{"title":"向量","authors":"José María Martínez Mediano, Tema","doi":"10.2307/j.ctv287sbks.6","DOIUrl":null,"url":null,"abstract":"1. Para a = (1, −2, 3) y b = (3, −1, 4), halla: a) b a + b) b a − 2 c) b a 3 + − d) b a c + = Solución: a) b a + = (1, −2, 3) + (3, −1, 4) = (4, –3, 7). b) b a − 2 = 2 · (1, −2, 3) – (3, −1, 4) = (2 – 3, –4 + 1, 6 – 4) = (–1, –3, 2). c) b a 3 + − = – (1, −2, 3) + 3 · (3, −1, 4) = (–1 + 9, 2 – 3, –3 + 12) = (8, –1, 9). d) b a c + = = ( ) ( ) ( ) + − − + = − + − 4 3 , 2 , 3 4 , 1 , 3 3 , 2 , 1 .","PeriodicalId":409897,"journal":{"name":"Introducción a la proporción y a los vectores","volume":"78 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"2020-07-15","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"1","resultStr":"{\"title\":\"Vectores\",\"authors\":\"José María Martínez Mediano, Tema\",\"doi\":\"10.2307/j.ctv287sbks.6\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"1. Para a = (1, −2, 3) y b = (3, −1, 4), halla: a) b a + b) b a − 2 c) b a 3 + − d) b a c + = Solución: a) b a + = (1, −2, 3) + (3, −1, 4) = (4, –3, 7). b) b a − 2 = 2 · (1, −2, 3) – (3, −1, 4) = (2 – 3, –4 + 1, 6 – 4) = (–1, –3, 2). c) b a 3 + − = – (1, −2, 3) + 3 · (3, −1, 4) = (–1 + 9, 2 – 3, –3 + 12) = (8, –1, 9). d) b a c + = = ( ) ( ) ( ) + − − + = − + − 4 3 , 2 , 3 4 , 1 , 3 3 , 2 , 1 .\",\"PeriodicalId\":409897,\"journal\":{\"name\":\"Introducción a la proporción y a los vectores\",\"volume\":\"78 1\",\"pages\":\"0\"},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2020-07-15\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"1\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Introducción a la proporción y a los vectores\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.2307/j.ctv287sbks.6\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Introducción a la proporción y a los vectores","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.2307/j.ctv287sbks.6","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 1
Vectores
1. Para a = (1, −2, 3) y b = (3, −1, 4), halla: a) b a + b) b a − 2 c) b a 3 + − d) b a c + = Solución: a) b a + = (1, −2, 3) + (3, −1, 4) = (4, –3, 7). b) b a − 2 = 2 · (1, −2, 3) – (3, −1, 4) = (2 – 3, –4 + 1, 6 – 4) = (–1, –3, 2). c) b a 3 + − = – (1, −2, 3) + 3 · (3, −1, 4) = (–1 + 9, 2 – 3, –3 + 12) = (8, –1, 9). d) b a c + = = ( ) ( ) ( ) + − − + = − + − 4 3 , 2 , 3 4 , 1 , 3 3 , 2 , 1 .