Cho hình chóp tứ giác đều S.ABCD có cạnh đáy bằng a . Gọi M, N lần lượt là trung điểm của SA và BC. Biết góc giữa MN và mặt phẳng (ABC) bằng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOnaiaaic % dacqGHWcaSaaa!395A! 60^\circ \) . Khoảng cách giữa hai đường thẳng BC và DM là
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai.png)
Gọi là trung điểm OA . Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaad2 % eacaqGVaGaae4laiaadofacaWGpbGaeyO0H4Taamysaiaad2eacqGH % LkIxdaqadaqaaiaadgeacaWGcbGaam4qaiaadseaaiaawIcacaGLPa % aaaaa!44F9! IM{\rm{//}}SO \Rightarrow IM \bot \left( {ABCD} \right)\) nên hình chiếu của MN lên (ABCD) là IN. Suy ra: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGnbGaamOtaiaadMeaaiaawkWaaiabg2da9iaaiAdacaaIWaGaeyiS % aalaaa!3D95! \widehat {MNI} = 60^\circ \)
Áp dụng định lí cô sin trong \(\Delta CIN\), ta có
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaad6 % eacqGH9aqpdaGcaaqaaiaadoeacaWGjbWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaam4qaiaad6eadaahaaWcbeqaaiaaikdaaaGccqGHsi % slcaaIYaGaam4qaiaadMeacaGGUaGaam4qaiaad6eacaGGUaGaae4y % aiaab+gacaqGZbGaaGinaiaaiwdacqGHWcaSaSqabaGccqGH9aqpda % GcaaqaamaabmaabaWaaSaaaeaacaaIZaGaamyyamaakaaabaGaaGOm % aaWcbeaaaOqaaiaaisdaaaaacaGLOaGaayzkaaWaaWbaaSqabeaaca % aIYaaaaOGaey4kaSYaaeWaaeaadaWcaaqaaiaadggaaeaacaaIYaaa % aaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaik % dadaWcaaqaaiaaiodacaWGHbWaaOaaaeaacaaIYaaaleqaaaGcbaGa % aGinaaaacaGGUaWaaSaaaeaacaWGHbaabaGaaGOmaaaacaGGUaWaaS % aaaeaadaGcaaqaaiaaikdaaSqabaaakeaacaaIYaaaaaWcbeaakiab % g2da9maalaaabaGaamyyamaakaaabaGaaGynaaWcbeaaaOqaaiaaik % dadaGcaaqaaiaaikdaaSqabaaaaaaa!6568! IN = \sqrt {C{I^2} + C{N^2} - 2CI.CN.{\rm{cos}}45^\circ } = \sqrt {{{\left( {\frac{{3a\sqrt 2 }}{4}} \right)}^2} + {{\left( {\frac{a}{2}} \right)}^2} - 2\frac{{3a\sqrt 2 }}{4}.\frac{a}{2}.\frac{{\sqrt 2 }}{2}} = \frac{{a\sqrt 5 }}{{2\sqrt 2 }}\)
Trong tam giác vuông MIN ta có.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiDaiaacg % gacaGGUbGaaGOnaiaaicdacqGHWcaScqGH9aqpdaWcaaqaaiaad2ea % caWGjbaabaGaamysaiaad6eaaaGaeyO0H4TaamytaiaadMeacqGH9a % qpcaWGjbGaamOtaiaac6cadaGcaaqaaiaaiodaaSqabaGccqGH9aqp % daWcaaqaaiaadggadaGcaaqaaiaaigdacaaI1aaaleqaaaGcbaGaaG % OmamaakaaabaGaaGOmaaWcbeaaaaGccqGH9aqpdaWcaaqaaiaadgga % daGcaaqaaiaaiodacaaIWaaaleqaaaGcbaGaaGinaaaacqGHshI3ca % WGtbGaam4taiabg2da9maalaaabaGaamyyamaakaaabaGaaG4maiaa % icdaaSqabaaakeaacaaIYaaaaaaa!5AA5! \tan 60^\circ = \frac{{MI}}{{IN}} \Rightarrow MI = IN.\sqrt 3 = \frac{{a\sqrt {15} }}{{2\sqrt 2 }} = \frac{{a\sqrt {30} }}{4} \Rightarrow SO = \frac{{a\sqrt {30} }}{2}\)
