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Jika ∣∣​a∣∣​=2, ∣∣​b∣∣​=3, dan sudut (a, b)=120∘, maka ∣∣​3a+2b∣∣​=....

Pertanyaan

Jika open vertical bar a with rightwards arrow on top close vertical bar equals 2open vertical bar b with rightwards arrow on top close vertical bar equals 3, dan sudut open parentheses a with rightwards arrow on top comma space b with rightwards arrow on top close parentheses equals 120 degree, maka open vertical bar 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top close vertical bar equals.... 

  1. 5

  2. 6

  3. 10

  4. 12

  5. 13

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah B.

Ingat!

  • Rumus untuk menentukan nilai cosinus antara vektor a with rightwards arrow on top dan vektor b with rightwards arrow on top adalah sebagai berikut:

cos space theta equals fraction numerator a with rightwards arrow on top space. space b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end fraction 

  • Rumus untuk menentukan panjang vektor r with rightwards arrow on top equals open parentheses table row x row y end table close parentheses adalah sebagai  berikut:

open vertical bar r with rightwards arrow on top close vertical bar equals square root of x squared plus y squared end root 

  • Rumus untuk menentukan hasil kali a with rightwards arrow on top space. space b with rightwards arrow on top jika diketahui vektor a with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses dan vektor b with rightwards arrow on top equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell end table close parentheses adalah sebagai berikut:

a with rightwards arrow on top times b with rightwards arrow on top equals x subscript 1 x subscript 2 plus y subscript 1 y subscript 2 

  • Rumus untuk perkalian skalar m dengan vektor a with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses adalah sebagai berikut:

m a with rightwards arrow on top equals m open parentheses table row cell x subscript 1 end cell row cell x subscript 2 end cell end table close parentheses equals open parentheses table row cell m x subscript 1 end cell row cell m x subscript 2 end cell end table close parentheses

Diketahui:open vertical bar a with rightwards arrow on top close vertical bar equals 2open vertical bar b with rightwards arrow on top close vertical bar equals 3, dan sudut open parentheses a with rightwards arrow on top comma space b with rightwards arrow on top close parentheses equals 120 degree.

Ditanya: open vertical bar 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top close vertical bar.

Jawab:

Dengan menggunakan rumus untuk menentukan nilai cosinus antara vektor a with rightwards arrow on top dan vektor b with rightwards arrow on top maka nilai dari a with rightwards arrow on top space. space b with rightwards arrow on top adalah sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator a with rightwards arrow on top space. space b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell row cell cos open parentheses 120 degree close parentheses end cell equals cell fraction numerator a with rightwards arrow on top space. space b with rightwards arrow on top over denominator 2 cross times 3 end fraction end cell row cell negative 1 half end cell equals cell fraction numerator a with rightwards arrow on top space. space b with rightwards arrow on top over denominator 6 end fraction end cell row blank left right double arrow cell a with rightwards arrow on top space. space b with rightwards arrow on top equals negative 1 half cross times 6 end cell row blank equals cell negative 3 end cell end table 

Misalkan vektor a with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses dan vektor b with rightwards arrow on top equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell end table close parentheses, dengan menggunakan rumus-rumus di atas, maka diperoleh hubungan sebagai berikut:

  • Untuk vektor a with rightwards arrow on top 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of x subscript 1 squared plus y subscript 1 squared end root end cell row 2 equals cell square root of x subscript 1 squared plus y subscript 1 squared end root end cell row cell 2 squared end cell equals cell x subscript 1 squared plus y subscript 1 squared end cell row 4 equals cell x subscript 1 squared plus y subscript 1 squared end cell row blank left right double arrow cell x subscript 1 squared plus y subscript 1 squared equals 4 end cell end table

  • Untuk vektor b with rightwards arrow on top

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of x subscript 2 squared plus y subscript 2 squared end root end cell row 3 equals cell square root of x subscript 2 squared plus y subscript 2 squared end root end cell row cell 3 squared end cell equals cell x subscript 2 squared plus y subscript 2 squared end cell row 9 equals cell x subscript 2 squared plus y subscript 2 squared end cell row blank left right double arrow cell x subscript 2 squared plus y subscript 2 squared equals 9 end cell end table 

  • Untuk vektor a with rightwards arrow on top space. space b with rightwards arrow on top 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top space. space b with rightwards arrow on top end cell equals cell x subscript 1 space x subscript 2 plus y subscript 1 space y subscript 2 end cell row cell negative 3 end cell equals cell x subscript 1 space x subscript 2 plus y subscript 1 space y subscript 2 end cell row blank left right double arrow cell x subscript 1 space x subscript 2 plus y subscript 1 space y subscript 2 equals negative 3 end cell end table 

  • Untuk vektor 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top 

 table attributes columnalign right center left columnspacing 0px end attributes row cell 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top end cell equals cell 3 open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses plus 2 open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 3 x subscript 1 end cell row cell 3 y subscript 1 end cell end table close parentheses plus open parentheses table row cell 2 x subscript 2 end cell row cell 2 y subscript 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 3 x subscript 1 plus 2 x subscript 2 end cell row cell 3 y subscript 1 plus 2 y subscript 2 end cell end table close parentheses end cell end table 

Dengan demikian, panjang vektor 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top adalah sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 3 x subscript 1 plus 2 x subscript 2 close parentheses squared plus open parentheses 3 y subscript 1 plus 2 y subscript 2 close parentheses squared end root end cell row blank equals cell square root of open parentheses 9 x subscript 1 squared plus 12 x subscript 1 x subscript 2 plus 4 x subscript 2 squared close parentheses plus open parentheses 9 y subscript 1 squared plus 12 y subscript 1 y subscript 2 plus 4 y subscript 2 squared close parentheses end root end cell row blank equals cell square root of 9 x subscript 1 squared plus 9 y subscript 1 squared plus 12 x subscript 1 x subscript 2 plus 12 y subscript 1 y subscript 2 plus 4 x subscript 2 squared plus 4 y subscript 2 squared end root end cell row blank equals cell square root of 9 open parentheses x subscript 1 squared plus y subscript 1 squared close parentheses plus 12 open parentheses x subscript 1 x subscript 2 plus y subscript 1 y subscript 2 close parentheses plus 4 open parentheses x subscript 2 squared plus y subscript 2 squared close parentheses end root end cell row blank equals cell square root of 9 open parentheses 4 close parentheses plus 12 open parentheses negative 3 close parentheses plus 4 open parentheses 9 close parentheses end root end cell row blank equals cell square root of 36 minus 36 plus 36 end root end cell row blank equals cell square root of 36 end cell row blank equals 6 end table 

Oleh karena itu, jawaban yang benar adalah B.

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