C\u00e2u 29:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (P) \u0111i qua \u0111i\u1ec3m A(1;0;1) v\u00e0 ch\u1ee9a tr\u1ee5c Ox<\/p>\n
A. x – 1 = 0\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. z – 1 = 0 \u00a0\u00a0\u00a0D. x + z – 1 = 0<\/p>\n
C\u00e2u 30:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t ph\u1eb3ng (P) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x + (m2<\/sup>\u00a0– 2m)y + (m – 1)z + m2<\/sup>\u00a0+ m = 0 , trong \u0111\u00f3 m l\u00e0 tham s\u1ed1. V\u1edbi nh\u1eefng gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m th\u00ec m\u1eb7t ph\u1eb3ng (P) song song v\u1edbi tr\u1ee5c Oy?<\/p>\n A. m=0\u00a0\u00a0\u00a0B. m=2\u00a0\u00a0\u00a0C. m=0 ho\u1eb7c m=2\u00a0\u00a0\u00a0D. m=1<\/p>\n C\u00e2u 31:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t ph\u1eb3ng (P) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh 2x – 3y + (2m – 4)z + m2<\/sup>\u00a0– m = 0 , trong \u0111\u00f3 m l\u00e0 tham s\u1ed1. V\u1edbi nh\u1eefng gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m th\u00ec (P) song song v\u1edbi tr\u1ee5c Oz?<\/p>\n A. m=2\u00a0\u00a0\u00a0B. m=0 \u00a0\u00a0\u00a0C. m=1\u00a0\u00a0\u00a0D. Kh\u00f4ng t\u1ed3n t\u1ea1i m<\/p>\n C\u00e2u 32:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho hai m\u1eb7t ph\u1eb3ng (P) v\u00e0 (Q) l\u1ea7n l\u01b0\u1ee3t c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 x + my + (m + 3)z + 1 = 0; x – y + 2z = 0 , trong \u0111\u00f3 m l\u00e0 tham s\u1ed1. V\u1edbi nh\u1eefng gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m th\u00ec m\u1eb7t ph\u1eb3ng (P) vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (Q)?<\/p>\n A. m=-1\u00a0\u00a0\u00a0B. m=0\u00a0\u00a0\u00a0C. m=-7\u00a0\u00a0\u00a0D. Kh\u00f4ng t\u1ed3n t\u1ea1i m<\/p>\n C\u00e2u 33:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho hai m\u1eb7t ph\u1eb3ng (P) v\u00e0 (Q) l\u1ea7n l\u01b0\u1ee3t c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 x – y + 2z = 0; 2x – 2y + (m2<\/sup>\u00a0+ 3m)z + m2<\/sup>\u00a0– m = 0 , trong \u0111\u00f3 m l\u00e0 tham s\u1ed1. V\u1edbi nh\u1eefng gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m th\u00ec hai m\u1eb7t ph\u1eb3ng (P) v\u00e0 (Q) song song?<\/p>\n A. m=1\u00a0\u00a0\u00a0B. m=-4\u00a0\u00a0\u00a0C. m=1 ho\u1eb7c m=-4\u00a0\u00a0\u00a0D. m=0<\/p>\n C\u00e2u 34:<\/b>\u00a034. Trong kh\u00f4ng gian Oxyz, cho ba m\u1eb7t ph\u1eb3ng (P), (Q), (R) l\u1ea7n l\u01b0\u1ee3t c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 (m2<\/sup>\u00a0+ m)x – (m + 2)y + z = 0; x + y + z = 0; 2x + y – z = 0 v\u00e0 , trong \u0111\u00f3 m l\u00e0 tham s\u1ed1. V\u1edbi nh\u1eefng gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m th\u00ec m\u1eb7t ph\u1eb3ng (P) \u0111\u1ed3ng th\u1eddi vu\u00f4ng g\u00f3c v\u1edbi c\u1ea3 hai m\u1eb7t ph\u1eb3ng (Q) v\u00e0 (R)?