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{"id":22114,"date":"2018-03-16T16:20:58","date_gmt":"2018-03-16T09:20:58","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=22114"},"modified":"2018-03-16T16:20:58","modified_gmt":"2018-03-16T09:20:58","slug":"hinh-hoc-chuong-3-bai-3-chuong-trinh-mat-phang","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-3-bai-3-chuong-trinh-mat-phang\/","title":{"rendered":"H\u00ecnh hoc – Ch\u01b0\u01a1ng 3 – B\u00e0i 3: Ch\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng"},"content":{"rendered":"

B\u00e0i 15 (trang 89 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng trong m\u1ed7i tr\u01b0\u1eddng h\u1ee3p sau:<\/p>\n

a) \u0110i qua ba \u0111i\u1ec3m M(2, 0, -1), N(1, -2, 3), P(0, 1, 2).<\/p>\n

b) \u0110i qua hai \u0111i\u1ec3m A(1, 1, -1), B(5, 2, 1) v\u00e0 song song v\u1edbi tr\u1ee5c Oz.<\/p>\n

c) \u0110i qua \u0111i\u1ec3m (3, 2, -1) v\u00e0 song song v\u1edbi m\u1eb7t ph\u1eb3ng c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh: x \u2013 5y + z=0<\/p>\n

d) \u0110i qua hai \u0111i\u1ec3m A(0; 1; 1), B(-1; 0; 2) v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng x – y + z + 1 = 0<\/p>\n

e) \u0110i qua \u0111i\u1ec3m M(a, b, c) (abc \u2260 0) v\u00e0 song song v\u1edbi m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed1.<\/p>\n

g) \u0110i qua \u0111i\u1ec3m G(1, 2, 3) v\u00e0 c\u1eaft c\u00e1c tr\u1ee5c t\u1ecda \u0111\u1ed9 t\u1ea1i c\u00e1c \u0111i\u1ec3m A, B, C sao cho G l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c ABC<\/p>\n

h) \u0110i qua \u0111i\u1ec3m H(2, 1, 1) v\u00e0 c\u1eaft tr\u1ee5c t\u1ecda \u0111\u1ed9 t\u1ea1i c\u00e1c \u0111i\u1ec3m A, B, C sao cho H l\u00e0 trung tr\u1ef1c t\u00e2m tam gi\u00e1c ABC.<\/p>\n

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

a) M\u1eb7t ph\u1eb3ng (MNP) nh\u1eadn vect\u01a1 [MN\u2192<\/i>,MP\u2192<\/i>\u00a0] l\u00e0 vect\u01a1 ph\u00e1p tuy\u1ebfn. ta c\u00f3\u00a0MN\u2192<\/i>=(1,-2,4),MP\u2192<\/i>=(-2,1,3) n\u00ean [\u00a0MN\u2192<\/i>,MP\u2192<\/i>\u00a0]=(-10,-5,-5). V\u1eady mp(MNP) \u0111i qua M(2, 0, -1) v\u00e0 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0 (-10, -5, -5) n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: -10(x-2)-5y-5(z+1)=0 <=> 2x + y + z \u2013 3 = 0<\/p>\n

b) V\u00ec m\u1eb7t ph\u1eb3ng \u0111i qua AB v\u00e0 song song v\u1edbi OZ n\u00ean n\u00f3 c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0\u00a0n\u2192<\/i>=[AB\u2192<\/i>,k\u2192<\/i>], v\u1edbi\u00a0AB\u2192<\/i>=(4,1,2),k\u2192<\/i>=(0,0,1) n\u00ean\u00a0n\u2192<\/i>=(1,-4,0)<\/p>\n

V\u1eady m\u1eb7t ph\u0103ng c\u1ea7n t\u00ecm \u0111i qua A(1, 1, -1) v\u00e0 c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0\u00a0n\u2192<\/i>=(1,-4,0) n\u00ean ta c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: 1(x-1)-4(y-1)+0(z+1)=0<\/p>\n

