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HomeLớp 12Toán lớp 12 - Sách Kết nối tri thứcBài 8. Biểu thức toạ độ của các phép toán vectơBài tập nâng cao
{"common":{"save":0,"post_id":"7593","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5681","post_id":"7593","mon_id":"1159285","chapter_id":"1159392","question":"<p>Trong không gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz, cho 2 \u0111i\u1ec3m <span class=\"math-tex\">$B(1;2;-3),C(7;4;-2)$<\/span>. N\u1ebfu \u0111i\u1ec3m E th\u1ecfa mãn \u0111\u1eb3ng th\u1ee9c <span class=\"math-tex\">$\\overrightarrow{CE}=2\\overrightarrow{EB}$<\/span> thì t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m E là:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\bigg(3;\\dfrac{8}{3};-\\dfrac{8}{3}\\bigg)$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\bigg(\\dfrac{8}{3};3;-\\dfrac{8}{3}\\bigg)$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\bigg(3;3;-\\dfrac{8}{3}\\bigg)$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\bigg(1;2;\\dfrac{1}{3}\\bigg)$<\/span>"],"correct":"1","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$\\bigg(3;\\dfrac{8}{3};-\\dfrac{8}{3}\\bigg)$<\/span>.<\/span><\/p><p>G\u1ecdi <span class=\"math-tex\">$E(x;y;z)$<\/span>.<\/p><p><span class=\"math-tex\">$\\overrightarrow{CE}=(x-7;y-4;z+2)$<\/span>, <span class=\"math-tex\">$2\\overrightarrow{EB}=(2-2x;4-2y;-6-2z)$<\/span>.<\/p><p><span class=\"math-tex\">$\\overrightarrow{CE}=2\\overrightarrow{EB}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}x-7=2-2x\\\\y-4=4-2y\\\\z+2=-6-2z\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}x=3\\\\y=\\dfrac{8}{3}\\\\z=-\\dfrac{8}{3}\\end{cases}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-09-06 02:00:11","option_type":"math","len":0},{"id":"5683","post_id":"7593","mon_id":"1159285","chapter_id":"1159392","question":"<p>Trong không gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz cho hình thang ABCD vuông t\u1ea1i A và B. Ba \u0111\u1ec9nh A(1;2;1), B(2;0;–1), C(6;1;0). Hình thang có di\u1ec7n tích b\u1eb1ng <span class=\"math-tex\">$6\\sqrt{2}$<\/span>. Gi\u1ea3 s\u1eed \u0111\u1ec9nh D(a;b;c). Tìm m\u1ec7nh \u0111\u1ec1 \u0111úng.<\/p>","options":["<strong>A.<\/strong> a + b + c = 6","<strong>B.<\/strong> a + b + c = 5","<strong>C.<\/strong> a + b + c = 8","<strong>D.<\/strong> a + b + c = 7"],"correct":"1","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> a + b + c = 6.<\/span><\/p><p><span class=\"svgedit\"><svg height=\"200\" width=\"260\"> <g><title><\/title><rect fill=\"#ffffaa\" height=\"202\" id=\"canvas_background\" width=\"262\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><path d=\"m65.66,43.91746l131,-37.99999l0,189.99999l-131,0l0,-152z\" fill=\"#aaffff\" id=\"svg_1\" stroke=\"#000\" stroke-width=\"1.5\" transform=\"rotate(90 131.1600036621094,100.91745758056642) \"><\/path> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"’Times New Roman’, Times, serif\" font-size=\"24\" font-weight=\"bold\" id=\"svg_2\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"12\" xml:space=\"preserve\" y=\"32.25\">A<\/text> <text fill=\"#000000\" font-family=\"’Times New Roman’, Times, serif\" font-size=\"24\" font-weight=\"bold\" id=\"svg_3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"11\" xml:space=\"preserve\" y=\"172.25\">B<\/text> <text fill=\"#000000\" font-family=\"’Times New