (Oyz) l\u1ea7n l\u01b0\u1ee3t c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0<\/p>\n

(a, b, -c); (a, -b, c) v\u00e0 (-a, b, c)<\/p>\n

B\u00e0i 6 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho hai \u0111i\u1ec3m A(x1<\/sub>,y1<\/sub>,z1<\/sub>) v\u00e0 B(x2<\/sub>,y2<\/sub>,z2<\/sub>). T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m M chia \u0111o\u1ea1n th\u1eb3ng AB theo t\u1ec9 s\u1ed1 K (t\u1ee9c\u00a0MA\u2192<\/i>=k (MB\u2192<\/i>) k \u2260 1<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Gi\u1ea3 s\u1eed M=(x,y,z) khi \u0111\u00f3\u00a0MA\u2192<\/i>=(x1<\/sub>-x,y1<\/sub>-y,z1<\/sub>-z) v\u00e0 k\u00a0MB\u2192<\/i>=(k(x2<\/sub>-x);k(y2<\/sub>-y);k(z2<\/sub>-z))<\/p>\n

\u0110\u1ec3\u00a0MA\u2192<\/i>=k\u00a0MB\u2192<\/i>\u00a0th\u00ec<\/p>\n

\n
\u00a0\"\"<\/div>\n
\n

B\u00e0i 7 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho h\u00ecnh b\u00ecnh h\u00e0nh ABCD, bi\u1ebft A(-3, -2, 0), B(3, -3, 1), C(5, 0, 2). T\u00ecm t\u1ecda \u0111\u1ed9 \u0111\u1ec9nh D v\u00e0 g\u00f3c gi\u1eefa hai vect\u01a1\u00a0AC\u2192<\/i>\u00a0v\u00e0\u00a0BD\u2192<\/i><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

G\u1ecdi D = (x, y, z), \u0111\u1ec3 ABCD l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh th\u00ec\u00a0AD\u2192<\/i>=BC\u2192<\/i><\/p>\n

Ta c\u00f3:\u00a0AD\u2192<\/i>=(x+3,y+2,z),BC\u2192<\/i>=(2,3,1)<\/p>\n

Khi \u0111\u00f3, ta c\u00f3\u00a0BD\u2192<\/i>=(-4,4,0)v\u00e0\u00a0AC\u2192<\/i>=(8,2,2)<\/p>\n

Suy ra<\/p>\n

\"\"<\/p>\n

B\u00e0i 8 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

a) T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m M thu\u1ed9c Ox sao cho M c\u00e1ch \u0111\u1ec1u hai \u0111i\u1ec3m A(1, 2, 3) v\u00e0 B(-3, -3, 2)<\/p>\n

b) Cho ba \u0111i\u1ec3m A(2, 0, 4) v\u00e0 B(4, \u221a3, 5) v\u00e0 C(sin 5t, cos 3t, sin 3t). T\u00ecm t \u0111\u1ec3 AB vu\u00f4ng g\u00f3c v\u1edbi OC (O l\u00e0 g\u1ed1c t\u1ecda \u0111\u1ed9).<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) G\u1ecdi M = (a, 0, 0) thu\u1ed9c Ox th\u1ecfa m\u00e3n MA = MB.<\/p>\n

Ta c\u00f3: MA2<\/sup>=(1-a)2<\/sup>+4+9=a2<\/sup>-2a+14<\/p>\n

MB2<\/sup>=(3+a)2<\/sup>+9+4=a2<\/sup>+6a+22<\/p>\n

\u0110\u1ec3 MA = MB th\u00ec MA2<\/sup>=MB2<\/sup>\u00a0<=> a2<\/sup>-2a+14=a2<\/sup>+6a+22 <=> a = -1<\/p>\n

V\u1eady M = (-1, 0, 0) l\u00e0 \u0111i\u1ec3m c\u1ea7n t\u00ecm.<\/p>\n

b) Ta c\u00f3\u00a0AB\u2192<\/i>=(2,\u221a3,1),OC\u2192<\/i>=(sin\u20615t+\u221a3 cos\u20613t+sin\u20613t=0<\/p>\n