<=> x \u2013 4y + 3 = 0<\/p>\n

c) V\u00ec m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm song song v\u1edbi mp: x \u2013 5y + z = 0, n\u00ean n\u00f3 c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng: x \u2013 5y + z + D = 0, m\u00e0 m\u1eb7t ph\u1eb3ng n\u00e0y l\u1ea1i \u0111i qua \u0111i\u1ec3m (3, 2, -1) n\u00ean ta c\u00f3:<\/p>\n

3-5.2+(-1)+D=0 <=> D = 8<\/p>\n

V\u1eady Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng song song v\u1edbi nhau th\u00ec c\u00f3 c\u00f9ng vect\u01a1 ph\u00e1p tuy\u1ebfn, nen m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0 n \u20d7=(1,-5,1) n\u00ean n\u00f3 c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: 1(x-3)-5(y-2)+1(z+1)=0 <=> x-5y+z+8=0<\/p>\n

d) V\u00ec m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm \u0111i qua AB v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng: x \u2013 y + z + 1 = 0 n\u00ean c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0\u00a0n\u2192<\/i>=[AB\u2192<\/i>,n1<\/sub>\u2192<\/i>], v\u1edbi\u00a0AB\u2192<\/i>=(-1,-1,1) v\u00e0\u00a0n1<\/sub>\u2192<\/i>=(1,-1,1) l\u00e0 vect\u01a1 ph\u00e1p tuy\u1ebfn c\u1ee7a m\u1eb7t ph\u1eb3ng: x-y+z+1=0. Suy ra\u00a0n\u2192<\/i>=(0,2,2).<\/p>\n

V\u1eady ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm l\u00e0: y + z \u2013 2 = 0<\/p>\n

e) N\u1ebfu m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm song song v\u1edbi mp(Oxy) th\u00ec n\u00f3 c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0\u00a0n\u2192<\/i>=(0,0,1), m\u1eb7t kh\u00e1c m\u1eb7t ph\u1eb3ng n\u00e0y \u0111i qua \u0111i\u1ec3m M(a, b, c) n\u00ean c\u00f3 ph\u01b0\u01a1ng t\u00ecnh l\u00e0: z \u2013 c = 0.<\/p>\n

T\u01b0\u01a1ng t\u1ef1, n\u1ebfu m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm \u0111i qua M(a, b, c) v\u00e0 song song v\u1edbi mp(Oxz) th\u00ec c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh: y \u2013 b = 0.<\/p>\n

g) Gi\u1ea3 s\u1eed 3 giao \u0111i\u1ec3m A, B, C c\u1ee7a m\u1eb7t ph\u1eb3ng v\u1edbi 3 tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0 A(a, 0, 0), B(0, b, 0), C(0, 0, c). v\u00ec G(1, 2, 3) l\u00e0 tr\u1ecdng r\u00e2m c\u1ee7a \u0394ABC n\u00ean ta c\u00f3:<\/p>\n

\"\"<\/p>\n

n\u00ean ta c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh mp(ABC) theo \u0111o\u1ea1n ch\u1eafn l\u00e0 :<\/p>\n

\"\"<\/p>\n

h) Gi\u1ea3 s\u1eed 3 giao \u0111i\u1ec3m A, B, C c\u1ee7a m\u1eb7t ph\u1eb3ng v\u1edbi 3 tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0: A(a, 0, 0); B(0, b, 0); C(0, 0, c). v\u00ec H(2, 1, 1) l\u00e0 tr\u1ef1c t\u00e2m \u0394ABC n\u00ean.<\/p>\n

\"\"<\/p>\n

khi \u0111\u00f3, ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (ABC) vi\u1ebft theo \u0111o\u1ea1n ch\u1eafn l\u00e0:<\/p>\n

 <\/p>\n

\"\"<\/p>\n

M\u1eb7t kh\u00e1c, m\u1eb7t ph\u1eb3ng n\u00e0y \u0111i qua H(2, 1, 1) n\u00ean ta c\u00f3:<\/p>\n

2.2 + 1 + 1 = 2a <=> a = 3<\/p>\n

V\u1eady Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm l\u00e0: 2x+y+z-6=0<\/p>\n

