Trong không gian Oxyz, cho mặt cầu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGtbaacaGLOaGaayzkaaGaaiOoamaabmaabaGaamiEaiabgkHiTiaa % igdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadMhacqGHRaWkcaaIYaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSYaaeWaaeaacaWG6bGaeyOeI0IaaG4maa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaikda % caaI3aaaaa!4CB7! \left( S \right):{\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z - 3} \right)^2} = 27\). Gọi \((\alpha)\) là mặt phẳng đi qua hai điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaaGimaiaacUdacaaIWaGaai4oaiabgkHiTiaaisdaaiaawIca % caGLPaaacaGGSaGaamOqamaabmaabaGaaGOmaiaacUdacaaIWaGaai % 4oaiaaicdaaiaawIcacaGLPaaaaaa!438D! A\left( {0;0; - 4} \right),B\left( {2;0;0} \right)\) và cắt (S) theo giao tuyến là đường tròn (C). Xét các khối nón có đỉnh là tâm của (S) và đáy là ( C ). Biết rằng khi thể tích của khối nón lớn nhất thì mặt phẳng \((\alpha)\) có phương trình dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiaadI % hacqGHRaWkcaWGIbGaamyEaiabgkHiTiaadQhacqGHRaWkcaWGKbGa % eyypa0JaaGimaaaa!4014! ax + by - z + d = 0\). Tính P = a + b + c.

A. -4

B. 8

C. 0

D. 4

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Mặt cầu có tâm I(1;-2;3) và bán kính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9iaaiodadaGcaaqaaiaaiodaaSqabaaaaa!3965! R = 3\sqrt 3 \).

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaadI % eacqGH9aqpcaWGObGaeyO0H4TaamisaiaadgeadaahaaWcbeqaaiaa % ikdaaaGccqGHsislcaWGObWaaWbaaSqabeaacaaIYaaaaOGaeyypa0 % JaaGOmaiaaiEdacqGHsislcaWGObWaaWbaaSqabeaacaaIYaaaaaaa % !4677! IH = h \Rightarrow H{A^2} - {h^2} = 27 - {h^2}\)

Thể tích khối nón đỉnh I đáy là đường tròn ( C) là : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaiodaaaGaeqiWdaNaamisaiaadgea % daahaaWcbeqaaiaaikdaaaGccaGGUaGaamiAaiabg2da9maalaaaba % GaaGymaaqaaiaaiodaaaGaeqiWda3aaeWaaeaacaaIYaGaaG4naiab % gkHiTiaadIgadaahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaaca % WGObGaeyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacqaHapaCdaqa % daqaaiaaikdacaaI3aGaamiAaiabgkHiTiaadIgadaahaaWcbeqaai % aaiodaaaaakiaawIcacaGLPaaaaaa!5555! V = \frac{1}{3}\pi H{A^2}.h = \frac{1}{3}\pi \left( {27 - {h^2}} \right)h = \frac{1}{3}\pi \left( {27h - {h^3}} \right)\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOvayaafa % Gaeyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacqaHapaCdaqadaqa % aiaaikdacaaI3aGaeyOeI0IaaG4maiaadIgadaahaaWcbeqaaiaaik % daaaaakiaawIcacaGLPaaacqGH9aqpcaaIWaGaeyi1HSTaamiAaiab % g2da9iaaiodaaaa!4881! V' = \frac{1}{3}\pi \left( {27 - 3{h^2}} \right) = 0 \Leftrightarrow h = 3\)

Từ đó suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyi1HSTaamiAaiabg2da % 9iaaiodacqGHshI3caWGKbWaaeWaaeaacaWGjbGaai4oamaabmaaba % GaeqySdegacaGLOaGaayzkaaaacaGLOaGaayzkaaGaeyypa0JaaG4m % aaaa!4A2B! {V_{\max }} \Leftrightarrow h = 3 \Rightarrow d\left( {I;\left( \alpha \right)} \right) = 3\).

