{"common":{"save":0,"post_id":"8308","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6134","post_id":"8308","mon_id":"1159285","chapter_id":"1159382","question":"<p>Trong không gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz cho m\u1eb7t ph\u1eb3ng (α) có ph\u01b0\u01a1ng trình <span class=\"math-tex\">$2x+y-z-1=0$<\/span> và m\u1eb7t c\u1ea7u (S) có ph\u01b0\u01a1ng trình <span class=\"math-tex\">$(x-1)^2+(y-1)^2+(z+2)^2=4$<\/span>. Xác \u0111\u1ecbnh bán kính r c\u1ee7a \u0111\u01b0\u1eddng tròn là giao tuy\u1ebfn c\u1ee7a m\u1eb7t ph\u1eb3ng (α) và m\u1eb7t c\u1ea7u (S).<br \/>.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{42}}{3}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{3}}{3}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{15}}{3}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{7}}{3}$<\/span>"],"correct":"2","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{3}}{3}$<\/span>.<\/span><\/p><p>M\u1eb7t c\u1ea7u (S) có tâm <span class=\"math-tex\">$I(1;1;-2)$<\/span> và bán kính R = 2.<\/p><p>G\u1ecdi d là kho\u1ea3ng cách t\u1eeb tâm I \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (α).<\/p><p>Ta có <span class=\"math-tex\">$d=d(I;(\\alpha))=\\dfrac{|2+1+2-1|}{\\sqrt{4+1+1}}=\\dfrac{2\\sqrt{6}}{3}$<\/span>.<\/p><p>Khi \u0111ó ta có: <span class=\"math-tex\">$r=\\sqrt{R^2-d^2}=\\dfrac{2\\sqrt{3}}{3}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 16:32:11","option_type":"math","len":0},{"id":"6137","post_id":"8308","mon_id":"1159285","chapter_id":"1159382","question":"<p>Trong không gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng <span class=\"math-tex\">$\\Delta:\\dfrac{x+2}{-1}=\\dfrac{y}{1}=\\dfrac{z-3}{-1}$<\/span> và m\u1eb7t c\u1ea7u (S): <span class=\"math-tex\">$x^2+y^2+z^2-2x-4y+6z-67=0$<\/span>. S\u1ed1 \u0111i\u1ec3m chung c\u1ee7a <span class=\"math-tex\">$\\Delta$<\/span> và (S) là<\/p>","options":["<strong>A.<\/strong> 3","<strong>B.<\/strong> 0","<strong>C.<\/strong> 1","<strong>D.<\/strong> 2"],"correct":"4","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> 2.<\/span><\/p><p>\u0110\u01b0\u1eddng th\u1eb3ng <span class=\"math-tex\">$\\Delta$<\/span> \u0111i qua M(–2 ; 0 ; 3) và có VTCP <span class=\"math-tex\">$\\overrightarrow{u}=(-1;1;-1)$<\/span>.<\/p><p>M\u1eb7t c\u1ea7u (S) có tâm I(1 ; 2 ; –3) và bán kính R = 9.<\/p><p>Ta có <span class=\"math-tex\">$\\overrightarrow{MI}=(3;2;-6)$<\/span> và <span class=\"math-tex\">$[\\overrightarrow{u};\\overrightarrow{MI}]=(-4;-9;-5)$<\/span>.<\/p><p>⇒ <span class=\"math-tex\">$d(I;\\Delta)=\\dfrac{[\\overrightarrow{u};\\overrightarrow{MI}]}{|\\overrightarrow{u}|}=\\dfrac{\\sqrt{366}}{3}$<\/span>.<\/p><p>Vì <span class=\"math-tex\">$d(I,\\Delta)< R$<\/span> nên <span class=\"math-tex\">$\\Delta$<\/span> c\u1eaft m\u1eb7t c\u1ea7u (S) t\u1ea1i hai \u0111i\u1ec3m phân bi\u1ec7t.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 16:40:50","option_type":"txt","len":0},{"id":"6140","post_id":"8308","mon_id":"1159285","chapter_id":"1159382","question":"<p>Trong không gian Oxyz, cho \u0111\u01b0\u1eddng th\u1eb3ng <span class=\"math-tex\">$\\Delta:\\dfrac{x}{2}=\\dfrac{y-1}{1}=\\dfrac{z-2}{-1}$<\/span> và m\u1eb7t c\u1ea7u (S): <span class=\"math-tex\">$x^2+y^2+z^2-2x+4z+1=0$<\/span>. S\u1ed1 \u0111i\u1ec3m chung c\u1ee7a <span class=\"math-tex\">$\\Delta$<\/span> và (S) là:<\/p>","options":["<strong>A.<\/strong> 0","<strong>B.<\/strong> 1","<strong>C.<\/strong> 2","<strong>D.<\/strong> 3"],"correct":"1","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> 0.<\/span><\/p><p>\u0110\u01b0\u1eddng th\u1eb3ng <span class=\"math-tex\">$\\Delta$<\/span> \u0111i qua <span class=\"math-tex\">$M(0;1;2)$<\/span> và có VTCP <span class=\"math-tex\">$\\overrightarrow{u}=(2;1;-1)$<\/span>.<\/p><p>M\u1eb7t c\u1ea7u (S) có tâm <span class=\"math-tex\">$I(1;0;-2)$<\/span> và bán kính R = 2.<\/p><p>Ta có <span class=\"math-tex\">$\\overrightarrow{MI}=(1;-1;-4)$<\/span> và <span class=\"math-tex\">$[\\overrightarrow{u};\\overrightarrow{MI}]=(-5;7;-3)$<\/span>.<\/p><p><span class=\"math-tex\">$d(I;\\Delta)=\\dfrac{[\\overrightarrow{u},\\overrightarrow{MI}]}{|\\overrightarrow{u}|}=\\dfrac{\\sqrt{498}}{6}$<\/span>.<\/p><p>Vì <span class=\"math-tex\">$d(I;\\Delta)>R$<\/span> nên <span class=\"math-tex\">$\\Delta$<\/span> không c\u1eaft m\u1eb7t c\u1ea7u (S).<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-05 16:52:40","option_type":"txt","len":0}]}