B\u00e0i 16 (trang 89 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

X\u00e9t c\u00e1c v\u1ecb tr\u1ecb t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a m\u1ed7i c\u1eb7p ph\u1eb3ng cho b\u1edfi c\u00e1c Ph\u01b0\u01a1ng tr\u00ecnh sau.<\/p>\n

a) x+2y-z+5=0 v\u00e0 2x+3y-7z-4=0<\/p>\n

b) x-2y+z-3=0 v\u00e0 2x-y+4z-2=0<\/p>\n

c) x+y+z-1=0 v\u00e0 2x+2y+2z+3=0<\/p>\n

d) 3x-2y+3z+5=0 v\u00e0 9x-6y-9z-5=0<\/p>\n

e) x-y+2z-4=0 v\u00e0 10x-10y+20z-40=0<\/p>\n

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

a) Hai m\u1eb7t ph\u1eb3ng c\u1eaft nhau, v\u00ec 1: 2: (-1) \u2260 2: 3: (-7)<\/p>\n

b) Hai m\u1eb7t ph\u1eb3ng c\u1eaft nhau, v\u00ec: 1: (-2): 1 \u2260 2: (-1): 4<\/p>\n

c) Hai m\u1eb7t ph\u1eb3ng song song, v\u00ec: 1\/2=1\/2=1\/2 \u2260 -1\/3<\/p>\n

d) Hai m\u1ea1t ph\u1eb3ng c\u1eaft nhau, v\u00ec: 3: (-2): 3 \u2260 9: (-6): (-9)<\/p>\n

e) Hai m\u1eb7t ph\u1eb3ng trung nhau, v\u00ec: 1\/10=-1\/(-10)=2\/20=-4\/(-40).<\/p>\n

B\u00e0i 17 (trang 89 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b>\u00a0X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a m v\u00e0 n \u0111\u1ec3 m\u1ed7i c\u1eb7p sau \u0111\u00e2y song song.<\/p>\n

a) 2x+ny+2z+3=0 v\u00e0 mx+ny+2z+8=0<\/p>\n

b) 2x+y+mz-2=0 v\u00e0 x+ny+2z+7=0<\/p>\n

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

a) \u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 hai m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho song song v\u1edbi nhau l\u00e0:<\/p>\n

\"\"<\/p>\n

b) \u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 hai m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho song song v\u1edbi nhau l\u00e0:<\/p>\n

\"\"<\/p>\n

B\u00e0i 18 (trang 90 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho hai m\u1eb7t ph\u1eb3ng c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u1ea7n l\u01b0\u1ee3t l\u00e0:<\/p>\n

2x-my+3z-6+m=0 v\u00e0 (m+3)x-2y+(5m+1)z-1=0<\/p>\n

V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m th\u00ec:<\/p>\n

a) Hai m\u1eb7t ph\u1eb3ng song song.<\/p>\n

b) Hai m\u1eb7t ph\u1eb3ng tr\u00f9ng nhau.<\/p>\n

c) Hai m\u1eb7t ph\u1eb3ng \u0111\u00f3 c\u1eaft nhau.<\/p>\n

d) Hai m\u1eb7t ph\u1eb3ng vu\u00f4ng g\u00f3c.<\/p>\n

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

C\u00e1c h\u1ec7 s\u1ed1 c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng: 2x-my+3z-6+m=0 l\u00e0: A = 2, B = -m, C = 3, D = m \u2013 6<\/p>\n

C\u00e1c h\u1ec7 s\u1ed1 c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng l\u00e0: (m+3)x-2y+(5m+1)z-1=0: A\u2019 = m + 3; B\u2019 = -2, C\u2019 = 5m + 1; d\u2019 = -10<\/p>\n

a) \u0110\u1ec3 hai m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho song song v\u1edbi nhau l\u00e0:<\/p>\n

\"\"<\/p>\n

H\u1ec7 n\u00e0y v\u00f4 nghi\u1ec7m, n\u00ean kh\u00f4ng c\u00f3 m \u0111\u1ec3 hai m\u1eb7t ph\u1eb3ng song song.<\/p>\n

b) \u0110\u1ec3 hai m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho tr\u1eebng v\u1edbi nhau l\u00e0:<\/p>\n

