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Diketahui vektor a=⎝⎛​−2p22​​⎠⎞​ dengan p∈ bilangan real dan vektor b=⎝⎛​112​​⎠⎞​. Jika a dan b membentuk sudut 60∘, maka kosinus sudut antara vektor a dan a+b adalah ....

Pertanyaan

Diketahui vektor a with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row p row cell 2 square root of 2 end cell end table close parentheses dengan p element of bilangan real dan vektor b with rightwards arrow on top equals open parentheses table row 1 row 1 row cell square root of 2 end cell end table close parentheses. Jika a with rightwards arrow on top space dan space b with rightwards arrow on top membentuk sudut 60 degree, maka kosinus sudut antara vektor a with rightwards arrow on top dan a with rightwards arrow on top plus b with rightwards arrow on top adalah ....

  1. 12 over 4 square root of 7

  2. 5 over 2 square root of 7

  3. 5 over 4 square root of 7

  4. 5 over 14 square root of 7

  5. 2 over 7 square root of 7

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah D.

Diketahui bahwa, vektor a with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row p row cell 2 square root of 2 end cell end table close parentheses dengan p element of bilangan real dan vektor b with rightwards arrow on top equals open parentheses table row 1 row 1 row cell square root of 2 end cell end table close parentheses. vektora with rightwards arrow on top space dan space b with rightwards arrow on topmembentuk sudut 60 degree, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 60 degree end cell equals cell fraction numerator open parentheses table row cell negative 2 end cell row p row cell 2 square root of 2 end cell end table close parentheses times open parentheses table row 1 row 1 row cell square root of 2 end cell end table close parentheses over denominator square root of open parentheses negative 2 close parentheses squared plus p squared plus open parentheses 2 square root of 2 close parentheses squared end root square root of 1 squared plus 1 squared plus open parentheses square root of 2 close parentheses squared end root end fraction end cell row cell 1 half end cell equals cell fraction numerator negative 2 plus p plus 4 over denominator square root of 4 plus p squared plus 8 end root square root of 1 plus 1 plus 2 end root end fraction end cell row blank equals cell fraction numerator 2 plus p over denominator square root of 12 plus p squared end root square root of 4 end fraction end cell row blank equals cell fraction numerator 2 plus p over denominator 2 square root of 12 plus p squared end root end fraction end cell row cell square root of 12 plus p squared end root end cell equals cell 2 plus p end cell row cell 12 plus p squared end cell equals cell 4 plus 4 p plus p squared end cell row cell 4 p minus 8 end cell equals 0 row cell 4 p end cell equals 8 row p equals 2 end table

Sehingga

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

sudut antara vektor a with rightwards arrow on top dan a with rightwards arrow on top plus b with rightwards arrow on top adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator open parentheses table row cell negative 2 end cell row 2 row cell 2 square root of 2 end cell end table close parentheses times open parentheses table row cell negative 1 end cell row 3 row cell 3 square root of 2 end cell end table close parentheses over denominator square root of open parentheses negative 2 close parentheses squared plus 2 squared plus open parentheses 2 square root of 2 close parentheses squared end root square root of open parentheses negative 1 close parentheses squared plus 3 squared plus open parentheses 3 square root of 2 close parentheses squared end root end fraction end cell row blank equals cell fraction numerator 2 plus 6 plus 12 over denominator square root of 4 plus 4 plus 8 end root square root of 1 plus 9 plus 18 end root end fraction end cell row blank equals cell fraction numerator 20 over denominator 4 square root of 28 end fraction end cell row blank equals cell fraction numerator 5 over denominator square root of 28 end fraction end cell row blank equals cell fraction numerator 5 over denominator 2 square root of 7 end fraction end cell row blank equals cell 5 over 14 square root of 7 end cell end table

Oleh karena itu, jawaban yang benar adalah D.

65

5.0 (4 rating)

Ryan Muhamad Saputra

Ini yang aku cari! Makasih ❤️

najaja

Makasih ❤️

Pertanyaan serupa

Diketahui vektor u=⎝⎛​121​⎠⎞​ dan v=⎝⎛​0−10​⎠⎞​. Jika θ adalah sudut antara u dan v, nilai cos θ adalah ....

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