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamOqaiaadoeacaGGSaGaamiraiaad2eaaiaawIcacaGLPaaa % cqGH9aqpcaWGKbWaaeWaaeaacaWGcbGaam4qaiaacYcadaqadaqaai % aadofacaWGbbGaamiraaGaayjkaiaawMcaaaGaayjkaiaawMcaaiab % g2da9iaadsgadaqadaqaaiaad6eacaGGSaWaaeWaaeaacaWGtbGaam % yqaiaadseaaiaawIcacaGLPaaaaiaawIcacaGLPaaacqGH9aqpcaaI % YaGaamizamaabmaabaGaam4taiaacYcadaqadaqaaiaadofacaWGbb % GaamiraaGaayjkaiaawMcaaaGaayjkaiaawMcaaiabg2da9iaaikda % caWGKbWaaeWaaeaacaWGpbGaaiilamaabmaabaGaam4uaiaadkeaca % WGdbaacaGLOaGaayzkaaaacaGLOaGaayzkaaaaaa!6222! d\left( {BC,DM} \right) = d\left( {BC,\left( {SAD} \right)} \right) = d\left( {N,\left( {SAD} \right)} \right) = 2d\left( {O,\left( {SAD} \right)} \right) = 2d\left( {O,\left( {SBC} \right)} \right)\)
Kẻ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadw % eacqGHLkIxcaWGtbGaamOtaiabgkDiElaad+eacaWGfbGaeyyPI41a % aeWaaeaacaWGtbGaamOqaiaadoeaaiaawIcacaGLPaaaaaa!448A! OE \bot SN \Rightarrow OE \bot \left( {SBC} \right)\)
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaam4taiaacYcadaqadaqaaiaadofacaWGcbGaam4qaaGaayjk % aiaawMcaaaGaayjkaiaawMcaaiabg2da9iaad+eacaWGfbaaaa!407E! d\left( {O,\left( {SBC} \right)} \right) = OE\) mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaam4taiaadweadaahaaWcbeqaaiaaikdaaaaaaOGaeyyp % a0ZaaSaaaeaacaaIXaaabaGaam4taiaadofadaahaaWcbeqaaiaaik % daaaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaam4taiaad6eadaah % aaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaacaaI0aaabaGaaG % 4maiaaicdacaWGHbWaaWbaaSqabeaacaaIYaaaaaaakiabgUcaRmaa % laaabaGaaGinaaqaaiaadggadaahaaWcbeqaaiaaikdaaaaaaOGaey % ypa0ZaaSaaaeaacaaI2aGaaGOmaaqaaiaaigdacaaI1aGaamyyamaa % CaaaleqabaGaaGOmaaaaaaGccqGHshI3caWGpbGaamyraiabg2da9m % aalaaabaGaamyyamaakaaabaGaaGymaiaaiwdaaSqabaaakeaadaGc % aaqaaiaaiAdacaaIYaaaleqaaaaaaaa!59C6! \frac{1}{{O{E^2}}} = \frac{1}{{O{S^2}}} + \frac{1}{{O{N^2}}} = \frac{4}{{30{a^2}}} + \frac{4}{{{a^2}}} = \frac{{62}}{{15{a^2}}} \Rightarrow OE = \frac{{a\sqrt {15} }}{{\sqrt {62} }}\)
Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamOqaiaadoeacaGGSaGaamiraiaad2eaaiaawIcacaGLPaaa % cqGH9aqpcaaIYaGaam4taiaadweacqGH9aqpdaWcaaqaaiaaikdaca % WGHbWaaOaaaeaacaaIXaGaaGynaaWcbeaaaOqaamaakaaabaGaaGOn % aiaaikdaaSqabaaaaOGaeyypa0ZaaOaaaeaadaWcaaqaaiaaiodaca % aIWaaabaGaaG4maiaaigdaaaaaleqaaOGaamyyaaaa!4AA8! d\left( {BC,DM} \right) = 2OE = \frac{{2a\sqrt {15} }}{{\sqrt {62} }} = \sqrt {\frac{{30}}{{31}}} a\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 1
