C\u00e2u 9:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M(0;1;-1), n\u1eb1m trong m\u1eb7t ph\u1eb3ng (P): x + 2y + z – 1 = 0 v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n
<\/p>\n
C\u00e2u 10:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho d l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m , v\u1edbi m l\u00e0 tham s\u1ed1, v\u00e0 song song v\u1edbi hai m\u1eb7t ph\u1eb3ng (Oxy), (Oxz). Trong nh\u1eefng kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai?<\/p>\n
A. T\u1ed3n t\u1ea1i m \u0111\u1ec3 d \u0111i qua g\u1ed1c t\u1ecda \u0111\u1ed9<\/p>\n
B. d c\u00f3 m\u1ed9t vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u00e0:\u00a0u\u2192<\/i>\u00a0= (1; 0; 0)<\/p>\n
C. Ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a d l\u00e0: x = t, y = -3, z = 4<\/p>\n
D. \u0110\u01b0\u1eddng th\u1eb3ng d n\u1eb1m trong hai m\u1eb7t ph\u1eb3ng: (P): y + 3 = 0, (Q): z – 4 = 0<\/p>\n
C\u00e2u 11:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M(2;-1;1) v\u00e0 song song v\u1edbi hai m\u1eb7t ph\u1eb3ng (P): x + y + z – 1 = 0 v\u00e0 (Q): x – 3y – 2z + 1 = 0 . Trong nh\u1eefng kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai?<\/p>\n
A. Hai vect\u01a1 (1;1;1) v\u00e0 (1;-3;-2) \u0111\u1ec1u vu\u00f4ng g\u00f3c v\u1edbi vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d<\/p>\n
B. Ph\u01b0\u01a1ng tr\u00ecnh tham s\u1ed1 c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0: x = 2 + t, y = -1 + 3t, z = 1 – 4t<\/p>\n
C. \u0110\u01b0\u1eddng th\u1eb3ng d \u0111i qua g\u1ed1c t\u1ecda \u0111\u1ed9<\/p>\n
D. Ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0:<\/p>\n
<\/p>\n
C\u00e2u 12:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh tham s\u1ed1 c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0 giao tuy\u1ebfn c\u1ee7a hai m\u1eb7t ph\u1eb3ng c\u1eaft nhau: (P): x + 2y – z + 1 = 0, (Q): x + y + 2z + 3 = 0<\/p>\n
A. d: x = -5 – 5t, y = 2 + 3t, z = t \u00a0\u00a0\u00a0C. d: x = -5 + 5t, y = 2 + 3t, z = t<\/p>\n
B. d: x = -5 – 5t, y = 2 – 3t, z = t \u00a0\u00a0\u00a0D. d: x = 5t, y = 3 – 3t, z = -t<\/p>\n
C\u00e2u 13:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0 giao tuy\u1ebfn c\u1ee7a hai m\u1eb7t ph\u1eb3ng c\u1eaft nhau (P): x + y – z + 3 = 0, (Q): 2x – y + 6z – 2 = 0. ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0:<\/p>\n
<\/p>\n
C\u00e2u 14:<\/b>\u00a0Cho tam gi\u00e1c ABC c\u00f3 A(1; 3; 5), B(-4; 0; -2), C(3; 9; 6) . G\u1ecdi G l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c ABC. Trong nh\u1eefng kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai?<\/p>\n
A. T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m G l\u00e0 (0;4;3)<\/p>\n
B. AG \u22a5 BC<\/p>\n
C. Ph\u01b0\u01a1ng tr\u00ecnh tham s\u1ed1 c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng OG l\u00e0: x = 0, y = 4t, z = 3t<\/p>\n
D. \u0110\u01b0\u1eddng th\u1eb3ng OG n\u1eb1m trong hai m\u1eb7t ph\u1eb3ng: (P): x = 0, (Q): 3y – 4z = 0<\/p>\n
C\u00e2u 15:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M(0;1;-1), vu\u00f4ng g\u00f3c v\u00e0 c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng \u0394: x = 1 – 4t, y = t, z = -1 + 4t<\/p>\n
<\/p>\n