\"\"<\/p>\n

\n
<\/div>\n<\/div>\n

c) Hai m\u1eb7t ph\u1eb3ng c\u1eaft nhau khi v\u00e0 ch\u1ec9 khi ch\u00fang kh\u00f4ng tr\u00f9ng nhau (v\u00ec theo c\u00e2u a, hai m\u1eb7t n\u00e0y kh\u00f4ng th\u1ec3 song song v\u1edbi nhau). Theo c\u00e2u b) ta suy ra gi\u00e1 tr\u1ecb m \u0111\u1ebb hai m\u1eb7t ph\u1eb3ng c\u1eaft nhau l\u00e0: m \u2260 1<\/p>\n

d) Hai vect\u01a1 ph\u00e1p tuy\u1ebfn c\u1ee7a hai m\u1eb7t ph\u1eb3ng l\u1ea7n l\u01b0\u1ee3t l\u00e0:\u00a0n1<\/sub>\u2192<\/i>(2,-m,3) v\u00e0\u00a0n2<\/sub>\u2192<\/i>(m+3),-2,5m+1)<\/p>\n

\u0110\u1ec3 hai m\u1eb7t ph\u1eb3ng vu\u00f4ng g\u00f3c th\u00ec\u00a0n1<\/sub>\u2192<\/i>\u22a5n2<\/sub>\u2192<\/i>\u00a0hay\u00a0n1<\/sub>\u2192<\/i>.n2<\/sub>\u2192<\/i>=0<\/p>\n

<=> 2(m+3)+2m+3(5m+1)=0 <=> m=-9\/19.<\/p>\n

V\u1eady m =-9\/19 l\u00e0 gi\u00e1 tr\u1ecb c\u1ea7n t\u00ecm.<\/p>\n

B\u00e0i 19 (trang 90 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b>\u00a0T\u00ecm t\u1eadp h\u1ee3p c\u00e1c di\u1ec3m c\u00e1ch \u0111\u1ec1u 2 m\u1eb7t ph\u1eb3ng (\u03b1) v\u00e0 (\u03b1’) trong m\u1ed7i tr\u01b0\u1eddng h\u1ee3p sau:<\/p>\n

a) (\u03b1): 2x-y+4z+5=0, (\u03b1’ ):3x+5y-z-1=0<\/p>\n

b) (\u03b1):2x+y-2z-1=0, (\u03b1’ ):6x-3y+2z=0<\/p>\n

c) (\u03b1):x+2y+z-1=0, (\u03b1’ ):x+2y+z+5=0<\/p>\n

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

a) G\u1ecdi \u0111i\u1ec3m M(x, y, z) l\u00e0 \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u (\u03b1) v\u00e0 (\u03b1’ ),khi \u0111\u00f3:<\/p>\n

\"\"<\/p>\n

V\u1eady qu\u1ef9 t\u00edch c\u00e1c \u0111i\u1ec3m M c\u00e1ch \u0111\u1ec1u 2 m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho l\u00e0 m\u1eb7t ph\u1eb3ng c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00e0 (2)<\/p>\n

b) C\u00e1ch gi\u1ea3i t\u01b0\u01a1ng t\u1ef1 c\u1ea7u a, ta c\u00f3 t\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m M c\u00e1ch \u0111\u1ec1u 2 m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh sau : -4x+16y-20z-1=0 v\u00e0 23x-2y-8z-13=0<\/p>\n

c) G\u1ecdi \u0111i\u1ec3m M(x, y, z) l\u00e0 \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u (\u03b1) v\u00e0 (\u03b1’ ),khi \u0111\u00f3:<\/p>\n

\"\"<\/p>\n

V\u1eady qu\u1ef9 t\u00edch \u0111i\u1ec3m M c\u1ea7n t\u00ecm c\u1ea7n t\u00ecm l\u00e0 m\u1eb7t ph\u1eb3ng Ph\u01b0\u01a1ng tr\u00ecnh x + 2y +z +2 = 0<\/p>\n

