Roboguru

Diketahui titik , , dan . Jika  mewakili  dan  mewakili , maka proyeksi ortogonal vektor  pada  adalah ....

Pertanyaan

Diketahui titik straight A open parentheses 2 comma space 7 comma space 8 close parenthesesstraight B open parentheses negative 1 comma space 1 comma space minus 1 close parentheses, dan straight C open parentheses 0 comma space 3 comma space 2 close parentheses. Jika u with rightwards arrow on top mewakili AB with rightwards arrow on top dan v with rightwards arrow on top mewakili BC with rightwards arrow on top, maka proyeksi ortogonal vektor u with rightwards arrow on top pada v with rightwards arrow on top adalah ....

  1. negative 3 i with rightwards arrow on top minus 6 j with rightwards arrow on top minus 9 k with rightwards arrow on top

  2. i with rightwards arrow on top plus 2 j with rightwards arrow on top plus 3 k with rightwards arrow on top

  3. 1 third i with rightwards arrow on top plus 2 over 3 j with rightwards arrow on top plus k with rightwards arrow on top

  4. negative 9 i with rightwards arrow on top minus 18 j with rightwards arrow on top minus 27 k with rightwards arrow on top

  5. 3 i with rightwards arrow on top plus 6 j with rightwards arrow on top plus 9 k with rightwards arrow on top

Pembahasan Soal:

Gunakan konsep proyeksi orthogonal vektor u with rightwards arrow on top pada v with rightwards arrow on top.

Proyeksi space ortogonal equals open parentheses fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar squared end fraction close parentheses v with rightwards arrow on top

Diketahui:
Titik straight A open parentheses 2 comma space 7 comma space 8 close parenthesesstraight B open parentheses negative 1 comma space 1 comma space minus 1 close parentheses, dan straight C open parentheses 0 comma space 3 comma space 2 close parentheses. Jika u with rightwards arrow on top mewakili AB with rightwards arrow on top dan v with rightwards arrow on top mewakili BC with rightwards arrow on top.
Akan ditentukan proyeksi ortogonal vektor u with rightwards arrow on top pada v with rightwards arrow on top.

Tentukan vektor u with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell u with rightwards arrow on top end cell equals cell AB with rightwards arrow on top end cell row blank equals cell straight B minus straight A end cell row blank equals cell open parentheses negative 1 minus 2 comma space 1 minus 7 comma space minus 1 minus 8 close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses negative 3 comma space minus 6 comma space minus 9 close parentheses end cell end table

Tentukan vektor v with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell v with rightwards arrow on top end cell equals cell BC with rightwards arrow on top end cell row blank equals cell straight C minus straight B end cell row blank equals cell open parentheses 0 minus open parentheses negative 1 close parentheses comma space 3 minus 1 comma space 2 minus open parentheses negative 1 close parentheses close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses 1 comma space 2 comma space 3 close parentheses end cell end table

Sehingga proyeksi ortogonal vektor u with rightwards arrow on top pada v with rightwards arrow on top dapat diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row Proyeksi equals cell open parentheses fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar squared end fraction close parentheses times v with rightwards arrow on top end cell row blank equals cell open parentheses fraction numerator open parentheses negative 3 close parentheses times 1 plus open parentheses negative 6 close parentheses times 2 plus open parentheses negative 9 close parentheses times 3 over denominator open parentheses square root of 1 squared plus 2 squared plus 3 squared end root close parentheses squared end fraction close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses fraction numerator negative 3 minus 12 minus 27 over denominator open parentheses square root of 1 plus 4 plus 9 end root close parentheses squared end fraction close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 42 over open parentheses square root of 14 close parentheses squared close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 42 over 14 close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell 3 cross times open parentheses 1 comma space 2 comma space 3 close parentheses end cell row Proyeksi equals cell open parentheses 3 comma space 6 comma space 9 close parentheses end cell end table

Sehingga diperoleh proyeksi ortogonalnya adalah open parentheses 3 comma space 6 comma space 9 close parentheses atau 3 i with rightwards arrow on top plus 6 j with rightwards arrow on top plus 9 k with rightwards arrow on top.

Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Fathoni

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika vektor , vektor , dan vektor proyeksi  pada  adalah , maka nilai  adalah ....

