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Diketahui vektor-vektor begin mathsize 14px style u with rightwards arrow on top equals 9 i with rightwards arrow on top plus a j with rightwards arrow on top plus b k with rightwards arrow on top end style dan begin mathsize 14px style v with rightwards arrow on top equals a i with rightwards arrow on top minus b j with rightwards arrow on top plus a k with rightwards arrow on top end style. Sudut antara vektor begin mathsize 14px style u with rightwards arrow on top end style dan vektor begin mathsize 14px style v with rightwards arrow on top end style adalah θ dengan begin mathsize 14px style cos space theta equals 6 over 11 end style. Proyeksi begin mathsize 14px style u with rightwards arrow on top end style pada begin mathsize 14px style v with rightwards arrow on top end style adalah begin mathsize 14px style p with rightwards arrow on top equals 4 i with rightwards arrow on top minus 2 j with rightwards arrow on top plus 4 k with rightwards arrow on top end style. Nilai b = ....

  1. begin mathsize 14px style square root of 2 end style

  2. 2

  3. begin mathsize 14px style 2 square root of 2 end style

  4. 4

  5. begin mathsize 14px style 4 square root of 2 end style

S. Intan

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Pembahasan

begin mathsize 14px style straight p with rightwards arrow on top space proyeksi space straight u with rightwards arrow on top space pada space straight v with rightwards arrow on top space maka space straight p with rightwards arrow on top space dan space straight v with rightwards arrow on top space kolinear comma space sehingga colon  straight p with rightwards arrow on top equals straight n. straight space straight v with rightwards arrow on top rightwards double arrow open parentheses table row 4 row cell negative 2 end cell row 4 end table close parentheses equals straight n straight space. straight space open parentheses table row straight a row cell negative straight b end cell row straight a end table close parentheses    Jadi comma space minus 2 equals straight n open parentheses negative straight b close parentheses rightwards double arrow straight n equals 2 over straight b space dan space 4 equals na rightwards double arrow straight n equals 4 over straight a  Sehingga colon  2 over straight b equals 4 over straight a rightwards double arrow straight a equals 2 straight b    cos invisible function application space straight theta equals fraction numerator straight u with rightwards arrow on top straight space. straight space straight v with rightwards arrow on top over denominator open vertical bar straight u with rightwards arrow on top close vertical bar open vertical bar straight v with rightwards arrow on top close vertical bar end fraction  6 over 11 equals fraction numerator 9 straight a plus ab minus ab over denominator square root of 9 squared plus straight a squared plus straight b squared end root. straight space square root of straight a squared plus left parenthesis negative straight b right parenthesis squared plus straight a squared end root end fraction      Substitusi space straight a equals 2 straight b  6 over 11 equals fraction numerator 18 straight b over denominator square root of 81 plus 5 straight b squared end root. straight space square root of 9 straight b squared end root end fraction  6 over 11 equals fraction numerator 6 over denominator square root of 81 plus 5 straight b squared end root end fraction  square root of 81 plus 5 straight b squared end root equals 11  81 plus 5 straight b squared equals 121  5 straight b squared equals 121 minus 81  5 straight b squared equals 40  straight b squared equals 8  straight b equals plus-or-minus 2 square root of 2    Jadi comma space nilai space straight b equals 2 square root of 2 end style

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Diketahui vektor-vektor u⇀=bi⇀+aj⇀​+9k⇀ dan v⇀=ai⇀−bj⇀​+ak⇀. Sudut antara vektor u⇀ dan v⇀ adalah θ dengan cosθ=116​. Proyeksi  pada  adalah p⇀​=4a⇀−2j⇀​+4k⇀. Nilai dari b = ...

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