Mặt phẳng \((\alpha)\) qua \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaaGimaiaacUdacaaIWaGaai4oaiabgkHiTiaaisdaaiaawIca % caGLPaaacaGGSaGaamOqamaabmaabaGaaGOmaiaacUdacaaIWaGaai % 4oaiaaicdaaiaawIcacaGLPaaaaaa!438D! A\left( {0;0; - 4} \right),B\left( {2;0;0} \right)\) và cách I một khoảng là 3.

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGUbWaaSbaaSqaamaabmaabaGaamiuaaGaayjkaiaawMcaaaqabaaa % kiaawEniaiabg2da9maabmaabaGaamyyaiaacUdacaWGIbGaai4oai % abgkHiTiaaigdaaiaawIcacaGLPaaacaGGSaWaa8HaaeaacaWGbbGa % amOqaaGaay51GaWaaeWaaeaacaaIYaGaai4oaiaaicdacaGG7aGaaG % inaaGaayjkaiaawMcaaiabgkDiEpaaFiaabaGaamOBamaaBaaaleaa % daqadaqaaiaadcfaaiaawIcacaGLPaaaaeqaaaGccaGLxdcacaGGUa % Waa8HaaeaacaWGbbGaamOqaaGaay51GaGaeyypa0JaaGOmaiaadgga % cqGHsislcaaI0aGaeyypa0JaaGimaiabgsDiBlaadggacqGH9aqpca % aIYaaaaa!627E! \overrightarrow {{n_{\left( P \right)}}} = \left( {a;b; - 1} \right),\overrightarrow {AB} \left( {2;0;4} \right) \Rightarrow \overrightarrow {{n_{\left( P \right)}}} .\overrightarrow {AB} = 2a - 4 = 0 \Leftrightarrow a = 2\)

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGqbaacaGLOaGaayzkaaGaaiOoaiaaikdacaWG4bGaey4kaSIaamOy % aiaadMhacqGHsislcaWG6bGaeyOeI0IaaGinaiabg2da9iaaicdaaa % a!42E6! \left( P \right):2x + by - z - 4 = 0\). Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacUdadaqadaqaaiabeg7aHbGaayjkaiaawMcaaaGa % ayjkaiaawMcaaiabg2da9maalaaabaWaaqWaaeaacaaIYaGaeyOeI0 % IaaGOmaiaadkgacqGHsislcaaI3aaacaGLhWUaayjcSdaabaWaaOaa % aeaacaaIYaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaaigdadaahaaWcbeqaaiaaikda % aaaabeaaaaGccqGH9aqpcaaIZaaaaa!4F1A! d\left( {I;\left( \alpha \right)} \right) = \frac{{\left| {2 - 2b - 7} \right|}}{{\sqrt {{2^2} + {b^2} + {1^2}} }} = 3\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaaIYaGaamOyaiabgUcaRiaaiwdaaiaawIcacaGLPaaadaah % aaWcbeqaaiaaikdaaaGccqGH9aqpcaaI5aWaaeWaaeaacaaI1aGaey % 4kaSIaamOyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiab % gsDiBlaaiwdacaWGIbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG % OmaiaaicdacaWGIbGaey4kaSIaaGOmaiaaicdacqGH9aqpcaaIWaGa % eyi1HSTaamOyaiabg2da9iaaikdacqGHshI3caWGHbGaey4kaSIaam % OyaiabgUcaRiaadsgacqGH9aqpcaaIYaGaey4kaSIaaGOmaiabgkHi % TiaaisdacqGH9aqpcaaIWaaaaa!63DA! \Leftrightarrow {\left( {2b + 5} \right)^2} = 9\left( {5 + {b^2}} \right) \Leftrightarrow 5{b^2} - 20b + 20 = 0 \Leftrightarrow b = 2 \Rightarrow a + b + d = 2 + 2 - 4 = 0\)

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