\u0110\u1ec3 AB\u22a5OC th\u00ec\u00a0AB\u2192<\/i>.OC\u2192<\/i>=0 <=>2 sin\u20615t+\u221a3 cos\u20613t+sin\u20613t=0<\/p>\n

\"\"<\/p>\n

V\u1edbi k, n thu\u1ed9c z<\/p>\n

\"\"<\/p>\n

B\u00e0i 9 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao)<\/b><\/p>\n

X\u00e9t s\u1ef1 \u0111\u1ed3ng ph\u1eb3ng c\u1ee7a ba vect\u01a1\u00a0u\u2192<\/i>,v\u2192<\/i>\u00a0v\u00e0\u00a0w\u2192<\/i>\u00a0trong m\u1ed7i tr\u01b0\u1eddng h\u1ee3p sau:<\/p>\n

a) u\u2192<\/i>(4,3,4);v\u2192<\/i>=(2; -1,2);w\u2192<\/i>(1,2,1)<\/p>\n

b) u\u2192<\/i>(1,-1,1),v\u2192<\/i>=(0,1,2),w\u2192<\/i>(4,2,3)<\/p>\n

c) u\u2192<\/i>(4,2,5),v\u2192<\/i>=(3,1,3),w\u2192<\/i>(2,0,1)<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\u0110\u1ec3 x\u00e9t t\u00ednh \u0111\u1ed3ng ph\u1eb3ng c\u1ee7a\u00a0u\u2192<\/i>,v\u2192<\/i>\u00a0v\u00e0\u00a0w\u2192<\/i>\u00a0ta x\u00e9t [u\u2192<\/i>,v\u2192<\/i>\u00a0].w\u2192<\/i><\/p>\n

N\u1ebfu [u\u2192<\/i>,v\u2192<\/i>\u00a0].w\u2192<\/i>=0 th\u00ec\u00a0u\u2192<\/i>,v\u2192<\/i>,w\u2192<\/i>\u00a0\u0111\u1ed3ng ph\u1eb3ng.<\/p>\n

a) Ta c\u00f3<\/p>\n

\"\"<\/p>\n

N\u00ean [u\u2192<\/i>,v\u2192<\/i>\u00a0].w\u2192<\/i>=10.1+0.2+(-10).1=0<\/p>\n

V\u1eady\u00a0u\u2192<\/i>,v\u2192<\/i>\u00a0v\u00e0\u00a0w\u2192<\/i>\u00a0\u0111\u1ed3ng ph\u1eb3ng<\/p>\n

b) u\u2192<\/i>,v\u2192<\/i>\u00a0v\u00e0\u00a0w\u2192<\/i>\u00a0kh\u00f4ng \u0111\u1ed3ng ph\u1eb3ng.<\/p>\n

c) u\u2192<\/i>,v\u2192<\/i>\u00a0v\u00e0\u00a0w\u2192<\/i>\u00a0\u0111\u1ed3ng ph\u1eb3ng.<\/p>\n

B\u00e0i 10 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b>\u00a0Cho ba \u0111i\u1ec3m A(1, 0, 0), B(0, 0, 1), C(2, 1, 1)<\/p>\n

a) Ch\u1ee9ng minh A, B, C kh\u00f4ng th\u1eb3ng h\u00e0ng.<\/p>\n

b) T\u00ecm t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m D \u0111\u1ec3 ABCD l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh.<\/p>\n

c) T\u00ednh \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c ABC.<\/p>\n

d) T\u00ednh \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c ABC k\u1ebb t\u1eeb \u0111\u1ec9nh A.<\/p>\n

e) T\u00ednh c\u00e1c g\u00f3c c\u1ee7a tam gi\u00e1c ABC.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) Ta c\u00f3\u00a0AB\u2192<\/i>=(-1,0,1),\u00a0AC\u2192<\/i>=(1,1,1), ta th\u1ea5y\u00a0AB\u2192<\/i>\u00a0v\u00e0\u00a0AC\u2192<\/i>\u00a0kh\u00f4ng c\u00f9ng Ph\u01b0\u01a1ng n\u00ean A, B, C kh\u00f4ng th\u1eb3ng h\u00e0ng.<\/p>\n

b) G\u1ecdi D = (x, y, z), ta c\u00f3\u00a0AD\u2192<\/i>=(x-1,y,z),BC\u2192<\/i>=(2,1,0). \u0110\u1ec3 ABCD l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh th\u00ec\u00a0AC\u2192<\/i>=BC\u2192<\/i>, hay<\/p>\n

