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HomeLớp 12Toán lớp 12 - Sách Kết nối tri thứcBài tập cuối chương 2. VECTƠ VÀ HỆ TRỤC TOẠ ĐỘ TRONG KHÔNG GIAN.Bài tập trung bình
{"common":{"save":0,"post_id":"7620","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5685","post_id":"7620","mon_id":"1159285","chapter_id":"1159392","question":"<p>Cho hai vect\u01a1 <span class=\"math-tex\">$\\overrightarrow{a},\\overrightarrow{b}$<\/span> có <span class=\"math-tex\">$(\\overrightarrow{a}+2\\overrightarrow{b})$<\/span> vuông góc v\u1edbi <span class=\"math-tex\">$(5\\overrightarrow{a}-4\\overrightarrow{b})$<\/span> và <span class=\"math-tex\">$|\\overrightarrow{a}|=|\\overrightarrow{b}|$<\/span>. Khi \u0111ó:<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{\\sqrt{2}}{2}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=0$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{\\sqrt{3}}{2}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{1}{2}$<\/span>"],"correct":"4","level":"2","hint":"<p><span class=\"math-tex\">$\\overrightarrow{a}\\bot\\overrightarrow{b}$<\/span> ⇔ <span class=\"math-tex\">$\\overrightarrow{a}.\\overrightarrow{b}=0$<\/span><\/p><p><span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{\\overrightarrow{a}.\\overrightarrow{b}}{|\\overrightarrow{a}|.|\\overrightarrow{b}|}$<\/span><\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> <span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{1}{2}$<\/span><\/span><\/p><p>Vì <span class=\"math-tex\">$(\\overrightarrow{a}+2\\overrightarrow{b})$<\/span> vuông góc v\u1edbi <span class=\"math-tex\">$(5\\overrightarrow{a}-4\\overrightarrow{b})$<\/span> nên<\/p><p><span class=\"math-tex\">$(\\overrightarrow{a}+2\\overrightarrow{b}).(5\\overrightarrow{a}-4\\overrightarrow{b})=0$<\/span> ⇔ <span class=\"math-tex\">$5\\overrightarrow{a}^2+6\\overrightarrow{a}.\\overrightarrow{b}-8\\overrightarrow{b}^2=0$<\/span> ⇔ <span class=\"math-tex\">$\\overrightarrow{a}.\\overrightarrow{b}=\\dfrac{-5\\overrightarrow{a}^2+8\\overrightarrow{b}^2}{6}$<\/span><\/p><p>Ta có: <span class=\"math-tex\">$|\\overrightarrow{a}|=|\\overrightarrow{b}|$<\/span> ⇔ <span class=\"math-tex\">$|\\overrightarrow{a}|^2=|\\overrightarrow{b}|^2$<\/span>. Suy ra <span class=\"math-tex\">$\\overrightarrow{a}.\\overrightarrow{b}=\\dfrac{3\\overrightarrow{a}^2}{6}$<\/span>.<\/p><p><span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{\\overrightarrow{a}.\\overrightarrow{b}}{|\\overrightarrow{a}|.|\\overrightarrow{b}|}=\\dfrac{3\\overrightarrow{a}^2}{6\\overrightarrow{a}^2}=\\dfrac{1}{2}$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-09-06 02:34:33","option_type":"math","len":0},{"id":"5686","post_id":"7620","mon_id":"1159285","chapter_id":"1159392","question":"<p>Cho hai vect\u01a1 <span class=\"math-tex\">$\\overrightarrow{a},\\overrightarrow{b}$<\/span> th\u1ecfa mãn: <span class=\"math-tex\">$|\\overrightarrow{a}|=4;|\\overrightarrow{b}|=3;|\\overrightarrow{a}-\\overrightarrow{b}|=4$<\/span>. G\u1ecdi α là góc gi\u1eefa hai vect\u01a1 <span class=\"math-tex\">$\\overrightarrow{a},\\overrightarrow{b}$<\/span>. Ch\u1ecdn kh\u1eb3ng \u0111\u1ecbnh <strong>\u0111úng<\/strong>?<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\cos\\alpha=\\dfrac{3}{8}$<\/span>","<strong>B.<\/strong> α = 30°","<strong>C.<\/strong> <span class=\"math-tex\">$\\cos\\alpha=\\dfrac{1}{3}$<\/span>","<strong>D.<\/strong> α = 60°"],"correct":"1","level":"2","hint":"<p><span class=\"math-tex\">$(\\overrightarrow{a}-\\overrightarrow{b})^2= |\\overrightarrow{a}|^2+|\\overrightarrow{b}|^2-2\\overrightarrow{a}.\\overrightarrow{b}$<\/span><\/p><p><span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{\\overrightarrow{a}.\\overrightarrow{b}}{|\\overrightarrow{a}|.|\\overrightarrow{b}|}$<\/span><\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$\\cos\\alpha=\\dfrac{3}{8}$<\/span><\/span><\/p><p><span class=\"math-tex\">$(\\overrightarrow{a}-\\overrightarrow{b})^2= |\\overrightarrow{a}|^2+|\\overrightarrow{b}|^2-2\\overrightarrow{a}.\\overrightarrow{b}$<\/span> ⇔ <span class=\"math-tex\">$\\overrightarrow{a}.\\overrightarrow{b}=\\dfrac{9}{2}$<\/span><\/p><p>Do \u0111ó <span class=\"math-tex\">$\\cos(\\overrightarrow{a},\\overrightarrow{b})=\\dfrac{\\overrightarrow{a}.\\overrightarrow{b}}{|\\overrightarrow{a}|.|\\overrightarrow{b}|}=\\dfrac{3}{8}$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-09-06 02:37:06","option_type":"math","len":0},{"id":"5687","post_id":"7620","mon_id":"1159285","chapter_id":"1159392","question":"<p><span class=\"math-tex\">$\\overrightarrow{u}$<\/span> và <span class=\"math-tex\">$\\overrightarrow{v}$<\/span> là 2 vect\u01a1 \u0111\u1ec1u khác <span class=\"math-tex\">$\\overrightarrow{0}$<\/span>. Khi \u0111ó <span class=\"math-tex\">$|\\overrightarrow{u}+2\\overrightarrow{v}|^2$<\/span> b\u1eb1ng<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\overrightarrow{u}^2+2\\overrightarrow{v}^2-4\\overrightarrow{u}.\\overrightarrow{v}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\overrightarrow{u}^2+4\\overrightarrow{v}^2+4\\overrightarrow{u}.\\overrightarrow{v}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\overrightarrow{u}^2+4\\overrightarrow{v}^2$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$4\\overrightarrow{u}.\\overrightarrow{v}(\\overrightarrow{u}-\\overrightarrow{v})$<\/span>"],"correct":"2","level":"2","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> <span class=\"math-tex\">$\\overrightarrow{u}^2+4\\overrightarrow{v}^2+4\\overrightarrow{u}.\\overrightarrow{v}$<\/span><\/span><\/p><p><span class=\"math-tex\">$|\\overrightarrow{u}+2\\overrightarrow{v}|^2=(\\overrightarrow{u}+2\\overrightarrow{v})^2=\\overrightarrow{u}^2+4\\overrightarrow{v}+4\\overrightarrow{u}.\\overrightarrow{v}$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-09-06 02:52:12","option_type":"math","len":0}]}