Trong không gian Oxyz, cho hình thoi ABCD với A(-1;2;1) ; B (2;3;2). Tâm I của hình thoi thuộc đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dadaWcaaqaaiaadIhacqGHRaWkcaaIXaaabaGaeyOeI0IaaGymaaaa % cqGH9aqpdaWcaaqaaiaadMhaaeaacqGHsislcaaIXaaaaiabg2da9m % aalaaabaGaamOEaiabgkHiTiaaikdaaeaacaaIXaaaaaaa!4421! d:\frac{{x + 1}}{{ - 1}} = \frac{y}{{ - 1}} = \frac{{z - 2}}{1}\). Tọa độ đỉnh D là
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Lời giải:
Báo saiGọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysamaabm % aabaGaeyOeI0IaaGymaiabgkHiTiaadshacaGG7aGaeyOeI0IaamiD % aiaacUdacaaIYaGaey4kaSIaamiDaaGaayjkaiaawMcaaiabgIGiol % aadsgacaGGUaWaa8HaaeaacaWGjbGaamyqaaGaay51GaGaeyypa0Za % aeWaaeaacaWG0bGaai4oaiaadshacqGHRaWkcaaIYaGaai4oaiabgk % HiTiaadshacqGHsislcaaIXaaacaGLOaGaayzkaaGaaiilamaaFiaa % baGaamysaiaadkeaaiaawEniaiabg2da9maabmaabaGaamiDaiabgU % caRiaaiodacaGG7aGaamiDaiabgUcaRiaaiodacaGG7aGaeyOeI0Ia % amiDaaGaayjkaiaawMcaaaaa!6281! I\left( { - 1 - t; - t;2 + t} \right) \in d.\overrightarrow {IA} = \left( {t;t + 2; - t - 1} \right),\overrightarrow {IB} = \left( {t + 3;t + 3; - t} \right)\)
Do ABCD là hình thoi nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaamyqaaGaay51GaGaaiOlamaaFiaabaGaamysaiaadkeaaiaa % wEniaiabg2da9iaaicdacqGHuhY2caaIZaGaamiDamaaCaaaleqaba % GaaGOmaaaakiabgUcaRiaaiMdacaWG0bGaey4kaSIaaGOnaiabg2da % 9iaaicdacqGHuhY2caWG0bGaeyypa0JaeyOeI0IaaGOmaiaacUdaca % WG0bGaeyypa0JaeyOeI0IaaGymaaaa!5465! \overrightarrow {IA} .\overrightarrow {IB} = 0 \Leftrightarrow 3{t^2} + 9t + 6 = 0 \Leftrightarrow t = - 2;t = - 1\).
Do C đối xứng A qua I và D đối xứng B qua I nên:
+) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iabgkHiTiaaigdacqGHshI3caWGjbWaaeWaaeaacaaIWaGaai4o % aiaaigdacaGG7aGaaGymaaGaayjkaiaawMcaaiabgkDiElaadoeada % qadaqaaiaaigdacaGG7aGaaGimaiaacUdacaaIXaaacaGLOaGaayzk % aaGaaiilaiaadseadaqadaqaaiabgkHiTiaaikdacaGG7aGaeyOeI0 % IaaGymaiaacUdacaaIWaaacaGLOaGaayzkaaaaaa!52E3! t = - 1 \Rightarrow I\left( {0;1;1} \right) \Rightarrow C\left( {1;0;1} \right),D\left( { - 2; - 1;0} \right)\)
+) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iabgkHiTiaaikdacqGHshI3caWGdbWaaeWaaeaacaaIZaGaai4o % aiaaikdacaGG7aGaeyOeI0IaaGymaaGaayjkaiaawMcaaiaacYcaca % WGebWaaeWaaeaacaaIWaGaai4oaiaaigdacaGG7aGaeyOeI0IaaGOm % aaGaayjkaiaawMcaaaaa!4A86! t = - 2 \Rightarrow C\left( {3;2; - 1} \right),D\left( {0;1; - 2} \right)\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 3