Trong không gian Oxyz, cho mặt cầu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGtbaacaGLOaGaayzkaaGaaiOoamaabmaabaGaamiEaiabgkHiTiaa % igdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkda % qadaqaaiaadMhacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSYaaeWaaeaacaWG6bGaeyOeI0IaaGymaa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaioda % daahaaWcbeqaaiaaikdaaaaaaa!4CE9! \left( S \right):{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 1} \right)^2} = {3^2}\) , mặt phẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGqbaacaGLOaGaayzkaaGaaiOoaiaadIhacqGHsislcaWG5bGaey4k % aSIaamOEaiabgUcaRiaaiodacqGH9aqpcaaIWaaaaa!4137! \left( P \right):x - y + z + 3 = 0\) và điểm N(1;0;-4) thuộc (P). Một đường thẳng \(\Delta\) đi qua N nằm trong (P) cắt (S) tại hai điểm A,B thỏa mãn AB =4. Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaqadaqaaiaaigdacaGG7aGaamOyaiaa % cUdacaWGJbaacaGLOaGaayzkaaGaaiilamaabmaabaGaam4yaiabg6 % da+iaaicdaaiaawIcacaGLPaaaaaa!441B! \overrightarrow u = \left( {1;b;c} \right),\left( {c > 0} \right)\) là một vecto chỉ phương của \(\Delta\), tổng b+c bằng

A. 1

B. 3

C. -1

D. 45

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Mặt cầu (S) có tâm I(1;2;1) bán kính R=3.

Do \(\Delta\)nằm trong (P) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1bWaaSbaaSqaaiabfs5aebqabaaakiaawEniaiaac6cadaWhcaqa % aiaad6gadaWgaaWcbaWaaeWaaeaacaWGqbaacaGLOaGaayzkaaaabe % aaaOGaay51GaGaeyypa0JaaGimaiabgsDiBlaaigdacqGHsislcaWG % IbGaey4kaSIaam4yaiabg2da9iaaicdacqGHuhY2caWGIbGaeyypa0 % Jaam4yaiabgUcaRiaaigdaaaa!512D! \overrightarrow {{u_\Delta }} .\overrightarrow {{n_{\left( P \right)}}} = 0 \Leftrightarrow 1 - b + c = 0 \Leftrightarrow b = c + 1\).

Mặt khác ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaaca % WGKbWaaeWaaeaacaWGjbGaai4oamaabmaabaGaamyqaiaadkeaaiaa % wIcacaGLPaaaaiaawIcacaGLPaaaaiaawUfacaGLDbaadaahaaWcbe % qaaiaaikdaaaGccqGHRaWkdaqadaqaamaalaaabaGaamyqaiaadkea % aeaacaaIYaaaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaki % abg2da9iaadkfadaahaaWcbeqaaiaaikdaaaGccqGHshI3caWGKbWa % aeWaaeaacaWGjbGaai4oamaabmaabaGaamyqaiaadkeaaiaawIcaca % GLPaaaaiaawIcacaGLPaaacqGH9aqpdaGcaaqaaiaadkfadaahaaWc % beqaaiaaikdaaaGccqGHsisldaqadaqaamaalaaabaGaamyqaiaadk % eaaeaacaaIYaaaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaa % aeqaaOGaeyypa0ZaaOaaaeaacaaI1aaaleqaaaaa!5C68! {\left[ {d\left( {I;\left( {AB} \right)} \right)} \right]^2} + {\left( {\frac{{AB}}{2}} \right)^2} = {R^2} \Rightarrow d\left( {I;\left( {AB} \right)} \right) = \sqrt {{R^2} - {{\left( {\frac{{AB}}{2}} \right)}^2}} = \sqrt 5 \)

