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Pertanyaan

Diberikan vektor a = ⎝ ⎛ ​ − 2 p 2 2 ​ ​ ⎠ ⎞ ​ dengan p ∈ Real dan vektor b = ⎝ ⎛ ​ 1 1 2 ​ ​ ⎠ ⎞ ​ . Jika a dan b membentuk sudut 6 0 ∘ , maka kosinus sudut antara vektor dan a + b adalah ....

Diberikan vektor  dengan  dan vektor . Jika  dan  membentuk sudut , maka kosinus sudut antara vektor a with rightwards arrow on top dan  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 

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Y. Fathoni

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

jawaban yang tepat adalah D.

Pembahasan

Operasi perkalian dot vektor dan vektor . Diketahui vektor dengan dan vektor . dan membentuk sudut . Akan ditentukan kosinus sudut antara vektor dan . Terlebih dahulu tentukan nilai pada vektor . Perhatikan perhitungan berikut. Karena dan membentuk sudut , maka diperoleh Diperoleh nilai , maka , kemudian tentukan Kemudian terlebih dahulutentukan , dan Sehinggakosinus sudut antara vektor dan dapat dihitung sebagai berikut. Diperoleh nilaikosinus sudut antara vektor dan adalah . Jadi, jawaban yang tepat adalah D.

Operasi perkalian dot vektor a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell row cell a subscript 3 end cell end table close parentheses dan vektor b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell row cell b subscript 3 end cell end table close parentheses.

table attributes columnalign right center left columnspacing 2px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos space theta end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3 end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses a subscript 1 close parentheses squared plus open parentheses a subscript 2 close parentheses squared plus open parentheses a subscript 3 close parentheses squared end root end cell end table

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 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. a with rightwards arrow on top dan b with rightwards arrow on top membentuk sudut 60 degree.

Akan ditentukan kosinus sudut antara vektor a with rightwards arrow on top dan a with rightwards arrow on top plus b with rightwards arrow on top.

Terlebih dahulu tentukan nilai p pada vektor a with rightwards arrow on top.

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3 end cell row blank equals cell negative 2 times 1 plus p times 1 plus 2 square root of 2 times square root of 2 end cell row blank equals cell negative 2 plus p plus 4 end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell p plus 2 end cell end table 

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 2 close parentheses squared plus open parentheses p close parentheses squared plus open parentheses 2 square root of 2 close parentheses squared end root end cell row blank equals cell square root of 4 plus p squared plus 8 end root end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of p squared plus 12 end root end cell row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 1 close parentheses squared plus open parentheses 1 close parentheses squared plus open parentheses square root of 2 close parentheses squared end root end cell row blank equals cell square root of 1 plus 1 plus 2 end root end cell row blank equals cell square root of 4 end cell row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals 2 end table

Karena a with rightwards arrow on top dan b with rightwards arrow on top membentuk sudut 60 degree, maka diperoleh

table attributes columnalign right center left columnspacing 2px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos space theta end cell row cell p plus 2 end cell equals cell square root of p squared plus 12 end root times 2 times cos space 60 degree end cell row cell p plus 2 end cell equals cell square root of p squared plus 12 end root times 2 times 1 half end cell row cell p plus 2 end cell equals cell square root of p squared plus 12 end root end cell row cell open parentheses p plus 2 close parentheses squared end cell equals cell open parentheses square root of p squared plus 12 end root close parentheses squared end cell row cell p squared plus 4 p plus 4 end cell equals cell p squared plus 12 end cell row cell p squared plus 4 p plus 4 minus p squared minus 12 end cell equals 0 row cell 4 p minus 8 end cell equals 0 row cell 4 p end cell equals 8 row p equals cell 8 over 4 end cell row p equals 2 end table

Diperoleh nilai p equals 2, maka a with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 2 row cell 2 square root of 2 end cell end table close parentheses, kemudian tentukan a with rightwards arrow on top plus b with rightwards arrow on top

a with rightwards arrow on top plus b with rightwards arrow on top equals 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 equals open parentheses table row cell negative 1 end cell row 3 row cell 3 square root of 2 end cell end table close parentheses

Kemudian terlebih dahulu tentukan a with rightwards arrow on top times open parentheses a with rightwards arrow on top plus b with rightwards arrow on top close parenthesesopen vertical bar a with rightwards arrow on top close vertical bar dan open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar

table attributes columnalign right center left columnspacing 2px end attributes row cell a with rightwards arrow on top times open parentheses a with rightwards arrow on top plus b with rightwards arrow on top close parentheses end cell equals cell negative 2 times negative 1 plus 2 times 3 plus 2 square root of 2 times 3 square root of 2 end cell row blank equals cell 2 plus 6 plus 12 end cell row cell a with rightwards arrow on top times open parentheses a with rightwards arrow on top plus b with rightwards arrow on top close parentheses end cell equals 20 row blank blank blank row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 2 close parentheses squared plus open parentheses 2 close parentheses squared plus open parentheses 2 square root of 2 close parentheses squared end root end cell row blank equals cell square root of 4 plus 4 plus 8 end root end cell row blank equals cell square root of 16 end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals 4 row blank blank blank row cell open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 1 close parentheses squared plus open parentheses 3 close parentheses squared plus open parentheses 3 square root of 2 close parentheses squared end root end cell row blank equals cell square root of 1 plus 9 plus 18 end root end cell row blank equals cell square root of 28 end cell row cell open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar end cell equals cell 2 square root of 7 end cell end table

Sehingga kosinus sudut antara vektor a with rightwards arrow on top dan a with rightwards arrow on top plus b with rightwards arrow on top dapat dihitung sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell a with rightwards arrow on top times open parentheses a with rightwards arrow on top plus b with rightwards arrow on top close parentheses end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top times b with rightwards arrow on top close vertical bar cos space theta end cell row 20 equals cell 4 times 2 square root of 7 times cos space theta end cell row 20 equals cell 8 square root of 7 times cos space theta end cell row cell cos space theta end cell equals cell fraction numerator 20 over denominator 8 square root of 7 end fraction end cell row blank equals cell fraction numerator 5 over denominator 2 square root of 7 end fraction cross times fraction numerator square root of 7 over denominator square root of 7 end fraction end cell row cell cos space theta end cell equals cell 5 over 14 square root of 7 end cell end table

Diperoleh nilai 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 5 over 14 square root of 7.

Jadi, jawaban yang tepat adalah D.

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