C\u00e2u 24:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho t\u1ee9 di\u1ec7n ABCD v\u1edbi A(2;-4;6) v\u00e0 ba \u0111i\u1ec3m B, C, D c\u00f9ng thu\u1ed9c m\u1eb7t ph\u1eb3ng (Oyz). G\u1ecdi M, N, P l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB, AC, AD. L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (MNP)<\/p>\n

A. x + 1 = 0\u00a0\u00a0\u00a0C. y + z – 1 = 0<\/p>\n

B. x – 1 = 0\u00a0\u00a0\u00a0D. x = 1 + t, y = -2, z = 3<\/p>\n

C\u00e2u 25:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng d v\u00e0 m\u1eb7t ph\u1eb3ng (P) l\u1ea7n l\u01b0\u1ee3t c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n

<\/p>\n

P): 2x + y – 3z – 4 = 0. Trong nh\u1eefng kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang?<\/p>\n

A. d n\u1eb1m trong m\u1eb7t ph\u1eb3ng (P)\u00a0\u00a0\u00a0C. d kh\u00f4ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (P)<\/p>\n

B. d song song v\u1edbi m\u1eb7t ph\u1eb3ng (P)\u00a0\u00a0\u00a0D. d c\u1eaft m\u1eb7t ph\u1eb3ng (P)<\/p>\n

C\u00e2u 26:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho t\u1ee9 di\u1ec7n ABCD v\u1edbi A(2;-4;6), B(1;1;1), C(0;3;0), D(0;0;3). Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh tham s\u1ed1 c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d ch\u1ee9a \u0111\u01b0\u1eddng cao AH c\u1ee7a t\u1ee9 di\u1ec7n ABCD<\/p>\n

A. x = 2 + t, y = -4 – t, z = 6 + t \u00a0\u00a0\u00a0C. x = 2 + t, y = -4 + t, z = 6 + t<\/p>\n

B. x = 1 + 2t, y = -1 -4t, z = 1 + 6t\u00a0\u00a0\u00a0D. x = 1 + 2t, y = 1 – 4t, z = 1 + 6t<\/p>\n

C\u00e2u 27:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau:d_{1<\/sub>: x = 1 + t, y = 1, z = 1 – t, d2<\/sub>: x = -t, y = 2 + t, z = 1. Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (P) ch\u1ee9a hai \u0111\u01b0\u1eddng th\u1eb3ng d1<\/sub>, d2<\/sub><\/p>\n}

A. x + y + z – 3 = 0 \u00a0\u00a0\u00a0C. x – y + z – 1 = 0<\/p>\n

B. x + y + z + 3 = 0 \u00a0\u00a0\u00a0D. x – y + z + 1 = 0<\/p>\n

C\u00e2u 28:<\/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 mx + y – 3z + 1 = 0; 4x – 2y + (n^{2<\/sup>\u00a0+ n)z – n = 0, trong \u0111\u00f3 m v\u00e0 n l\u00e0 hai tham s\u1ed1. V\u1edbi nh\u1eefng gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m v\u00e0 n th\u00ec hai m\u1eb7t ph\u1eb3ng (P) v\u00e0 (Q) song song v\u1edbi nhau<\/p>\n}

A. m=-2 v\u00e0 n=2 \u00a0\u00a0\u00a0C. m=-2 v\u00e0 n=2 ho\u1eb7c n=-3<\/p>\n

B. m=2 v\u00e0 n=-3 \u00a0\u00a0\u00a0D. m=-2 v\u00e0 n=-3<\/p>\n

C\u00e2u 29:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n

<\/p>\n

v\u00e0 m\u1eb7t ph\u1eb3ng (P): 2x – y + 2z = 0. Cho m\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m I thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng d, c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng 1 v\u00e0 ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t ph\u1eb3ng (P). T\u1ecda \u0111\u1ed9 t\u1ea5t c\u1ea3 c\u00e1c \u0111i\u1ec3m I c\u00f3 th\u1ec3 l\u00e0:<\/p>\n

