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Pertanyaan

Suatu segitiga PQR memiliki koordinat titik P ( 4 , 7 , 0 ) , Q ( 1 , 9 , 0 ) , dan R ( 6 , 10 , − 6 ) . Sudut QPR adalah yang dibentuk oleh vektor PQ ​ dan PR , maka besar sudut QPR adalah ...

Suatu segitiga PQR memiliki koordinat titik , dan . Sudut QPR adalah yang dibentuk oleh vektor  dan , maka besar sudut QPR adalah ...

  1. 1 third straight pi 

  2. 1 fourth straight pi 

  3. 2 over 3 straight pi 

  4. 5 over 6 straight pi 

  5. 1 half straight pi 

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L. Marlina

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

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Pembahasan

Sudut antara dan dapat kita cari dengan, Jadi, jawaban yang tepat adalah E.

stack P Q with rightwards arrow on top equals q with rightwards arrow on top minus p with rightwards arrow on top space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space stack P R with rightwards arrow on top equals r with rightwards arrow on top minus p with rightwards arrow on top space space space space space equals open parentheses table row 1 row 9 row 0 end table close parentheses minus open parentheses table row 4 row 7 row 0 end table close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space equals open parentheses table row 6 row 10 row cell negative 6 end cell end table close parentheses minus open parentheses table row 4 row 7 row 0 end table close parentheses space space space space space equals open parentheses table row cell negative 3 end cell row 2 row 0 end table close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open parentheses table row 2 row 3 row cell negative 6 end cell end table close parentheses 

Sudut antara stack P Q with rightwards arrow on top dan stack P R with rightwards arrow on top dapat kita cari dengan,

cos alpha equals fraction numerator stack P Q with rightwards arrow on top. stack P R with rightwards arrow on top over denominator open vertical bar stack P Q with rightwards arrow on top close vertical bar. open vertical bar stack P R with rightwards arrow on top close vertical bar end fraction space space space space space space space space equals fraction numerator open parentheses table row cell negative 3 end cell row 2 row 0 end table close parentheses. open parentheses table row 2 row 3 row cell negative 6 end cell end table close parentheses over denominator square root of open parentheses negative 3 close parentheses squared plus 2 squared plus 0 squared end root. square root of 2 squared plus 3 squared plus open parentheses negative 6 close parentheses squared end root end fraction space space space space space space space space equals fraction numerator open parentheses negative 3 close parentheses.2 plus 2.3 plus 0. open parentheses negative 6 close parentheses over denominator square root of 9 plus 4 plus 0 end root. square root of 4 plus 9 plus 36 end root end fraction space space space space space space space space equals fraction numerator negative 6 plus 6 plus 0 over denominator square root of 13. square root of 49 end fraction space space space space space space space space equals fraction numerator 0 over denominator 7 square root of 13 end fraction space space space space space space space space equals 0 

cos alpha equals 0 space space rightwards arrow space space alpha equals 90 degree space space space space space space space space space space space space space space space space space space space space space space space space space space equals 90 degree cross times fraction numerator straight pi over denominator 180 degree end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half straight pi 

Jadi, jawaban yang tepat adalah E.

 

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Diketahui vektor u = ⎝ ⎛ ​ 2 − 1 1 ​ ⎠ ⎞ ​ dan v → = ⎝ ⎛ ​ 7 0 1 ​ ⎠ ⎞ ​ . Tentukan sudut antara u dan v .

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