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Diketahui vektor a = ( 0 , − 1 , 3 ) dan b = ( − 2 , 0 , 1 ) . Tentukan: b. ∣ ∣ ​ b − a ∣ ∣ ​

Diketahui vektor . Tentukan:

b.     

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N. Puspita

Master Teacher

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didapatkan: didapatkan: jadi, hasil dari

begin mathsize 14px style b with rightwards arrow on top minus a with rightwards arrow on top end style didapatkan:

begin mathsize 14px style b with rightwards arrow on top minus a with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 0 row 1 end table close parentheses minus open parentheses table row 0 row cell negative 1 end cell row 3 end table close parentheses b with rightwards arrow on top minus a with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 1 row cell negative 2 end cell end table close parentheses end style 

begin mathsize 14px style open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar end style didapatkan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 2 close parentheses squared plus 1 squared plus open parentheses negative 2 close parentheses squared end root end cell row cell open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar end cell equals cell square root of 4 plus 1 plus 4 end root end cell row cell open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar end cell equals cell square root of 9 end cell row cell open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar end cell equals 3 end table end style 

jadi, hasil dari open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar space adalah space 3 

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