<\/p>\n A. m=1 \u00a0\u00a0\u00a0C. m= -3\/2<\/p>\n B. m=-1\u00a0\u00a0\u00a0D. m= -3\/2 ho\u1eb7c m=-1<\/p>\n C\u00e2u 35:<\/b>\u00a035. Trong kh\u00f4ng gian Oxyz, cho hai m\u1eb7t ph\u1eb3ng -mx + 3y + 2z + m – 6 = 0 v\u00e0 -2x + (5m + 1)y + (m + 3)z – 10 = 0. Hai m\u1eb7t ph\u1eb3ng n\u00e0y c\u1eaft nhau khi v\u00e0 ch\u1ec9 khi:<\/p>\n A. m \u2260 -4\u00a0\u00a0\u00a0B. m \u2260 -6\/5\u00a0\u00a0\u00a0C. m \u2260 1\u00a0\u00a0\u00a0D. M\u1ecdi m<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 29:<\/b><\/p>\n T\u1eeb gi\u1ea3 thi\u1ebft suy ra<\/p>\n <\/p>\n T\u1eeb \u0111\u00f3 suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (P) l\u00e0 : 1(y – 0) = 0 <=> y = 0<\/p>\n C\u00e2u 30:<\/b><\/p>\n Ta c\u00f3\u00a0np<\/sub>\u2192<\/i>\u00a0= (1; m2<\/sup>\u00a0– 2m; m – 1). M\u1eb7t ph\u1eb3ng (P) song song v\u1edbi tr\u1ee5c Oy khi v\u00e0 ch\u1ec9 khi<\/p>\n <\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n T\u1eeb \u0111\u00f3 ta \u0111\u01b0\u1ee3c m=2. V\u1eady \u0111\u00e1p \u00e1n B l\u00e0 \u0111\u00fang<\/p>\n <\/p>\n C\u00e2u 31:<\/b><\/p>\n M\u1eb7t ph\u1eb3ng (P) song song v\u1edbi tr\u1ee5c Oz khi v\u00e0 ch\u1ec9 khi<\/p>\n <\/p>\n C\u00e2u 32:<\/b><\/p>\n Ta c\u00f3:\u00a0np<\/sub>\u2192<\/i>\u00a0= (1; m; m + 3),\u00a0nQ<\/sub>\u2192<\/i>\u00a0= (1; -1; 2). Hai m\u1eb7t ph\u1eb3ng (P) v\u00e0 (Q) vu\u00f4ng g\u00f3c khi v\u00e0 ch\u1ec9 khi\u00a0np<\/sub>\u2192<\/i>.nQ<\/sub>\u2192<\/i>\u00a0= 0<\/p>\n <=> 1.1 + m.(-1) + (m + 3).2 = 0 <=> m + 7 = 0 <=> m = -7<\/p>\n C\u00e2u 33:<\/b><\/p>\n Hai m\u1eb7t ph\u1eb3ng (P) v\u00e0 (Q) song song v\u1edbi nhau khi v\u00e0 ch\u1ec9 khi t\u1ed3n t\u1ea1i m\u1ed9t s\u1ed1 th\u1ef1c k sao cho<\/p>\n <\/p>\n C\u00e2u 34:<\/b><\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n M\u1eb7t ph\u1eb3ng (P) \u0111\u1ed3ng th\u1eddi vu\u00f4ng g\u00f3c v\u1edbi c\u1ea3 hai m\u1eb7t ph\u1eb3ng (Q) v\u00e0 (R) khi v\u00e0 ch\u1ec9 khi<\/p>\n <\/p>\n C\u00e2u 35:<\/b><\/p>\n G\u1ecdi hai m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho l\u1ea7n l\u01b0\u1ee3t l\u00e0 (P) v\u00e0 (Q). Ta c\u00f3<\/p>\n <\/p>\n Hai vect\u01a1 n\u00e0y song song khi v\u00e0 ch\u1ec9 khi t\u1ed3n t\u1ea1i m\u1ed9t s\u1ed1 th\u1ef1c k sao cho<\/p>\n <\/p>\n <\/p>\n T\u1eeb \u0111\u00f3 suy ra hai m\u1eb7t ph\u1eb3ng (P) v\u00e0 (Q) c\u1eaft nhau khi v\u00e0 ch\u1ec9 khi hai vect\u01a1 ph\u00e1p tuy\u1ebfn c\u1ee7a ch\u00fang kh\u00f4ng song song, \u0111i\u1ec1u \u0111\u00f3 t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi m kh\u00e1c 1.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 29:\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (P) \u0111i qua \u0111i\u1ec3m A(1;0;1) v\u00e0 ch\u1ee9a tr\u1ee5c Ox A. x – 1 = 0\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. z – 1 = 0 \u00a0\u00a0\u00a0D. x + z – 1 = 0 C\u00e2u 30:\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t ph\u1eb3ng (P) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x […]<\/p>\n","protected":false},"author":3,"featured_media":27849,"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\n\n
\n 29-B<\/td>\n 30-B<\/td>\n 31-A<\/td>\n 32-C<\/td>\n 33-B<\/td>\n 34-A<\/td>\n 35-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n