\n

B\u00e0i 20 (trang 90 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

T\u00ecm kho\u1ea3ng c\u00e1ch gi\u1eeda hai m\u1eb7t ph\u1eb3ng:<\/p>\n

Ax+By+Cz+D=0 v\u00e0 Ax+By+Cz+D’=0 v\u1edbi D \u2260 D’<\/p>\n

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

Ta nh\u1eadn th\u1ea5y hai m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho song song v\u1edbi nhau, n\u00ean kho\u1ea3ng c\u00e1ch gi\u1eefa 2 m\u1eb7t ph\u1eb3ng l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb 1 \u0111i\u1ec3m M b\u1ea5t k\u00ec \u0111\u1ebfn m\u1eb7t ph\u1eb3ng kia.<\/p>\n

Gi\u1ea3 s\u1eed \u0111i\u1ec3m M(x0<\/sub>,y0<\/sub>,z0<\/sub>) thu\u1ed9c m\u1eb7t ph\u1eb3ng: Ax + By + Cz + D = 0 ta c\u00f3 kho\u1ea3ng c\u00e1ch c\u1ea7n t\u00ecm l\u00e0:<\/p>\n

\"\"<\/div>\n

\n

B\u00e0i 21 (trang 90 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

T\u00ecm \u0111i\u1ec3m M tr\u00ean tr\u1ee5c Oz trong m\u1ed7i tr\u01b0\u1eddng h\u1ee3p sau:<\/p>\n

a) M c\u00e1c \u0111\u1ec1u \u0111i\u1ec3m A(2, 3, 4) v\u00e0 m\u1eb7t ph\u1eb3ng: 2x + 3y + z \u2013 17 = 0<\/p>\n

b) M c\u00e1ch \u0111\u1ec1u hai m\u1eb7t ph\u1eb3ng x+y-z+1=0 v\u00e0 x-y+z+5=0<\/p>\n

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

V\u00ec M n\u1eb1m tr\u00ean tr\u1ee5c Oz n\u00ean c\u00f3 t\u1ecda \u0111\u1ed9 d\u1ea1ng: M = (0, 0, c)<\/p>\n

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

\"\"<\/p>\n

Kho\u1ea3ng c\u00e1ch t\u1eeb M \u0111\u1ebfn m\u1eb7t ph\u1eb3ng: 2x+3y+z-17=0 l\u00e0:<\/p>\n

\"\"<\/p>\n

theo b\u00e0i ra, ta c\u00f3: MA = h <=> MA2<\/sup>=h2<\/sup><\/p>\n

\"\"<\/p>\n

<=> c = 3. V\u1eady M = (0, 0, 3) l\u00e0 \u0111i\u1ec3m c\u1ea7n t\u00ecm.<\/p>\n

b) V\u00ec M(0, 0, c) c\u00e1ch \u0111\u1ec1u hai m\u1eb7t ph\u1eb3ng: x+y-z+1=0 v\u00e0 x-y+z+5=0 n\u00ean ta c\u00f3:<\/p>\n

\"\"<\/p>\n

<=>(-c+1)=\u00b1(c+5) <=> c = -2<\/p>\n

V\u1eady M = (0, 0, -2) l\u00e0 \u0111i\u1ec3m c\u1ea7n t\u00ecm.<\/p>\n

B\u00e0i 22 (trang 90 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho t\u1ee9 di\u1ec7n OABC c\u00f3 tam gi\u00e1c OAB, OBC, OCA l\u00e0 c\u00e1c tam gi\u00e1c vu\u00f4ng \u0111\u1ec9nh O, g\u1ecdi \u03b1,\u03b2,\u03b3 l\u1ea7n l\u01b0\u1ee3t l\u00e0 c\u00e1c g\u00f3c gi\u1eefa m\u1eb7t ph\u1eb3ng (ABC) v\u00e0 c\u00e1c m\u1eb7t ph\u1eb3ng (OBC), (OCA), (OAB). B\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u1ecda \u0111\u1ed9 h\u00e3y ch\u1ee9ng minh.<\/p>\n

a) Tam gi\u00e1c ABC c\u00f3 ba g\u00f3c nh\u1ecdn<\/p>\n

b) cos2<\/sup>\u2061\u03b1+cos2<\/sup>\u2061\u2061\u03b2+cos2<\/sup>\u2061\u03b3=1<\/p>\n

