C\u00e2u 33:<\/b>\u00a0T\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng ch\u00e9o nhau sau \u0111\u00e2y<\/p>\n

<\/p>\n

C\u00e2u 34:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n

<\/p>\n

v\u00e0 m\u1eb7t ph\u1eb3ng 2x – 2y + z + 3 = 0. T\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa d v\u00e0 (P)<\/p>\n

A. 0\u00a0\u00a0\u00a0B. 3\u00a0\u00a0\u00a0C. 1\u00a0\u00a0\u00a0D. 9<\/p>\n

C\u00e2u 35:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a m\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m l\u00e0 I(1;0;-1) v\u00e0 ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n

<\/p>\n

A. (x – 1)^{2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 81\u00a0\u00a0\u00a0C. (x + 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z – 1)2<\/sup>\u00a0= 81<\/p>\n}

B. (x – 1)^{2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 9 \u00a0\u00a0\u00a0D. (x – 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 3<\/p>\n}

C\u00e2u 36:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a m\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m l\u00e0 I(1;0;-1) v\u00e0 c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng<\/p>\n

<\/p>\n

theo m\u1ed9t d\u00e2y cung AB c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng 8<\/p>\n

A. (x – 1)^{2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 16 \u00a0\u00a0\u00a0C. (x – 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 25<\/p>\n}

B. (x – 1)^{2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 5\u00a0\u00a0\u00a0D. (x + 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z – 1)2<\/sup>\u00a0= 25<\/p>\n}

C\u00e2u 37:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho hai \u0111i\u1ec3m A(-2; -2; -4), M(1; 0; 0) . L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M, n\u1eb1m trong m\u1eb7t ph\u1eb3ng (P): x + y + z – 1 = 0 sao cho kho\u1ea3ng c\u00e1ch t\u1eeb A \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng d \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t<\/p>\n

<\/p>\n

C\u00e2u 38:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho hai \u0111i\u1ec3m A(2; 0; 1), B(8; 4; -5) v\u00e0 m\u1eb7t ph\u1eb3ng 2x + 2y – z + 1 = 0 . T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m M thu\u1ed9c m\u1eb7t ph\u1eb3ng (P) sao cho AM^{2<\/sup>\u00a0+ BM2<\/sup>\u00a0\u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t<\/p>\n}

A. M(1; -2; -1) B. M(9; 6; -5) C. M(1; -2; -5) D. \u0110\u00e1p \u00e1n kh\u00e1c<\/p>\n

C\u00e2u 39:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t c\u1ea7u (S) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: (x + 1)^{2<\/sup>\u00a0+ (y – 4)2<\/sup>\u00a0+ (z + 3)2<\/sup>\u00a0= 36 . S\u1ed1 m\u1eb7t ph\u1eb3ng (P) ch\u1ee9a tr\u1ee5c Ox v\u00e0 ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t c\u1ea7u (S) l\u00e0:<\/p>\n}

A. 0 B. 1 C. 2 D. V\u00f4 s\u1ed1<\/p>\n

C\u00e2u 40:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho ba \u0111i\u1ec3m A(0; 0; 0), B(1; 2; 3), C(2; 3; 1). G\u1ecdi D l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh A c\u1ee7a tam gi\u00e1c ABC. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sai?<\/p>\n

A. AD \u22a5 BC<\/p>\n

B. M\u1ed9t vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng AD l\u00e0:\u00a0AB\u2192<\/i>\u00a0+\u00a0AC\u2192<\/i><\/p>\n

C. M\u1ed9t vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng AD l\u00e0:<\/p>\n

<\/p>\n

D. M\u1ed9t vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng AD l\u00e0:\u00a0u_{AD<\/sub>\u2192<\/i>\u00a0= (1; 1; -2)<\/p>\n}