\"\"<\/p>\n

c) Chu vi \u0394ABC l\u00e0: P=AB+BC+AC=\u221a2+\u221a5+\u221a3<\/p>\n

Di\u1ec7n t\u00edch \u0394ABC l\u00e0:<\/p>\n

\"\"<\/p>\n

B\u00e0i 11 (trang 81 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b>\u00a0\\Cho b\u1ed1n \u0111i\u1ec3m A(1, 0, 0), B(0, 1, 0), C(0, 0, 1) v\u00e0 D(-2, 1, -2).<\/p>\n

a) Ch\u1ee9ng minh r\u1eb1ng A, B, C, D l\u00e0 b\u1ed1n \u0111\u1ec9nh c\u1ee7a m\u1ed9t h\u00ecnh t\u1ee9 di\u1ec7n.<\/p>\n

b) T\u00ednh g\u00f3c b\u1edfi c\u00e1c c\u1ea1nh \u0111\u1ed1i c\u1ee7a t\u1ee9 di\u1ec7n \u0111\u00f3. T\u00ednh th\u1ec3 t\u00edch t\u1ee9 di\u1ec7n ABCD v\u00e0 \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng cao c\u1ee7a t\u1ee9 di\u1ec7n k\u1ebb t\u1eeb \u0111\u1ec9nh A.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) Ta c\u00f3\u00a0AB\u2192<\/i>=(-1,1,0),AC\u2192<\/i>=(-1,0,1),AD\u2192<\/i>=(-3,1,2) n\u00ean ta c\u00f3: [AB\u2192<\/i>,AC\u2192<\/i>\u00a0]=(1,1,1), suy ra [AB\u2192<\/i>,AC\u2192<\/i>\u00a0].AD\u2192<\/i>=-4 \u2260 0. V\u1eady A, B, C, d kh\u00f4ng \u0111\u1ed3ng ph\u1eb3ng, hay A, B, C, D l\u00e0 b\u1ed1n \u0111\u1ec9nh c\u1ee7a t\u1ee9 di\u1ec7n.<\/p>\n

b)Ta c\u00f3:<\/p>\n

\"\"<\/p>\n

B\u00e0i 12 (trang 82 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho h\u00ecnh ch\u00f3p S.ABC c\u00f3 \u0111\u01b0\u1eddng cao SA = h, \u0111\u00e1y l\u00e0 tam gi\u00e1c ABC vu\u00f4ng t\u1ea1o C, AC = b, Bc = a. g\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AC v\u00e0 N l\u00e0 trung \u0111i\u1ec3m sao cho:<\/p>\n

\"\"<\/p>\n

a) T\u00ednh \u0111\u1ed9 d\u00e0i MN.<\/p>\n

b) T\u00ecm s\u1ef1 li\u00ean h\u1ec7 gi\u1eefa a, b, h \u0111\u1ec3 MN vu\u00f4ng g\u00f3c v\u1edbi SB.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz sao cho: g\u1ed1c t\u1ecda \u0111\u1ed9 O tr\u00f9ng v\u1edbi A, Ox l\u00e0 tia AC. Khi \u0111\u00f3, ta c\u00f3:<\/p>\n

A = (0, 0, 0), B(b, a, 0), C = (b, 0, 0), S = (0, 0, h) v\u00e0<\/p>\n

SB\u2192<\/i>=(b,a,-h)<\/p>\n

a) Ta c\u00f3<\/p>\n

\"Gi\u1ea3i<\/p>\n

b) \u0110\u1ec3 MN \u22a5 SB th\u00ec\u00a0MN\u2192<\/i>.SB\u2192<\/i>=0<\/p>\n

\"\"<\/p>\n

B\u00e0i 13 (trang 82 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

T\u00ecm t\u1ecda \u0111\u1ed9 t\u00e2m v\u00e0 b\u00e1n k\u00ednh m\u1eb7t c\u1ea7u sau \u0111\u00e2y:<\/p>\n

a) x2<\/sup>+y2<\/sup>+z2<\/sup>-8x+2y+1=0<\/p>\n

b) 3x2<\/sup>+3y2<\/sup>+3z2<\/sup>+6x-3y=15z-2=0<\/p>\n

c) 9x2<\/sup>+9y2<\/sup>+9z2<\/sup>-6x+18y+1=0<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) Ta c\u00f3: x2<\/sup>+y2<\/sup>+z2<\/sup>-8x+2y+1=0<\/p>\n

