Phương trình mặt phẳng

Home Lớp 12 Toán lớp 12 - Sách Kết nối tri thức Bài 14. Phương trình mặt phẳngBài tập nâng cao

{"common":{"save":0,"post_id":"8212","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6032","post_id":"8212","mon_id":"1159285","chapter_id":"1159382","question":"Trong không gian Oxyz cho \u0111i\u1ec3m M(1 ; 2 ; 3). Vi\u1ebft ph\u01b0\u01a1ng trình m\u1eb7t ph\u1eb3ng (P) \u0111i qua \u0111i\u1ec3m M và c\u1eaft các tr\u1ee5c t\u1ecda \u0111\u1ed9 Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t t\u1ea1i A, B, C sao cho M là tr\u1ecdng tâm c\u1ee7a tam giác ABC.<\/p>","options":["A.<\/strong> $(P):6x+3y+2z+18=0$<\/span>","B.<\/strong> $(P):6x+3y+2z+6=0$<\/span>","C.<\/strong> $(P):6x+3y+2z-18=0$<\/span>","D.<\/strong> $(P):6x+3y+2z-6=0$<\/span>"],"correct":"3","level":"3","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng trình theo \u0111o\u1ea1n ch\u1eafn và tr\u1ecdng tâm tam giác.<\/p>","answer":"Ch\u1ecdn C.<\/strong> $(P):6x+3y+2z-18=0$<\/span>.<\/span><\/p>Theo gi\u1ea3 thi\u1ebft A ∈ Ox, B ∈ Oy, C ∈ Oz nên ta có th\u1ec3 \u0111\u1eb7t A(a ; 0 ; 0), B(0 ; b ; 0) và C(0 ; 0 ; c).<\/p>Vì M(1 ; 2 ; 3) là tr\u1ecdng tâm tam giác ABC nên a = 3 ; b = 6 ; c = 9.<\/p>T\u1eeb \u0111ó ta có ph\u01b0\u01a1ng trình m\u1eb7t ph\u1eb3ng theo \u0111o\u1ea1n ch\u1eafn là:<\/p>$(P):\\dfrac{x}{3}+\\dfrac{y}{6}+\\dfrac{z}{9}=1$<\/span> hay $(P):6x+3y+2z-18=0$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-12-25 08:00:49","option_type":"math","len":0},{"id":"6034","post_id":"8212","mon_id":"1159285","chapter_id":"1159382","question":"Trong không gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz, cho \u0111i\u1ec3m G(1 ; 4 ; 3). M\u1eb7t ph\u1eb3ng nào sau \u0111ây c\u1eaft các tr\u1ee5c Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t t\u1ea1i A, B, C sao cho G là tr\u1ecdng tâm t\u1ee9 di\u1ec7n OABC?<\/p>","options":["A.<\/strong> $\\dfrac{x}{3}+\\dfrac{y}{12}+\\dfrac{z}{9}=1$<\/span>","B.<\/strong> $12x+3y+4z-48=0$<\/span>","C.<\/strong> $\\dfrac{x}{4}+\\dfrac{y}{16}+\\dfrac{z}{12}=0$<\/span>","D.<\/strong> $12x+3y+4z=0$<\/span>"],"correct":"2","level":"3","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng trình theo \u0111o\u1ea1n ch\u1eafn và tr\u1ecdng tâm t\u1ee9 di\u1ec7n.<\/p>","answer":"Ch\u1ecdn B.<\/strong> $12x+3y+4z-48=0$<\/span>.<\/span><\/p>M\u1eb7t ph\u1eb3ng (P) c\u1eaft tr\u1ee5c Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t t\u1ea1i A, B, C nên A(a ; 0 ; 0), B(0 ; b ; 0), C(0 ; 0 ; c).