Trong không gian Oxyz, cho điểm A(2;-3;4), đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHsislcaaIXaaabaGaaGOmaaaacqGH9aqp % daWcaaqaaiaadMhacqGHRaWkcaaIYaaabaGaaGymaaaacqGH9aqpda % WcaaqaaiaadQhaaeaacaaIYaaaaaaa!424A! d:\frac{{x - 1}}{2} = \frac{{y + 2}}{1} = \frac{z}{2}\) và mặt cầu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGtbaacaGLOaGaayzkaaGaaiOoamaabmaabaGaamiEaiabgkHiTiaa % iodaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadMhacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSYaaeWaaeaacaWG6bGaey4kaSIaaGymaa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaikda % caaIWaaaaa!4CB1! \left( S \right):{\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z + 1} \right)^2} = 20\). Mặt phẳng (P) chứa đường thẳng d thỏa mãn khoảng cách từ điểm A đến (P) lớn nhất. Mặt cầu (S) cắt (P) theo đường tròn có bán kính bằng:
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Lời giải:
Báo saiGọi H, K lần lượt là hình chiếu của A lên (P) và d ta có , khi đó mặt phẳng (P) chứa đường thẳng d thỏa mãn khoảng cách từ điểm A đến (P) lớn nhất ; (P) nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGbbGaam4saaGaay51Gaaaaa!393E! \overrightarrow {AK} \) là 1 VTPT.
Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaabm % aabaGaaGymaiabgUcaRiaaikdacaWG0bGaai4oaiabgkHiTiaaikda % cqGHRaWkcaWG0bGaai4oaiaaikdacaWG0baacaGLOaGaayzkaaGaey % icI4SaamizaiabgkDiEpaaFiaabaGaamyqaiaadUeaaiaawEniaiab % g2da9maabmaabaGaaGOmaiaadshacqGHsislcaaIXaGaai4oaiaads % hacqGHRaWkcaaIXaGaai4oaiaaikdacaWG0bGaeyOeI0IaaGinaaGa % ayjkaiaawMcaaaaa!57CA! K\left( {1 + 2t; - 2 + t;2t} \right) \in d \Rightarrow \overrightarrow {AK} = \left( {2t - 1;t + 1;2t - 4} \right)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1bWaaSbaaSqaaiaadsgaaeqaaaGccaGLxdcadaqadaqaaiaaikda % caGG7aGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3EFB! \overrightarrow {{u_d}} \left( {2;1;2} \right)\) là 1 VTCP của d
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3daWhcaqaaiaadgeacaWGlbaacaGLxdcacaGGUaWaa8HaaeaacaWG % 1bWaaSbaaSqaaiaadsgaaeqaaaGccaGLxdcacqGH9aqpcaaIWaGaey % i1HSTaaGinaiaadshacqGHsislcaaIYaGaey4kaSIaamiDaiabgUca % RiaaigdacqGHRaWkcaaI0aGaamiDaiabgkHiTiaaiIdacqGH9aqpca % aIWaGaeyi1HSTaaGyoaiaadshacqGHsislcaaI5aGaeyypa0JaaGim % aiabgsDiBlaadshacqGH9aqpcaaIXaaabaGaeyO0H4Taam4samaabm % aabaGaaG4maiaacUdacqGHsislcaaIXaGaai4oaiaaikdaaiaawIca % caGLPaaacqGHshI3daWhcaqaaiaadgeacaWGlbaacaGLxdcacqGH9a % qpdaqadaqaaiaaigdacaGG7aGaaGOmaiaacUdacqGHsislcaaIYaaa % caGLOaGaayzkaaaabaGaeyO0H49aaeWaaeaacaWGqbaacaGLOaGaay % zkaaGaaiOoaiaadIhacqGHsislcaaIZaGaey4kaSIaaGOmamaabmaa % baGaamyEaiabgUcaRiaaigdaaiaawIcacaGLPaaacqGHsislcaaIYa % WaaeWaaeaacaWG6bGaeyOeI0IaaGOmaaGaayjkaiaawMcaaiabg2da % 9iaaicdacqGHuhY2caWG4bGaey4kaSIaaGOmaiaadMhacqGHsislca % aIYaGaamOEaiabgUcaRiaaiodacqGH9aqpcaaIWaaaaaa!955D! \begin{array}{l} \Rightarrow \overrightarrow {AK} .\overrightarrow {{u_d}} = 0 \Leftrightarrow 4t - 2 + t + 1 + 4t - 8 = 0 \Leftrightarrow 9t - 9 = 0 \Leftrightarrow t = 1\\ \Rightarrow K\left( {3; - 1;2} \right) \Rightarrow \overrightarrow {AK} = \left( {1;2; - 2} \right)\\ \Rightarrow \left( P \right):x - 3 + 2\left( {y + 1} \right) - 2\left( {z - 2} \right) = 0 \Leftrightarrow x + 2y - 2z + 3 = 0 \end{array}\)
Mặt cầu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGtbaacaGLOaGaayzkaaGaaiOoamaabmaabaGaamiEaiabgkHiTiaa % iodaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadMhacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSYaaeWaaeaacaWG6bGaey4kaSIaaGymaa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaikda % caaIWaaaaa!4CB1! \left( S \right):{\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z + 1} \right)^2} = 20\) có tâm I ( 3;2;-1), bán kính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maakaaabaGaaGOmaiaaicdaaSqabaGccqGH9aqpcaaIYaWaaOaa % aeaacaaI1aaaleqaaaaa!3C08! R = \sqrt {20} = 2\sqrt 5 \)
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabg2 % da9iaadsgadaqadaqaaiaadMeacaGG7aWaaeWaaeaacaWGqbaacaGL % OaGaayzkaaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaadaabdaqaai % aaiodacqGHRaWkcaaIYaGaaiOlaiaaikdacqGHsislcaaIYaWaaeWa % aeaacqGHsislcaaIXaaacaGLOaGaayzkaaGaey4kaSIaaG4maaGaay % 5bSlaawIa7aaqaamaakaaabaGaaGymaiabgUcaRiaaisdacqGHRaWk % caaI0aaaleqaaaaakiabg2da9maalaaabaGaaGymaiaaikdaaeaaca % aIZaaaaiabg2da9iaaisdaaaa!55E8! d = d\left( {I;\left( P \right)} \right) = \frac{{\left| {3 + 2.2 - 2\left( { - 1} \right) + 3} \right|}}{{\sqrt {1 + 4 + 4} }} = \frac{{12}}{3} = 4\)
Gọi r là đường kính đường tròn giao tuyến của (P) và (S) ta có:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuamaaCa % aaleqabaGaaGOmaaaakiabg2da9iaadsgadaahaaWcbeqaaiaaikda % aaGccqGHRaWkcaWGYbWaaWbaaSqabeaacaaIYaaaaOGaeyi1HSTaam % OCaiabg2da9maakaaabaGaamOuamaaCaaaleqabaGaaGOmaaaakiab % gkHiTiaadsgadaahaaWcbeqaaiaaikdaaaaabeaakiabg2da9maaka % aabaGaaGOmaiaaicdacqGHsislcaaIXaGaaGOnaaWcbeaakiabg2da % 9iaaikdaaaa!4D33! {R^2} = {d^2} + {r^2} \Leftrightarrow r = \sqrt {{R^2} - {d^2}} = \sqrt {20 - 16} = 2\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 4