<=> x2<\/sup>+y2<\/sup>+z2<\/sup>-y+5z-2\/3=0<\/p>\n

N\u00ean m\u1eb7t c\u1ea7u c\u00f3 t\u00e2m la I(4, -1, 0) v\u00e0 b\u00e1n k\u00ednh R = 4.<\/p>\n

b) Ta c\u00f3: 3x2<\/sup>+3y2<\/sup>+3z2<\/sup>+6x-3y+15z-2=0<\/p>\n

\"\"<\/p>\n

c) Ta c\u00f3: 9x2<\/sup>+9y2<\/sup>+9z2<\/sup>-6x+18y+1=0<\/p>\n

\"\"<\/p>\n

N\u00ean m\u1eb7t c\u1ea7u c\u00f3 t\u00e2m l\u00e0 I = (1\/3; -1,0) v\u00e0 b\u00e1n k\u00ednh R = 1.<\/p>\n

B\u00e0i 14 (trang 82 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Trong m\u1ed7i tr\u01b0\u1eddng h\u1ee3p sau, vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u:<\/p>\n

a) \u0110i qua ba \u0111i\u1ec3m A(0, 8, 0), B(4, 6, 2), C(0, 12, 4) v\u00e0 c\u00f3 t\u00e2m n\u1eb1m tr\u00ean mp(Oyz).<\/p>\n

b) C\u00f3 b\u00e1n k\u00ednh b\u1eb1ng 2, ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t ph\u1eb3ng (Oyz) v\u00e0 c\u00f3 t\u00e2m n\u1eb1m tr\u00ean tia Ox.<\/p>\n

c) C\u00f3 t\u00e2m I(1, 2, 3) v\u00e0 ti\u1ebfp x\u00fac v\u1edbi mp(Oyz).<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) V\u00ec t\u00e2m m\u1eb7t c\u1ea7u n\u1eb1m tr\u00ean mp(Oyz) n\u00ean ta g\u1ecdi t\u00e2m m\u1eb7t c\u1ea7u l\u00e0 I = (0, b, c).<\/p>\n

V\u00ec c\u1ea7u \u0111i qua A, B, C n\u00ean ta c\u00f3 h\u1ec7:<\/p>\n

\"\"<\/p>\n

V\u1eady Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u l\u00e0: x2<\/sup>+(x-7)2<\/sup>+(z-5)2<\/sup>=26<\/p>\n

b) V\u00ec t\u00e2m m\u1eb7t c\u1ea7u m\u1eb7t c\u1ea7u n\u1eb1m tr\u00ean Ox n\u00ean ta g\u1ecdi t\u00e2m m\u1eb7t c\u1ea7u l\u00e0 I(a, 0, 0). V\u00ec m\u1eb7t c\u1ea7u ti\u1ebfp x\u00fac v\u1edbi (Oyz) n\u00ean b\u00e1n k\u00ednh R = d(I, (Oyz)= |a| theo b\u00e0i ra ta c\u00f3 a = 2.<\/p>\n

V\u1eady Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u l\u00e0: (x-2)2<\/sup>+y2<\/sup>+z2<\/sup>=4<\/p>\n

c) V\u00ec m\u1eb7t c\u1ea7u ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t ph\u1eb3ng (Oyz) v\u00e0 c\u00f3 t\u00e2m I(1, 2, 3) n\u00ean ta c\u00f3 b\u00e1n k\u00ednh m\u1eb7t c\u1ea7u l\u00e0: R = d(I, (Oyz)) = 1<\/p>\n

