B\u00e0i 1 (trang 68 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba vect\u01a1:\u00a0a\u2192\u00a0= (2; -5; 3),\u00a0b\u2192\u00a0= (0; 2; -1),\u00a0c\u2192\u00a0= (1; 7; 2)<\/strong><\/span><\/p>\n a) T\u00ednh t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1\u00a0d\u2192\u00a0= 4a\u2192\u00a0– 1\/3\u00a0b\u2192\u00a0+ 3c\u2192<\/strong><\/span><\/p>\n b) T\u00ednh t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1\u00a0e\u2192\u00a0=\u00a0a\u2192\u00a0– 4b\u2192\u00a0– 2c\u2192<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n a) Ta c\u00f3: 4a\u2192\u00a0= (8; -20; 12)<\/p>\n -1\/3\u00a0b\u2192\u00a0= (0; -2\/3 ; 1\/3)<\/p>\n 3c\u2192\u00a0= (3; 21; 6)<\/p>\n V\u1eady\u00a0d\u2192\u00a0= 4a\u2192\u00a0– 1\/3\u00a0b\u2192\u00a0+ 3c\u2192\u00a0= (11; 1\/3; 55\/3)<\/p>\n b) Ta c\u00f3: 4b\u2192\u00a0= (0; -8; 4)<\/p>\n -2c\u2192\u00a0= (-2; -14; -4)<\/p>\n V\u1eady\u00a0e\u2192\u00a0=\u00a0a\u2192\u00a0– 4b\u2192\u00a0– 2c\u2192\u00a0= (0; -27; 3)<\/p>\n B\u00e0i 2 (trang 68 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba \u0111i\u1ec3m A(1; -1; 1), B(0; 1; 2), C(1;0;1). T\u00ecm t\u1ecda \u0111\u1ed9 tr\u1ecdng t\u00e2m G c\u1ee7a tam gi\u00e1c ABC.<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n \u00a0<\/p>\n B\u00e0i 3 (trang 68 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho h\u00ecnh h\u1ed9p ABCD.A’B’C’D’ bi\u1ebft A(1; 0; 1), B(2; 1; 2), D(1; -1; 1), C'(4; 5; -5). T\u00ednh t\u1ecda \u0111\u1ed9 c\u00e1c \u0111\u1ec9nh c\u00f2n l\u1ea1i c\u1ee7a h\u00ecnh h\u1ed9p.<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n B\u00e0i 4 (trang 68 SGK H\u00ecnh h\u1ecdc 12):\u00a0T\u00ednh:<\/strong><\/span><\/p>\n <\/p>\n B\u00e0i 5 (trang 68 SGK H\u00ecnh h\u1ecdc 12):\u00a0T\u00ecm t\u00e2m v\u00e0 b\u00e1n k\u00ednh c\u1ee7a c\u00e1c m\u1eb7t c\u1ea7u sau \u0111\u00e2y:<\/strong><\/span><\/p>\n a)x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0\u2013 8x \u2013 2y + 1 = 0<\/strong><\/span><\/p>\n b)3x2<\/sup>\u00a0+ 3y2<\/sup>\u00a0+ 3z2<\/sup>\u2013 6x + 8y + 15z \u2013 3 = 0<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n B\u00e0i 6 (trang 68 SGK H\u00ecnh h\u1ecdc 12):\u00a0L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u trong hai tr\u01b0\u1eddng h\u1ee3p sau \u0111\u00e2y:<\/strong><\/span><\/p>\n a)C\u00f3 \u0111\u01b0\u1eddng k\u00ednh AB v\u1edbi A(4; -3; 7), B(2; 1; 3)<\/strong><\/span><\/p>\n b)\u0110i qua \u0111i\u1ec3m A(5; -2; 1) v\u00e0 c\u00f3 t\u00e2m C(3; -3; 1)<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" B\u00e0i 1 (trang 68 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba vect\u01a1:\u00a0a\u2192\u00a0= (2; -5; 3),\u00a0b\u2192\u00a0= (0; 2; -1),\u00a0c\u2192\u00a0= (1; 7; 2) a) T\u00ednh t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1\u00a0d\u2192\u00a0= 4a\u2192\u00a0– 1\/3\u00a0b\u2192\u00a0+ 3c\u2192 b) T\u00ednh t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1\u00a0e\u2192\u00a0=\u00a0a\u2192\u00a0– 4b\u2192\u00a0– 2c\u2192 L\u1eddi gi\u1ea3i: a) Ta c\u00f3: 4a\u2192\u00a0= (8; -20; 12) -1\/3\u00a0b\u2192\u00a0= (0; -2\/3 ; 1\/3) 3c\u2192\u00a0= (3; 21; 6) V\u1eady\u00a0d\u2192\u00a0= […]<\/p>\n","protected":false},"author":3,"featured_media":20384,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1298],"tags":[1377,1356,1355],"yoast_head":"\n