C\u00e2u 16:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M v\u00e0 c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u00e0\u00a0u\u2192<\/i>\u00a0; cho \u0111\u01b0\u01a1ng th\u1eb3ng d\u2019 \u0111i qua \u0111i\u1ec3m M\u2019 v\u00e0 c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u00e0\u00a0u’\u2192<\/i>\u00a0th\u1ecfa m\u00e3n [u\u2192<\/i>,\u00a0u’\u2192<\/i>].MM’\u2192<\/i>\u00a0= 0 . Trong nh\u1eefng k\u1ebft lu\u1eadn d\u01b0\u1edbi \u0111\u00e2y, k\u1ebft lu\u1eadn n\u00e0o sai?<\/p>\n
A. d v\u00e0 d\u2019 ch\u00e9o nhau\u00a0\u00a0\u00a0C. d v\u00e0 d\u2019 c\u00f3 th\u1ec3 c\u1eaft nhau<\/p>\n
B. d v\u00e0 d\u2019 c\u00f3 th\u1ec3 song song v\u1edbi nhau\u00a0\u00a0\u00a0D. d v\u00e0 d\u2019 c\u00f3 th\u1ec3 tr\u00f9ng nhau<\/p>\n
H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 10:<\/b><\/p>\n T\u1eeb gi\u1ea3 thi\u1ebft ta suy ra \u0111\u01b0\u1eddng th\u1eb3ng d song song v\u1edbi tr\u1ee5c Ox. K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec3m O thu\u1ed9c Ox, ta suy ra \u0111\u01b0\u1eddng th\u1eb3ng d kh\u00f4ng th\u1ec3 \u0111i qua \u0111i\u1ec3m O v\u1edbi m\u1ecdi m. V\u1eady A l\u00e0 kh\u1eb3ng \u0111\u1ecbnh sai.<\/p>\n C\u00e2u 11:<\/b><\/p>\n N\u1ebfu \u0111\u01b0\u1eddng th\u1eb3ng d \u0111i qua g\u1ed1c t\u1ecda \u0111\u1ed9 O th\u00ec \u0111\u01b0\u1eddng th\u1eb3ng d c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u00e0\u00a0OM\u2192<\/i>\u00a0= (1; -1; 1) v\u00e0 vect\u01a1 n\u00e0y s\u1ebd vu\u00f4ng g\u00f3c v\u1edbi vect\u01a1 ph\u00e1p tuy\u1ebfn\u00a0np<\/sub>\u2192<\/i>= (1; 1; 1) c\u1ee7a m\u1eb7t ph\u1eb3ng (P). \u0110i\u1ec1u n\u00e0y sai, do\u00a0ud<\/sub>\u2192<\/i>.up<\/sub>\u2192<\/i>\u00a0= 2.1 – 1.1 + 1.1 = 2 \u2260 0<\/p>\n V\u1eady kh\u1eb3ng \u0111\u1ecbnh C l\u00e0 sai.<\/p>\n C\u00e2u 12:<\/b><\/p>\n T\u1ecda \u0111\u1ed9 c\u00e1c \u0111i\u1ec3m thu\u1ed9c d l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh :<\/p>\n <\/p>\n \u0110\u1eb7t z = t, thay v\u00e0o h\u1ec7 tr\u00ean ta \u0111\u01b0\u1ee3c :<\/p>\n <\/p>\n C\u00e2u 14:<\/b><\/p>\n Ta th\u1ea5y A l\u00e0 kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang, t\u1eeb \u0111\u00f3 suy ra C v\u00e0 D c\u0169ng l\u00e0 nh\u1eefng kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang. V\u1eady B l\u00e0 kh\u1eb3ng \u0111\u1ecbnh sai.<\/p>\n C\u00e2u 16:<\/b><\/p>\n T\u1eeb gi\u1ea3 thi\u1ebft ta suy ra hai \u0111\u01b0\u1eddng th\u1eb3ng d v\u00e0 d\u2019 \u0111\u1ed3ng ph\u1eb3ng, do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh A l\u00e0 sai.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 9:\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M(0;1;-1), n\u1eb1m trong m\u1eb7t ph\u1eb3ng (P): x + 2y + z – 1 = 0 v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng C\u00e2u 10:\u00a0Trong kh\u00f4ng gian Oxyz, cho d l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m , v\u1edbi m l\u00e0 […]<\/p>\n","protected":false},"author":3,"featured_media":27345,"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 9-A<\/td>\n 10-A<\/td>\n 11-C<\/td>\n 12-A<\/td>\n 13-B<\/td>\n 14-B<\/td>\n 15-A<\/td>\n 16-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n