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

33-B<\/td>\n

34-B<\/td>\n

35-B<\/td>\n

36-C<\/td>\n

37-A<\/td>\n

38-A<\/td>\n

39-A<\/td>\n

40-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 35:<\/b><\/p>\n \u0110\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M(6 ;1 ;0) v\u00e0 c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u00e0\u00a0ud<\/sub>\u2192<\/i>\u00a0= (4; -1; -1). Ta c\u00f3 :<\/p>\n <\/p>\n Do \u0111\u01b0\u1eddng th\u1eb3ng d ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t c\u1ea7u (S) n\u00ean (S) c\u00f3 b\u00e1n k\u00ednh l\u00e0 :<\/p>\n <\/p>\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u00e0 : (x – 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 9<\/p>\n \n \n C\u00e2u 36:<\/b><\/p>\n \u0110\u01b0\u1eddng th\u1eb3ng d \u0111i qua \u0111i\u1ec3m M(-2 ;3 ;2) v\u00e0 c\u00f3 vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng l\u00e0\u00a0ud<\/sub>\u2192<\/i>\u00a0= (-4; 1; 1) Ta c\u00f3 :<\/p>\n<\/div>\n<\/div>\n <\/p>\n Kho\u1ea3ng c\u00e1ch t\u1eeb I \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng d l\u00e0 :<\/p>\n <\/p>\n Do d c\u1eaft (S) theo d\u00e2y cung AB c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng 8 n\u00ean ta c\u00f3:<\/p>\n <\/p>\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u00e0: (x – 1)2<\/sup>\u00a0+ y2<\/sup>\u00a0+ (z + 1)2<\/sup>\u00a0= 25<\/p>\n C\u00e2u 37:<\/b><\/p>\n G\u1ecdi H l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a A tr\u00ean d. Ta c\u00f3: d(A; d) = AH \u2264 AM = \u221a29<\/p>\n D\u1ea5u b\u1eb1ng x\u1ea3y ra khi v\u00e0 ch\u1ec9 khi H tr\u00f9ng M, ngh\u0129a l\u00e0 d vu\u00f4ng g\u00f3c v\u1edbi AM.<\/p>\n T\u1eeb \u0111\u00f3 ta \u0111\u01b0\u1ee3c<\/p>\n <\/p>\n V\u1eady d c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n <\/p>\n C\u00e2u 39:<\/b><\/p>\n M\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m I(-1;4;-3) v\u00e0 c\u00f3 b\u00e1n k\u00ednh R = 6. G\u1ecdi H l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a I tr\u00ean tr\u1ee5c Ox. Ta c\u00f3 H(-1;0;0) v\u00e0 IH=5.<\/p>\n G\u1ecdi K l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a I tr\u00ean m\u1eb7t ph\u1eb3ng (P). Ta c\u00f3<\/p>\n d(I; (P)) = IK \u2264 IH = 5 < R = 6<\/p>\n Do \u0111\u00f3 m\u1eb7t ph\u1eb3ng (P) lu\u00f4n c\u1eaft m\u1eb7t c\u1ea7u (S) theo m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. V\u1eady kh\u00f4ng t\u1ed3n t\u1ea1i m\u1eb7t ph\u1eb3ng (P) ch\u1ee9a Ox v\u00e0 ti\u1ebfp x\u00fac v\u1edbi (S)<\/p>\n C\u00e2u 40:<\/b><\/p>\n Ta th\u1ea5y tam gi\u00e1c ABC c\u00e2n t\u1ea1i \u0111\u1ec9nh A, do \u0111\u00f3 c\u00e1c kh\u1eb3ng \u0111\u1ecbnh A, B v\u00e0 C \u0111\u1ec1u \u0111\u00fang. V\u1eady kh\u1eb3ng \u0111\u1ecbnh D sai.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 33:\u00a0T\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng ch\u00e9o nhau sau \u0111\u00e2y C\u00e2u 34:\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng v\u00e0 m\u1eb7t ph\u1eb3ng 2x – 2y + z + 3 = 0. T\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa d v\u00e0 (P) A. 0\u00a0\u00a0\u00a0B. 3\u00a0\u00a0\u00a0C. 1\u00a0\u00a0\u00a0D. 9 C\u00e2u 35:\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc c\u1ee7a […]<\/p>\n","protected":false},"author":3,"featured_media":27202,"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