Roman’, Times, serif\" font-size=\"24\" font-weight=\"bold\" id=\"svg_4\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"191\" xml:space=\"preserve\" y=\"33.25\">D<\/text> <text fill=\"#000000\" font-family=\"’Times New Roman’, Times, serif\" font-size=\"24\" font-weight=\"bold\" id=\"svg_5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"231\" xml:space=\"preserve\" y=\"171.25\">C<\/text> <rect fill=\"#aaffff\" height=\"14\" id=\"svg_6\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\" width=\"14\" x=\"36\" y=\"35.25\"><\/rect> <rect fill=\"#aaffff\" height=\"14\" id=\"svg_7\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\" width=\"14\" x=\"36\" y=\"152.25\"><\/rect> <\/g> <\/svg><\/span><\/p><p><span class=\"math-tex\">$\\overrightarrow{AB}=(1;-2;-2)$<\/span> ⇒ <span class=\"math-tex\">$AB=3$<\/span><\/p><p><span class=\"math-tex\">$\\overrightarrow{BC}=(4;1;1)$<\/span> ⇒ <span class=\"math-tex\">$BC=3\\sqrt{2}$<\/span><\/p><p><span class=\"math-tex\">$\\overrightarrow{AD}=(a-1;b-2;c-1)$<\/span><\/p><p>Theo gi\u1ea3 thi\u1ebft ABCD là hình thang vuông t\u1ea1i A và B và có di\u1ec7n tích b\u1eb1ng <span class=\"math-tex\">$6\\sqrt{2}$<\/span><\/p><p>⇔ <span class=\"math-tex\">$\\dfrac{1}{2}AB(AD+BC)=6\\sqrt{2}$<\/span> ⇔ <span class=\"math-tex\">$\\dfrac{1}{2}.3.(AD+3\\sqrt{2})=6\\sqrt{2}$<\/span> ⇔ <span class=\"math-tex\">$AD=\\sqrt{2}$<\/span> ⇒ <span class=\"math-tex\">$BC=3AD$<\/span>.<\/p><p>Do ABCD là hình thang vuông t\u1ea1i A và B nên <span class=\"math-tex\">$\\overrightarrow{BC}=3.\\overrightarrow{AD}$<\/span> <\/p><p>⇔ <span class=\"math-tex\">$\\begin{cases}4=3(a-1)\\\\1=3(b-2)\\\\1=3(c-1)\\end{cases}$<\/span> ⇔ <span class=\"math-tex\">$\\begin{cases}a=\\dfrac{7}{3}\\\\b=\\dfrac{7}{3}\\\\c=\\dfrac{4}{3}\\end{cases}$<\/span> ⇒ a + b + c = 6.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-09-06 02:23:38","option_type":"txt","len":2},{"id":"5684","post_id":"7593","mon_id":"1159285","chapter_id":"1159392","question":"<p>Trong không gian Oxyz, cho hình h\u1ed9p ABCD.A'B'C'D' có <span class=\"math-tex\">$A(1;0;1)$<\/span>, <span class=\"math-tex\">$B(2;1;2)$<\/span>, <span class=\"math-tex\">$D(1;-1;1)$<\/span>, <span class=\"math-tex\">$C^\\prime(4;5;-5)$<\/span>. Tính t\u1ecda \u0111\u1ed9 \u0111\u1ec9nh A' c\u1ee7a hình h\u1ed9p.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$A^\\prime(4;6;-5)$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$A^\\prime(2;0;2)$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$A^\\prime(3;5;-6)$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$A^\\prime(3;4;-6)$<\/span>"],"correct":"3","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> <span class=\"math-tex\">$A^\\prime(3;5;-6)$<\/span>.<\/span><\/p><p>Theo quy t\u1eafc hình h\u1ed9p ta có <span class=\"math-tex\">$\\overrightarrow{AB}+\\overrightarrow{AD}+\\overrightarrow{AA^\\prime}=\\overrightarrow{AC^\\prime}$<\/span>.<\/p><p>Suy ra <span class=\"math-tex\">$\\overrightarrow{AA^\\prime}=\\overrightarrow{AC^\\prime}-\\overrightarrow{AB}-\\overrightarrow{AD}$<\/span>.<\/p><p>L\u1ea1i có <span class=\"math-tex\">$\\overrightarrow{AC^\\prime}=(3;5;-6)$<\/span>, <span class=\"math-tex\">$\\overrightarrow{AB}=(1;1;1)$<\/span>, <span class=\"math-tex\">$\\overrightarrow{AD}=(0;-1;0)$<\/span><\/p><p>Do \u0111ó <span class=\"math-tex\">$\\overrightarrow{AA^\\prime}=(2;5;-7)$<\/span>.<\/p><p>Suy ra <span class=\"math-tex\">$A^\\prime(3;5;-6)$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-09-06 02:33:36","option_type":"math","len":0}]}