V\u1eady Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u l\u00e0: (x-1)2<\/sup>+(y-2)2<\/sup>+(z-3)2<\/sup>=1<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 1 (trang 80 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Trong h\u1ec7 t\u1ecda \u0111\u1ed9 (0,\u00a0i\u2192,j\u2192,k\u2192) cho c\u00e1c vect\u01a1 u\u2192=i\u2192-2j\u2192;v\u2192=3i\u2192+5(j\u2192–k\u2192\u00a0);w\u2192=2i\u2192–k\u2192+3j\u2192 a) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1 \u0111\u00f3; b) T\u00ecm c\u00f4sin c\u1ee7a c\u00e1c g\u00f3c (v\u2192,i\u2192),(v\u2192,j\u2192\u00a0)v\u00e0 (v\u2192,k\u2192) c) T\u00ednh c\u00e1c t\u00edch v\u00f4 h\u01b0\u1edbng\u00a0u\u2192,v\u2192,u\u2192.w\u2192,v\u2192.w\u2192 L\u1eddi gi\u1ea3i: \u00a0 B\u00e0i 2 (trang 80 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Trong h\u1ec7 […]<\/p>\n","protected":false},"author":3,"featured_media":22136,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1302],"tags":[1392,1393],"yoast_head":"\nH\u00ecnh h\u1ecdc - Ch\u01b0\u01a1ng 3 - B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian<\/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\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"H\u00ecnh h\u1ecdc - Ch\u01b0\u01a1ng 3 - B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 1 (trang 80 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Trong h\u1ec7 t\u1ecda \u0111\u1ed9 (0,\u00a0i\u2192,j\u2192,k\u2192) cho c\u00e1c vect\u01a1 u\u2192=i\u2192-2j\u2192;v\u2192=3i\u2192+5(j\u2192–k\u2192\u00a0);w\u2192=2i\u2192–k\u2192+3j\u2192 a) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1 \u0111\u00f3; b) T\u00ecm c\u00f4sin c\u1ee7a c\u00e1c g\u00f3c (v\u2192,i\u2192),(v\u2192,j\u2192\u00a0)v\u00e0 (v\u2192,k\u2192) c) T\u00ednh c\u00e1c t\u00edch v\u00f4 h\u01b0\u1edbng\u00a0u\u2192,v\u2192,u\u2192.w\u2192,v\u2192.w\u2192 L\u1eddi gi\u1ea3i: \u00a0 B\u00e0i 2 (trang 80 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Trong h\u1ec7 […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/\" \/>\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=\"2018-03-16T09:49:36+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/1-117.png\" \/>\n\t<meta property=\"og:image:width\" content=\"490\" \/>\n\t<meta property=\"og:image:height\" content=\"509\" \/>\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=\"9 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/\",\"url\":\"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/\",\"name\":\"H\u00ecnh h\u1ecdc - Ch\u01b0\u01a1ng 3 - B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian\",\"isPartOf\":{\"@id\":\"https:\/\/lop12.edu.vn\/#website\"},\"datePublished\":\"2018-03-16T09:49:36+00:00\",\"dateModified\":\"2018-03-16T09:49:36+00:00\",\"author\":{\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\"},\"breadcrumb\":{\"@id\":\"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/lop12.edu.vn\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"H\u00ecnh h\u1ecdc – Ch\u01b0\u01a1ng 3 – B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/lop12.edu.vn\/#website\",\"url\":\"https:\/\/lop12.edu.vn\/\",\"name\":\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12\",\"description\":\"\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/lop12.edu.vn\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\",\"name\":\"H\u00e0 Trang\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"caption\":\"H\u00e0 Trang\"},\"url\":\"https:\/\/lop12.edu.vn\/author\/trangvth2\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"H\u00ecnh h\u1ecdc - Ch\u01b0\u01a1ng 3 - B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-1-he-toa-do-trong-khong-gian\/","og_locale":"en_US","og_type":"article","og_title":"H\u00ecnh h\u1ecdc - Ch\u01b0\u01a1ng 3 - B\u00e0i 1: H\u1ec7 t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian","og_description":"B\u00e0i 1 (trang 80 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Trong h\u1ec7 t\u1ecda \u0111\u1ed9 (0,\u00a0i\u2192,j\u2192,k\u2192) cho c\u00e1c vect\u01a1 u\u2192=i\u2192-2j\u2192;v\u2192=3i\u2192+5(j\u2192–k\u2192\u00a0);w\u2192=2i\u2192–k\u2192+3j\u2192 a) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1 \u0111\u00f3; 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