Tổng và hiệu của hai vectơ - Bài tập trung bình số 2

Home Lớp 10 Toán lớp 10 - Sách Kết nối tri thức Bài 8: Tổng và hiệu của hai vectơBài tập trung bình

{"common":{"save":0,"post_id":"6396","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6217","mon_id":"1158767","chapter_id":"1158861","post_id":"6396","question":"Cho tam giác $ABC$<\/span>  có $AB=AC=a$<\/span>   và  $\\widehat{A}=120^o$<\/span>  . Tính  $|\\overrightarrow{AB}+\\overrightarrow{AC}|$<\/span>  <\/p>","options":["A. $|\\overrightarrow{AB}+\\overrightarrow{AC}|=a\\sqrt3$<\/span>","B. $|\\overrightarrow{AB}+\\overrightarrow{AC}|=a$<\/span>","C. $|\\overrightarrow{AB}+\\overrightarrow{AC}|={a\\over2}$<\/span>"],"correct":"2","level":"2","hint":"","answer":"\u0110áp án B<\/strong><\/span><\/p><svg height=\"246\" width=\"383\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"248\" id=\"canvas_background\" width=\"385\" 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> <image height=\"246\" id=\"svg_1\" width=\"383\" x=\"0\" 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y=\"0\"><\/image> <\/g> <\/svg><\/span><\/p>$|\\overrightarrow{AB}+\\overrightarrow{AC}|=2|\\overrightarrow{AM}|=2AM=2.AB.\\cos60^o=2.a.{1\\over2}=a$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2023-01-04 03:54:09","option_type":"math","len":0},{"id":"6218","mon_id":"1158767","chapter_id":"1158861","post_id":"6396","question":"Cho tam giác  $ABC$<\/span> v\u1edbi $AB=3,AC=4,\\angle BAC=120^o$<\/span>  . Tính \u0111\u1ed9 dài c\u1ee7a véc t\u01a1  $\\overrightarrow{\\nu}=\\overrightarrow{AB}+\\overrightarrow{AC}$<\/span><\/p>","options":["A. $|\\overrightarrow{\\nu}|=2\\sqrt{13}$<\/span>","B. $|\\overrightarrow{\\nu}|=\\sqrt{13}$<\/span>","C. $|\\overrightarrow{\\nu}|={\\sqrt{13}\\over2}$<\/span>","D. $|\\overrightarrow{\\nu}|=\\sqrt{37}$<\/span>"],"correct":"2","level":"2","hint":"","answer":"\u0110áp án: B<\/strong><\/span><\/p><svg height=\"206\" width=\"384\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"208\" id=\"canvas_background\" width=\"386\" 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> <image height=\"206\" id=\"svg_1\" width=\"384\" x=\"0\" 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\" y=\"0\"><\/image> <\/g> <\/svg><\/span><\/p>$\\overrightarrow{\\nu}=\\overrightarrow{AB}+\\overrightarrow{AC}\\Rightarrow |\\overrightarrow{\\nu}|=|\\overrightarrow{AB}+\\overrightarrow{AC}|=|\\overrightarrow{AD}|=2AM$<\/span><\/p>$BC^2=AB^2+AC^2-2AB.AC.\\cos120^o=3^2+4^2-2.3.4.\\left(1\\over2\\right)=37$<\/span><\/p>$AM^2=={AB^2+AC^2\\over2}-{BC^2\\over4}\\Rightarrow AM={\\sqrt{13}\\over2}\\Rightarrow AD=2AM=\\sqrt{13}$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2023-01-04 04:04:23","option_type":"math","len":0},{"id":"6219","mon_id":"1158767","chapter_id":"1158861","post_id":"6396","question":"Cho hình thoi $ABCD$<\/span> có $AC=2a$<\/span>  và $BD=a$<\/span>  . Tính  $|\\overrightarrow{AC}+\\overrightarrow{BD}|$<\/span><\/p>","options":["A. $|\\overrightarrow{AC}+\\overrightarrow{BD}|=3a$<\/span>","B. $|\\overrightarrow{AC}+\\overrightarrow{BD}|=a\\sqrt3$<\/span>","C. $|\\overrightarrow{AC}+\\overrightarrow{BD}|=a\\sqrt5$<\/span>","D. $|\\overrightarrow{AC}+\\overrightarrow{BD}|=5a$<\/span>"],"correct":"3","level":"2","hint":"","answer":"\u0110áp án: C<\/strong><\/span><\/p><svg height=\"253\" width=\"379\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"255\" id=\"canvas_background\" width=\"381\" 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> <image height=\"253\" id=\"svg_1\" width=\"379\" x=\"0\" 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\" y=\"0\"><\/image> <\/g> <\/svg><\/span><\/p>G\u1ecdi $O$<\/span> là giao \u0111i\u1ec3m c\u1ee7a $AC,BD$<\/span><\/p>$\\overrightarrow{AC}+\\overrightarrow{BD}=2\\overrightarrow{OC}+2\\overrightarrow{OD}=2\\left(\\overrightarrow{OC}+\\overrightarrow{OD}\\right)=4\\overrightarrow{OI}=2\\overrightarrow{BC}$<\/span><\/p>$\\Rightarrow|\\overrightarrow{AC}+\\overrightarrow{BD}|=2BC=2\\sqrt{a^2+{a\\over2}}=a\\sqrt5$<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"147","test":"0","date":"2023-01-04 04:15:38","option_type":"math","len":0}]}