A. I_{1<\/sub>(5; 11; 2)\u00a0\u00a0\u00a0C. I2<\/sub>(3; 7; 1) ho\u1eb7c I3<\/sub>(-3; -5; -2)<\/p>\n}

B. I_{2<\/sub>(3; 7; 1)\u00a0\u00a0\u00a0D. I1<\/sub>(5; 11; 2) ho\u1eb7c I4<\/sub>(-1; -1; -1)<\/p>\n}

C\u00e2u 30:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t c\u1ea7u (S) \u0111i qua 3 \u0111i\u1ec3m O, A(2;0;0), B(0;2;0) v\u00e0 t\u00e2m thu\u1ed9c m\u1eb7t ph\u1eb3ng (P): x + y + z – 3 = 0<\/p>\n

A. (x – 1)^{2<\/sup>\u00a0+ (y – 1)2<\/sup>\u00a0+ (z – 1)2<\/sup>\u00a0= 3 \u00a0\u00a0\u00a0C. (x – 1)2<\/sup>\u00a0+ (y – 1)2<\/sup>\u00a0+ (z – 1)2<\/sup>\u00a0= 9<\/p>\n}

B. (x + 1)^{2<\/sup>\u00a0+ (y + 1)2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 3\u00a0\u00a0\u00a0D. (x + 1)2<\/sup>\u00a0+ (y + 1)2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 9<\/p>\n}

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n

24-B<\/td>\n

25-D<\/td>\n

26-C<\/td>\n

27-A<\/td>\n

28-D<\/td>\n

29-D<\/td>\n

30-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 25:<\/b><\/p>\n Ta c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0 :\u00a0ud<\/sub>\u2192<\/i>\u00a0= (2; 1; -3), \u0111\u1ed3ng th\u1eddi\u00a0nP<\/sub>\u2192<\/i>\u00a0= (2; 1; -3) =\u00a0ud<\/sub>\u2192<\/i>\u00a0c\u0169ng l\u00e0 vect\u01a1 ph\u00e1p tuy\u1ebfn c\u1ee7a m\u1eb7t ph\u1eb3ng (P). Do \u0111\u00f3 \u0111\u01b0\u1eddng th\u1eb3ng d vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (P), suy ra \u0111\u01b0\u1eddng th\u1eb3ng d c\u1eaft m\u1eb7t ph\u1eb3ng (P).<\/p>\n C\u00e2u 28:<\/b><\/p>\n Hai m\u1eb7t ph\u1eb3ng \u0111\u00e3 cho 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 C\u00e2u 29:<\/b><\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh tham s\u1ed1 c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0 : d: x = 1 +2 t, y = 3+ 4t, z = t<\/p>\n Ta c\u00f3 I \u2208 d => I(1 + 2t, 3 + 4t, t). V\u00ec (S) ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t ph\u1eb3ng (P) n\u00ean ta c\u00f3 :<\/p>\n <\/p>\n C\u00e2u 30:<\/b><\/p>\n G\u1ecdi I(a,b,c) l\u00e0 t\u00e2m c\u1ee7a m\u1eb7t c\u1ea7u (S). Ta c\u00f3:<\/p>\n <\/p>\n => I(1; 1; 1); R = OI = \u221a3<\/p>\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u00e0: (x – 1)2<\/sup>\u00a0+ (y – 1)2<\/sup>\u00a0+ (z – 1)2<\/sup>\u00a0= 3<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 24:\u00a0Trong kh\u00f4ng gian Oxyz, cho t\u1ee9 di\u1ec7n ABCD v\u1edbi A(2;-4;6) v\u00e0 ba \u0111i\u1ec3m B, C, D c\u00f9ng thu\u1ed9c m\u1eb7t ph\u1eb3ng (Oyz). G\u1ecdi M, N, P l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB, AC, AD. L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (MNP) A. x + 1 = 0\u00a0\u00a0\u00a0C. y + z – 1 = 0 B. […]<\/p>\n","protected":false},"author":3,"featured_media":27096,"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