Lại có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacUdacqqHuoaraiaawIcacaGLPaaacqGH9aqpdaWc % aaqaamaaemaabaWaamWaaeaadaWhcaqaaiaadMeacaWGobGaaiOlaa % Gaay51GaWaa8HaaeaacaWG1bWaaSbaaSqaaiabfs5aebqabaaakiaa % wEniaaGaay5waiaaw2faaaGaay5bSlaawIa7aaqaamaaemaabaWaa8 % HaaeaacaWG1bWaaSbaaSqaaiabfs5aebqabaaakiaawEniaaGaay5b % SlaawIa7aaaacqGH9aqpdaWcaaqaamaaemaabaWaaeWaaeaacaaIYa % Gaam4yaiabgkHiTiaaiwdacaWGIbGaai4oaiaaiwdacaGG7aGaeyOe % I0IaaGOmaaGaayjkaiaawMcaaaGaay5bSlaawIa7aaqaamaaemaaba % WaaeWaaeaacaaIXaGaai4oaiaadkgacaGG7aGaam4yaaGaayjkaiaa % wMcaaaGaay5bSlaawIa7aaaacqGH9aqpdaWcaaqaamaakaaabaWaae % WaaeaacaaIYaGaam4yaiabgkHiTiaaiwdacaWGIbaacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaaiMdaaSqaba % aakeaadaGcaaqaaiaaigdacqGHRaWkcaWGIbWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaam4yamaaCaaaleqabaGaaGOmaaaaaeqaaaaaki % abg2da9maakaaabaGaaGynaaWcbeaaaaa!7A43! d\left( {I;\Delta } \right) = \frac{{\left| {\left[ {\overrightarrow {IN.} \overrightarrow {{u_\Delta }} } \right]} \right|}}{{\left| {\overrightarrow {{u_\Delta }} } \right|}} = \frac{{\left| {\left( {2c - 5b;5; - 2} \right)} \right|}}{{\left| {\left( {1;b;c} \right)} \right|}} = \frac{{\sqrt {{{\left( {2c - 5b} \right)}^2} + 29} }}{{\sqrt {1 + {b^2} + {c^2}} }} = \sqrt 5 \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaS % aaaeaadaqadaqaaiaaikdacaWGJbGaeyOeI0IaaGynaiaadogacqGH % sislcaaI1aaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey % 4kaSIaaGOmaiaaiMdaaeaacaaIXaGaey4kaSYaaeWaaeaacaWGJbGa % ey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaki % abgUcaRiaadogadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaaGyn % aiabgsDiBpaabmaabaGaaG4maiaadogacqGHRaWkcaaI1aaacaGLOa % GaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaaiMda % cqGH9aqpcaaI1aWaaeWaaeaacaaIYaGaamiEamaaCaaaleqabaGaaG % OmaaaakiabgUcaRiaaikdacaWGJbGaey4kaSIaaGOmaaGaayjkaiaa % wMcaaiabgsDiBlaadogadaahaaWcbeqaaiaaikdaaaGccqGHsislca % aIYaGaaGimaiaadogacqGHsislcaaI0aGaaGinaiabg2da9iaaicda % cqGHuhY2daWabaabaeqabaGaam4yaiabg2da9iaaikdacaaIYaaaba % Gaam4yaiabg2da9iabgkHiTiaaikdaaaGaay5waaaaaa!78E3! \Leftrightarrow \frac{{{{\left( {2c - 5c - 5} \right)}^2} + 29}}{{1 + {{\left( {c + 1} \right)}^2} + {c^2}}} = 5 \Leftrightarrow {\left( {3c + 5} \right)^2} + 29 = 5\left( {2{x^2} + 2c + 2} \right) \Leftrightarrow {c^2} - 20c - 44 = 0 \Leftrightarrow \left[ \begin{array}{l} c = 22\\ c = - 2 \end{array} \right.\) Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yaiabg6 % da+iaaicdacqGHshI3caWGJbGaeyypa0JaaGOmaiaaikdacaGG7aGa % amOyaiabg2da9iaaikdacaaIZaGaeyO0H4TaamOyaiabgUcaRiaado % gacqGH9aqpcaaI0aGaaGynaaaa!4A16! c > 0 \Rightarrow c = 22;b = 23 \Rightarrow b + c = 45\)

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