<\/p>Vì G là tr\u1ecdng tâm t\u1ee9 di\u1ec7n OABC nên <\/p>$\\begin{cases}x_G=\\dfrac{x_A+x_B+x_C+x_O}{4}=\\dfrac{a}{4}\\\\ y_G=\\dfrac{y_A+y_B+y_C+y_O}{4}=\\dfrac{b}{4}\\\\ z_G=\\dfrac{z_A+z_B+z_C+z_O}{4}=\\dfrac{c}{4}\\end{cases}$<\/span> hay $\\begin{cases}a=4\\\\b=16\\\\c=12\\end{cases}$<\/span>.<\/p>Khi \u0111ó m\u1eb7t ph\u1eb3ng (P) có ph\u01b0\u01a1ng trình là $\\dfrac{x}{4}+\\dfrac{y}{16}+\\dfrac{z}{12}=1$<\/span> hay $12x+3y+4z-48=0$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-12-25 08:04:14","option_type":"math","len":0},{"id":"6039","post_id":"8212","mon_id":"1159285","chapter_id":"1159382","question":"Vi\u1ebft ph\u01b0\u01a1ng trình m\u1eb7t ph\u1eb3ng (α) \u0111i qua M(2 ; 1 ; –3), bi\u1ebft (α) c\u1eaft tr\u1ee5c Ox, Oy, Oz l\u1ea7n l\u01b0\u1ee3t t\u1ea1i A, B, C sao cho tam giác ABC nh\u1eadn M làm tr\u1ef1c tâm.<\/p>","options":["A.<\/strong> $(\\alpha):2x+5y+z-6=0$<\/span>","B.<\/strong> $(\\alpha):2x+y-6z-23=0$<\/span>","C.<\/strong> $(\\alpha):2x+y-3z-14=0$<\/span>","D.<\/strong> $(\\alpha):3x+4y+3z-1=0$<\/span>"],"correct":"3","level":"3","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng trình theo \u0111o\u1ea1n ch\u1eafn và tr\u1ef1c tâm tam giác.<\/p>","answer":"Ch\u1ecdn C.<\/strong> $(\\alpha):2x+y-3z-14=0$<\/span>.<\/span><\/p>Gi\u1ea3 s\u1eed A(a ; 0 ; 0), B(0 ; b ; 0), C(0 ; 0 ; c) v\u1edbi a $\\ne$<\/span> 0, b $\\ne$<\/span> 0, c $\\ne$<\/span> 0.<\/p>Khi \u0111ó m\u1eb7t ph\u1eb3ng (α) có d\u1ea1ng $\\dfrac{x}{a}+\\dfrac{y}{b}+\\dfrac{z}{c}=1$<\/span>.<\/p>Vì M ∈ (α) nên $\\dfrac{2}{a}+\u200b\u200b\\dfrac{1}{b}-\\dfrac{3}{c}=1\u200b\u200b$<\/span> (1).<\/p>Ta có $\\overrightarrow{AM}=(2-a;1;-3)$<\/span>, $\\overrightarrow{BM}=(2;1-b;-3)$<\/span>, $\\overrightarrow{BC}=(0;-b;c)$<\/span>, $\\overrightarrow{AC}=(-a;0;c)$<\/span>.<\/p>Do M là tr\u1ef1c tâm tam giác ABC nên<\/p>$\\begin{cases}\\overrightarrow{AM}.\\overrightarrow{BC}=0\\\\\\overrightarrow{BM}.\\overrightarrow{AC}=0\\end{cases}$<\/span> ⇒ $\\begin{cases}-b-3c=0\\\\-2a-3c=0\\end{cases}$<\/span> ⇔ $\\begin{cases}b=-3c\\\\a=-\\dfrac{3c}{2}\\end{cases}$<\/span> (2).<\/p>Thay (2) vào (1) ta \u0111\u01b0\u1ee3c $-\\dfrac{4}{3c}-\\dfrac{1}{3c}-\\dfrac{3}{c}=1$<\/span> hay $c=-\\dfrac{14}{3}$<\/span>.<\/p>Suy ra $b=14$<\/span>, $a=7$<\/span>.<\/p>V\u1eady $(\\alpha):\\dfrac{x}{7}+\\dfrac{y}{14}-\\dfrac{3z}{14}=1$<\/span> hay $(\\alpha):2x+y-3z-14=0$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-12-25 08:31:43","option_type":"math","len":0}]}