Trong không gian Oxyz, cho hai đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaGabaabaeqabaGaamiEaiabg2da9iabgkHiTiaaigdacqGHsisl % caaIYaGaamiDaaqaaiaadMhacqGH9aqpcaWG0baabaGaamOEaiabg2 % da9iabgkHiTiaaigdacqGHRaWkcaaIZaGaamiDaaaacaGL7baacaGG % SaGaamizaiaacEcacaGG6aWaaiqaaqaabeqaaiaadIhacqGH9aqpca % aIYaGaey4kaSIaamiDaiaacEcaaeaacaWG5bGaeyypa0JaeyOeI0Ia % aGymaiabgUcaRiaaikdacaWG0bGaai4jaaqaaiaadQhacqGH9aqpcq % GHsislcaaIYaGaamiDaiaacEcaaaGaay5Eaaaaaa!5DF5! d:\left\{ \begin{array}{l} x = - 1 - 2t\\ y = t\\ z = - 1 + 3t \end{array} \right.,d':\left\{ \begin{array}{l} x = 2 + t'\\ y = - 1 + 2t'\\ z = - 2t' \end{array} \right.\) và mặt phẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGqbaacaGLOaGaayzkaaGaaiOoaiaadIhacqGHRaWkcaWG5bGaey4k % aSIaamOEaiabgUcaRiaaikdacqGH9aqpcaaIWaaaaa!412C! \left( P \right):x + y + z + 2 = 0\) . Đường thẳng vuông góc với mặt phẳng (P) và cắt hai đường thẳng d,d' có phương trình là:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaeyOeI0IaaG4maaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaeyOeI0IaaGymaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG6b % Gaey4kaSIaaGOmaaqaaiaaigdaaaaaaa!424B! \frac{{x - 3}}{1} = \frac{{y - 1}}{1} = \frac{{z + 2}}{1}\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaeyOeI0IaaGymaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaeyOeI0IaaGymaaqaaiabgkHiTiaaigdaaaGaeyypa0ZaaSaaae % aacaWG6bGaeyOeI0IaaGymaaqaaiabgkHiTiaaisdaaaaaaa!4430! \frac{{x - 1}}{1} = \frac{{y - 1}}{{ - 1}} = \frac{{z - 1}}{{ - 4}}\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaey4kaSIaaGOmaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaey4kaSIaaGymaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG6b % GaeyOeI0IaaGymaaqaaiaaigdaaaaaaa!423E! \frac{{x + 2}}{1} = \frac{{y + 1}}{1} = \frac{{z - 1}}{1}\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaey4kaSIaaGymaaqaaiaaikdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaeyOeI0IaaGymaaqaaiaaikdaaaGaeyypa0ZaaSaaaeaacaWG6b % GaeyOeI0IaaGinaaqaaiaaikdaaaaaaa!424E! \frac{{x + 1}}{2} = \frac{{y - 1}}{2} = \frac{{z - 4}}{2}\)

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Gọi \(\Delta\) là đường thẳng cần tìm

Giả sử \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2 % da9iabfs5aejabgMIihlaadsgacqGHshI3caWGbbWaaeWaaeaacqGH % sislcaaIXaGaeyOeI0IaaGOmaiaadshacaGG7aGaamiDaiaacUdacq % GHsislcaaIXaGaey4kaSIaaG4maiaadshaaiaawIcacaGLPaaaaaa!4B5A! A = \Delta \cap d \Rightarrow A\left( { - 1 - 2t;t; - 1 + 3t} \right)\)

(P) nhận \(\overrightarrow n \left( {1;1;1} \right)\) là 1 VTCP

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaey % yPI41aaeWaaeaacaWGqbaacaGLOaGaayzkaaGaeyO0H49aa8Haaeaa % caWGbbGaamOqaaGaay51Gaaaaa!4107! \Delta \bot \left( P \right) \Rightarrow \overrightarrow {AB} \) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGUbaacaGLxdcaaaa!389B! \overrightarrow n \) là 2 vec tơ cùng phương.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3caaIYaGaamiDaiabgUcaRiaadshacaGGNaGaey4kaSIaaG4maiab % g2da9iabgkHiTiaadshacqGHRaWkcaaIYaGaamiDaiaacEcacqGHRa % WkcaaIXaGaeyypa0JaeyOeI0IaaG4maiaadshacqGHsislcaaIYaGa % amiDaiaacEcacqGHRaWkcaaIXaGaeyi1HS9aaiqaaqaabeqaaiaaio % dacaWG0bGaeyOeI0IaamiDaiaacEcacqGHRaWkcaaI0aGaeyypa0Ja % aGimaaqaaiaaikdacaWG0bGaey4kaSIaaGinaiaadshacaGGNaGaey % OeI0IaaGOmaiabg2da9iaaicdaaaGaay5EaaGaeyi1HS9aaiqaaqaa % beqaaiaadshacqGH9aqpcqGHsislcaaIXaaabaGaamiDaiaacEcacq % GH9aqpcaaIXaaaaiaawUhaaaqaaiabgkDiElaadgeadaqadaqaaiaa % igdacaGG7aGaeyOeI0IaaGymaiaacUdacqGHsislcaaI0aaacaGLOa % GaayzkaaGaaiilaiaadkeadaqadaqaaiaaiodacaGG7aGaaGymaiaa % cUdacqGHsislcaaIYaaacaGLOaGaayzkaaGaeyO0H49aa8Haaeaaca % WGbbGaamOqaaGaay51GaGaeyypa0ZaaeWaaeaacaaIYaGaai4oaiaa % ikdacaGG7aGaaGOmaaGaayjkaiaawMcaaiaac+cacaGGVaWaaeWaae % aacaaIXaGaai4oaiaaigdacaGG7aGaaGymaaGaayjkaiaawMcaaaaa % aa!90F3! \begin{array}{l} \Rightarrow 2t + t' + 3 = - t + 2t' + 1 = - 3t - 2t' + 1 \Leftrightarrow \left\{ \begin{array}{l} 3t - t' + 4 = 0\\ 2t + 4t' - 2 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} t = - 1\\ t' = 1 \end{array} \right.\\ \Rightarrow A\left( {1; - 1; - 4} \right),B\left( {3;1; - 2} \right) \Rightarrow \overrightarrow {AB} = \left( {2;2;2} \right)//\left( {1;1;1} \right) \end{array}\)

Vậy phương trình đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaai % OoamaalaaabaGaamiEaiabgkHiTiaaiodaaeaacaaIXaaaaiabg2da % 9maalaaabaGaamyEaiabgkHiTiaaigdaaeaacaaIXaaaaiabg2da9m % aalaaabaGaamOEaiabgUcaRiaaikdaaeaacaaIXaaaaaaa!446F! \Delta :\frac{{x - 3}}{1} = \frac{{y - 1}}{1} = \frac{{z + 2}}{1}\)

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