C\u00e2u 17:<\/b>\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n
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A. C\u1eaft nhau\u00a0\u00a0\u00a0B. song song\u00a0\u00a0\u00a0C. ch\u00e9o nhau\u00a0\u00a0\u00a0D. tr\u00f9ng nhau<\/p>\n
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C\u00e2u 18:<\/b>\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n
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A. C\u1eaft nhau\u00a0\u00a0\u00a0B. song song\u00a0\u00a0\u00a0C. ch\u00e9o nhau\u00a0\u00a0\u00a0D. tr\u00f9ng nhau<\/p>\n
C\u00e2u 19:<\/b>\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n
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A. C\u1eaft nhau\u00a0\u00a0\u00a0B. song song\u00a0\u00a0\u00a0C. ch\u00e9o nhau\u00a0\u00a0\u00a0D. tr\u00f9ng nhau<\/p>\n
C\u00e2u 20:<\/b>\u00a0T\u00ecm t\u1ea5t c\u1ea3 c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a a \u0111\u1ec3 hai \u0111\u01b0\u1eddng th\u1eb3ng sau ch\u00e9o nhau :<\/p>\n
d1<\/sub>: x = 1 + at, y = t, z = -1 + 2t, d2<\/sub>: x = 1 – t’, y = 2 + 2t’, z = 3 – t’<\/p>\n A. a > 0 \u00a0\u00a0\u00a0B. a \u2260 -4\/3\u00a0\u00a0\u00a0C. a \u2260 0 \u00a0\u00a0\u00a0D. a = 0<\/p>\n C\u00e2u 21:<\/b>\u00a0T\u00ecm t\u1ea5t c\u1ea3 c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a a \u0111\u1ec3 hai \u0111\u01b0\u1eddng th\u1eb3ng sau vu\u00f4ng g\u00f3c :<\/p>\n d1<\/sub>: x = 1 – t, y = 1 + 2t, z = 3 + at, d2<\/sub>: x = a + at, y = -1 + t, z = -2 + 2t<\/p>\n A. a=-2\u00a0\u00a0\u00a0B. a=2\u00a0\u00a0\u00a0C. a \u2260 2 \u00a0\u00a0\u00a0D. Kh\u00f4ng t\u1ed3n t\u1ea1i a<\/p>\n C\u00e2u 22:<\/b>\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d: x = 1 + 2t, y = 1 – t, z = 1 – t v\u00e0 m\u1eb7t ph\u1eb3ng (P): x + y + z – 3 = 0 l\u00e0:<\/p>\n A. d \u2282 (P)\u00a0\u00a0\u00a0B. c\u1eaft nhau\u00a0\u00a0\u00a0C. song song\u00a0\u00a0\u00a0D. \u0110\u00e1p \u00e1n kh\u00e1c<\/p>\n C\u00e2u 23:<\/b>\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d: x = 2 + 4t, y = 3 + t, z = -5t v\u00e0 m\u1eb7t ph\u1eb3ng (P): x + y + z – 3 = 0 l\u00e0 :<\/p>\n A. d \u2282 (P)\u00a0\u00a0\u00a0B. c\u1eaft nhau\u00a0\u00a0\u00a0C. song song\u00a0\u00a0\u00a0D. \u0110\u00e1p \u00e1n kh\u00e1c<\/p>\n C\u00e2u 24:<\/b>\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n <\/p>\n v\u00e0 m\u1eb7t ph\u1eb3ng (P): x + y + z – 10 = 0 l\u00e0 :<\/p>\n A. d \u2282 (P)\u00a0\u00a0\u00a0B. c\u1eaft nhau \u00a0\u00a0\u00a0C. song song\u00a0\u00a0\u00a0D. \u0110\u00e1p \u00e1n kh\u00e1c<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 17:<\/b><\/p>\n Do hai vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00e3 cho c\u00f9ng ph\u01b0\u01a1ng n\u00ean c\u00e1c \u0111\u00e1p \u00e1n A v\u00e0 C l\u00e0 sai. Trong hai \u0111\u00e1p \u00e1n c\u00f2n l\u1ea1i, ta th\u1ea5y \u0111i\u1ec3m (1 ;2 ;3) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng c\u00f2n l\u1ea1i. V\u1eady hai \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00e3 cho tr\u00f9ng nhau.<\/p>\n C\u00e2u 20:<\/b><\/p>\n Hai \u0111\u01b0\u1eddng th\u1eb3ng d1<\/sub>, d2<\/sub>\u00a0l\u1ea7n l\u01b0\u1ee3t \u0111i qua hai \u0111i\u1ec3m M1<\/sub>(1; 0; -1), M2<\/sub>(1; 2; 3) v\u00e0 c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0<\/p>\n <\/p>\n Hai \u0111\u01b0\u1eddng th\u1eb3ng ch\u00e9o nhau khi v\u00e0 ch\u1ec9 khi :<\/p>\n <\/p>\n <=> -5.0 + (a – 2).2 + (2a + 1).4 \u2260 0 <=> 10a \u2260 0 <=> a \u2260 0<\/p>\n C\u00e2u 23:<\/b><\/p>\n \u0110\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M(2 ;3 ;0) v\u00e0 c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u00e0\u00a0ud<\/sub>\u2192<\/i>\u00a0= (4; 1; -5), m\u1eb7t ph\u1eb3ng (P) c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0\u00a0uP<\/sub>\u2192<\/i>\u00a0= (1; 1; 1). Ta c\u00f3 :<\/p>\n <\/p>\n Suy ra \u0111\u01b0\u1eddng th\u1eb3ng d song song v\u1edbi m\u1eb7t ph\u1eb3ng (P).<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 17:\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng A. C\u1eaft nhau\u00a0\u00a0\u00a0B. song song\u00a0\u00a0\u00a0C. ch\u00e9o nhau\u00a0\u00a0\u00a0D. tr\u00f9ng nhau C\u00e2u 18:\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng A. C\u1eaft nhau\u00a0\u00a0\u00a0B. song song\u00a0\u00a0\u00a0C. ch\u00e9o nhau\u00a0\u00a0\u00a0D. tr\u00f9ng nhau C\u00e2u 19:\u00a0V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng A. C\u1eaft nhau\u00a0\u00a0\u00a0B. song song\u00a0\u00a0\u00a0C. ch\u00e9o nhau\u00a0\u00a0\u00a0D. tr\u00f9ng nhau C\u00e2u […]<\/p>\n","protected":false},"author":3,"featured_media":27255,"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 17-D<\/td>\n 18-A<\/td>\n 19-C<\/td>\n 20-C<\/td>\n 21-A<\/td>\n 22-A<\/td>\n 23-C<\/td>\n 24-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n