Pembahasan Soal:

Dengan menerapkan rumus proyeksi ortogonal vektor, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell r with rightwards arrow on top end cell equals cell open parentheses fraction numerator p with rightwards arrow on top times q with rightwards arrow on top over denominator open vertical bar q with rightwards arrow on top close vertical bar squared end fraction close parentheses times q with rightwards arrow on top end cell row cell open parentheses table row cell 8 over 9 end cell row cell 4 over 9 end cell row cell fraction numerator negative 8 over denominator 9 end fraction end cell end table close parentheses end cell equals cell open parentheses fraction numerator 2 times 4 plus x times 2 plus open parentheses negative 3 close parentheses open parentheses negative 4 close parentheses over denominator 4 squared plus 2 squared plus open parentheses negative 4 close parentheses squared end fraction close parentheses open parentheses table row 4 row 2 row cell negative 4 end cell end table close parentheses end cell row cell open parentheses table row cell 8 over 9 end cell row cell 4 over 9 end cell row cell fraction numerator negative 8 over denominator 9 end fraction end cell end table close parentheses end cell equals cell open parentheses fraction numerator 8 plus 2 x plus 12 over denominator 16 plus 4 plus 16 end fraction close parentheses open parentheses table row 4 row 2 row cell negative 4 end cell end table close parentheses end cell row cell 2 over 9 up diagonal strike open parentheses table row 4 row 2 row cell negative 4 end cell end table close parentheses end strike end cell equals cell fraction numerator 20 plus 2 x over denominator 36 end fraction up diagonal strike open parentheses table row 4 row 2 row cell negative 4 end cell end table close parentheses end strike end cell row cell 2 over 9 end cell equals cell fraction numerator 20 plus 2 x over denominator 36 end fraction end cell row cell 180 plus 18 x end cell equals 72 row cell 18 x end cell equals cell 72 minus 180 end cell row blank equals cell negative 108 end cell row x equals cell negative 6 end cell row cell x squared end cell equals cell open parentheses negative 6 close parentheses squared end cell row blank equals 36 end table 

Jadi, nilai x squared adalah 36.

Dengan demikian, jawaban yang tepat adalah E.

Roboguru

Berapa besar sudut orthogonal?

Pembahasan Soal:

Besar sudut orthogonal adalah begin mathsize 14px style 90 degree end style.

Roboguru

Berapa jumlah besaran? Sebutkan!

Pembahasan Soal:

Besaran dibagi menjadi 2, diantaranya:

  1. Besaran Skalar, yaitu besaran yang hanya mempunyai nilai.
  2. Besaran Vektor, yaitu besaran yang mempunyai nilai dan arah.
Roboguru

Diketahui  dan . Maka vektor proyeksi ortogonal dari vektor  pada vektor  adalah ....

Pembahasan Soal:

Misalkan vektor proyeksi ortogonal dari vektor begin mathsize 14px style a with rightwards arrow on top end style terhadap vektor begin mathsize 14px style b with rightwards arrow on top end style adalah begin mathsize 14px style p with rightwards arrow on top end style. Oleh karena itu, didapatkan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top end cell equals cell open parentheses fraction numerator a with rightwards arrow on top ⋅ b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar squared end fraction close parentheses b with rightwards arrow on top end cell row blank equals cell open parentheses fraction numerator open parentheses table row cell negative sign 4 end cell row cell negative sign 3 end cell row 5 end table close parentheses ⋅ open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses over denominator open parentheses square root of open parentheses negative sign 3 close parentheses squared plus 3 squared plus open parentheses negative sign 1 close parentheses squared end root close parentheses squared end fraction close parentheses open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell open parentheses fraction numerator open parentheses negative sign 4 close parentheses ⋅ open parentheses negative sign 3 close parentheses plus open parentheses negative sign 3 close parentheses ⋅ 3 plus 5 ⋅ open parentheses negative sign 1 close parentheses over denominator open parentheses square root of 9 plus 9 plus 1 end root close parentheses squared end fraction close parentheses open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell open parentheses fraction numerator 12 minus sign 9 minus sign 5 over denominator open parentheses square root of 19 close parentheses squared end fraction close parentheses open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell fraction numerator negative sign 2 over denominator 19 end fraction open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 6 over 19 end cell row cell negative sign 6 over 19 end cell row cell 2 over 19 end cell end table close parentheses end cell end table end style   
 

Jadi, jawaban yang tepat adalah D.

Roboguru

Diketahui vektor-vektor dan . Sudut antara vektor  dan vektor  adalah θ dengan . Proyeksi  pada  adalah . Nilai b = ....

Pembahasan Soal:

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

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

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