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Diketahui vektor-vektor . Cosinus sudut antara vektor  dan (2a−b) adalah ….

Pertanyaan

Diketahui vektor-vektor begin mathsize 14px style straight a with rightwards arrow on top equals open parentheses table row 3 row cell negative 1 end cell row 2 end table close parentheses space dan space straight b with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row 4 row 2 end table close parentheses end style. Cosinus sudut antara vektor begin mathsize 14px style open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end style dan begin mathsize 14px style left parenthesis 2 straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis end style adalah ….

  1. begin mathsize 14px style 1 fifth end style 

  2. begin mathsize 14px style negative 2 over 5 end style 

  3. begin mathsize 14px style 1 over 11 end style 

  4. begin mathsize 14px style 2 over 11 end style 

  5. begin mathsize 14px style negative 2 over 11 end style 

S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.space space  

Pembahasan

begin mathsize 14px style straight a with rightwards arrow on top plus straight b with rightwards arrow on top equals open parentheses table row 3 row cell negative 1 end cell row 2 end table close parentheses plus open parentheses table row cell negative 3 end cell row 4 row 2 end table close parentheses equals open parentheses table row 0 row 3 row 4 end table close parentheses 2 straight a with rightwards arrow on top minus straight b with rightwards arrow on top equals open parentheses table row 6 row cell negative 2 end cell row 4 end table close parentheses minus open parentheses table row cell negative 3 end cell row 4 row 2 end table close parentheses equals open parentheses table row 9 row cell negative 6 end cell row 2 end table close parentheses end style 

Untuk mengetahui besar sudutnya gunakan rumus

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application straight alpha end cell equals cell fraction numerator open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses times open parentheses 2 straight a with rightwards arrow on top minus straight c with rightwards arrow on top close parentheses over denominator open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar open vertical bar 2 straight a with rightwards arrow on top minus straight c with rightwards arrow on top close vertical bar end fraction end cell row cell cos invisible function application straight alpha end cell equals cell fraction numerator open parentheses table row 0 row 3 row 4 end table close parentheses times open parentheses table row 9 row cell negative 6 end cell row 2 end table close parentheses over denominator square root of 0 squared plus 3 squared plus 4 squared end root times square root of 9 squared plus open parentheses negative 6 close parentheses squared plus 2 squared end root end fraction end cell row cell cos invisible function application straight alpha end cell equals cell fraction numerator 0 minus 18 plus 8 over denominator square root of 0 plus 9 plus 16 end root times square root of 81 plus 36 plus 4 end root end fraction end cell row cell cos invisible function application straight alpha end cell equals cell fraction numerator negative 10 over denominator square root of 25 times square root of 121 end fraction end cell row cell cos invisible function application straight alpha end cell equals cell negative fraction numerator 10 over denominator 5 times 11 end fraction end cell row cell cos invisible function application straight alpha end cell equals cell negative 2 over 11 end cell end table end style 

Jadi, jawaban yang tepat adalah E.space space  

102

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