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Diketahui vektor − vektor u = a i + 9 j ​ + b k dan v = − b i + a j ​ + a k . Sudut antara u dan v adalah θ dengan cos θ = 11 6 ​ . Proyeksi u pada v adalah p = − 2 i + 4 j ​ + 4 k . Nilai b = ...

  1. square root of 2

  2. 2

  3. 2 square root of 2

  4. 4

  5. 4 square root of 2

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F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

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Pembahasan

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 searah comma space sehingga colon  straight p with rightwards arrow on top equals straight n. straight v with rightwards arrow on top rightwards double arrow open parentheses table row cell negative 2 end cell row 4 row 4 end table close parentheses equals straight n. open parentheses table row cell negative straight b end cell row straight a row straight a end table close parentheses  Jadi comma space minus 2 equals straight n left parenthesis negative straight b right parenthesis 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 space equals 4 over straight a rightwards double arrow straight a equals 2 straight b    cos invisible function application straight theta equals fraction numerator straight u with rightwards arrow on top space. space straight v with rightwards arrow on top over denominator vertical line straight u with rightwards arrow on top vertical line vertical line straight v with rightwards arrow on top vertical line end fraction  6 over 11 equals fraction numerator negative ab plus 9 straight a plus ab over denominator square root of straight a squared plus 9 squared plus straight b squared end root. square root of left parenthesis negative straight b right parenthesis squared plus straight a squared plus straight a squared end root end fraction  6 over 11 equals fraction numerator 9 straight a over denominator square root of 81 plus straight a squared plus straight b squared end root. square root of 2 straight a squared plus straight b squared end root end fraction  Substitusi space straight a equals 2 straight b  6 over 11 equals fraction numerator 9 left parenthesis 2 straight b right parenthesis over denominator square root of 81 plus left parenthesis 2 straight b right parenthesis squared plus straight b squared end root. square root of 2 left parenthesis 2 straight b right parenthesis squared plus straight b squared end root end fraction  6 over 11 equals fraction numerator 18 straight b over denominator square root of 81 plus 5 straight b squared end root. 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 space yang space memenuhi space adalah space straight b equals 2 square root of 2.

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Diketahui vektor a = ⎝ ⎛ ​ x 1 2 ​ ⎠ ⎞ ​ dan b = ⎝ ⎛ ​ 2 3 − 1 ​ ⎠ ⎞ ​ dan panjang proyeksi vektor a pada vektor b adalah 7 1 ​ 14 ​ . Sudut antara dan adalah α , maka cos α = ...

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