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

\"\"<\/p>\n

Ch\u1ecdn h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz sao cho: O = (0, 0, 0); A = (a, 0, 0); B = (0, b, 0); C = (0, 0, c)<\/p>\n

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

AB\u2192<\/i>=(-a,b,0),AC\u2192<\/i>=(a,0,c) n\u00ean<\/p>\n

\"\"<\/p>\n

T\u01b0\u01a1ng t\u1ef1, ta c\u00f3 g\u00f3c ACB v\u00e0 g\u00f3c ABC g\u00f3c nh\u1ecdn.<\/p>\n

V\u1eady \u0394ABC c\u00f3 ba g\u00f3c nh\u1ecdn (\u0111pcm)<\/p>\n

b) M\u1eb7t ph\u1eb3ng (ABC) c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0:\u00a0n\u2192<\/i>=[AB\u2192<\/i>,AC\u2192<\/i>\u00a0]=(bc,ac,ab).<\/p>\n

C\u00e1c m\u1eb7t ph\u1eb3ng (OBC), (OAC), (OAB) l\u1ea7n l\u01b0\u1ee3t c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0<\/p>\n

n1<\/sub>\u2192<\/i>=(1,0,0),\u00a0n2<\/sub>\u2192<\/i>=(0,1,0),n3<\/sub>\u2192<\/i>=(0,0,1) n\u00ean ta c\u00f3:<\/p>\n

\"\"<\/p>\n

B\u00e0i 23 (trang 90 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Vi\u1ebft Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng song song v\u1edbi m\u1eb7t ph\u1eb3ng 4x + 3y \u2013 12z + 1 = 0 v\u00e0 ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t c\u1ea7u Ph\u01b0\u01a1ng tr\u00ecnh: x2<\/sup>+y2<\/sup>+z2<\/sup>-2x-4y-6z-2=0<\/p>\n

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

M\u1eb7t c\u1ea7u: x2<\/sup>+y2<\/sup>+z2<\/sup>-2x-4y-6z-2=0 c\u00f3 t\u00e2m I(1, 2, 3), b\u00e1n k\u00ednh R = 4. M\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm song song v\u1edbi m\u1eb7t ph\u1eb3ng 4x+3y-12z+1=0 n\u00ean Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng: 4x+3y-12z+D=0 (\u03b1)<\/p>\n

V\u00ec mp(\u03b1) ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t c\u1ea7m t\u00e2m I(1, 2, 3), b\u00e1n k\u00ednh R = 4 n\u00ean ta c\u00f3:<\/p>\n

d(I,\u03b1)=R<\/p>\n

\"\"<\/p>\n

V\u1eady m\u1eb7t ph\u1eb3ng c\u1ea7n t\u00ecm c\u00f3 Ph\u01b0\u01a1ng t\u00ecnh l\u00e0: 4x+3y-12z+78=0 ho\u1eb7c 4x+3y-12z-26=0<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 15 (trang 89 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng trong m\u1ed7i tr\u01b0\u1eddng h\u1ee3p sau: a) \u0110i qua ba \u0111i\u1ec3m M(2, 0, -1), N(1, -2, 3), P(0, 1, 2). b) \u0110i qua hai \u0111i\u1ec3m A(1, 1, -1), B(5, 2, 1) v\u00e0 song song v\u1edbi tr\u1ee5c Oz. c) \u0110i qua \u0111i\u1ec3m […]<\/p>\n","protected":false},"author":3,"featured_media":22115,"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 hoc - Ch\u01b0\u01a1ng 3 - B\u00e0i 3: Ch\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng<\/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-3-chuong-trinh-mat-phang\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"H\u00ecnh hoc - 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