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Sudut antara vektor a dan b sama dengan =150∘. Jika ∣∣​b∣∣​=6 dan a.b=−9 maka a.(b−a)=...

Pertanyaan

Sudut antara vektor a with bar on top dan b with bar on top sama dengan equals 150 degree. Jika open vertical bar b with bar on top close vertical bar equals 6 dan a with bar on top. b with bar on top equals negative 9 maka a with bar on top. left parenthesis b with bar on top minus a with bar on top right parenthesis equals...  

  1. negative 15 

  2. negative 12 

  3. negative 9 plus square root of 3 

  4. negative 6 

  5. negative 3 

Pembahasan Soal:

Diketahui:

  • alpha equals 150 degree (sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top)
  • open vertical bar b with bar on top close vertical bar equals 6
  • a. b equals negative 9 

Ingat!

  • Perkalian skalar vektor a with rightwards arrow on top dengan vektor b with rightwards arrow on top open parentheses table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards arrow on top end cell end table close parentheses didefinisikan sebagai hasil kali panjang vektor a with rightwards arrow on top dan panjang vektor b with rightwards arrow on top dengan kosinus sudut yang diapit oleh vektor a with rightwards arrow on top dan vektor b with rightwards arrow on top atau table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell stack open vertical bar a close vertical bar with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open vertical bar b with rightwards arrow on top close vertical bar end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cos end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank alpha end table.
  • a with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar squared 

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top. 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 space cos space alpha end cell row cell negative 9 end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar cross times 6 cross times cos space 150 degree end cell row cell negative 9 end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar cross times 6 cross times open parentheses negative 1 half square root of 3 close parentheses end cell row cell negative 9 end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar cross times open parentheses negative 3 square root of 3 close parentheses end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell fraction numerator negative 9 over denominator negative 3 square root of 3 end fraction end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell fraction numerator 3 over denominator square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of 3 space satuan space panjang end cell end table 

Sehingga:

 table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top. left parenthesis b with rightwards arrow on top minus a with rightwards arrow on top right parenthesis end cell equals cell a with rightwards arrow on top. b with rightwards arrow on top minus a with rightwards arrow on top. a with rightwards arrow on top end cell row blank equals cell a with rightwards arrow on top. b with rightwards arrow on top minus open vertical bar a with rightwards arrow on top close vertical bar squared end cell row blank equals cell negative 9 minus open parentheses square root of 3 close parentheses squared end cell row blank equals cell negative 9 minus 3 end cell row blank equals cell negative 12 end cell end table 

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Hajrianti

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Sudut antara vektor a dan b sama dengan =120∘. Jika ∣a∣=4 dan ∣b∣=2 maka a.(a+b)=...

Pembahasan Soal:

Diketahui:

  • alpha equals 120 degree (sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top)
  • open vertical bar a close vertical bar equals 4
  • open vertical bar b close vertical bar equals 2 

Ingat!

  • Perkalian skalar vektor a with rightwards arrow on top dengan vektor b with rightwards arrow on top open parentheses table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards arrow on top end cell end table close parentheses didefinisikan sebagai hasil kali panjang vektor a with rightwards arrow on top dan panjang vektor b with rightwards arrow on top dengan kosinus sudut yang diapit oleh vektor a with rightwards arrow on top dan vektor b with rightwards arrow on top atau table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell stack open vertical bar a close vertical bar with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open vertical bar b with rightwards arrow on top close vertical bar end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cos end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank alpha end table.
  • a with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar squared 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top. left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top right parenthesis end cell equals cell a with rightwards arrow on top. a with rightwards arrow on top plus a with rightwards arrow on top. b with rightwards arrow on top end cell row blank equals cell open vertical bar a with rightwards arrow on top close vertical bar squared plus open vertical bar a close vertical bar open vertical bar b close vertical bar space cos space alpha end cell row blank equals cell 4 squared plus 4 cross times 2 cross times cos space 120 degree end cell row blank equals cell 16 plus 4 cross times 2 cross times open parentheses negative 1 half close parentheses end cell row blank equals cell 16 minus 4 end cell row blank equals 12 end table 

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Jika a dan b adalah sembarang vektor dari ∣a∣ menyatakan panjang vektor , maka (a+b)∙(a+b)=…

Pembahasan Soal:

Ingat kembali rumus berikut.

  • top enclose a bullet top enclose a equals open vertical bar top enclose a close vertical bar squared 
  • top enclose a bullet top enclose b equals open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar times cos space alpha 
  • top enclose a bullet top enclose b equals top enclose b bullet top enclose a
  • table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses a with bar on top plus b with bar on top close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bullet end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses a with bar on top plus b with bar on top close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bullet end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bullet end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bullet end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bullet end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with bar on top end cell end table
  • table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar p with bar on top plus q with bar on top close vertical bar squared end cell equals cell open vertical bar p with bar on top close vertical bar squared plus open vertical bar q with bar on top close vertical bar squared plus 2 open vertical bar space p with bar on top close vertical bar times open vertical bar stack space q with bar on top close vertical bar times cos space alpha end cell end table  

Dari rumus diatas diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a with bar on top plus b with bar on top close parentheses bullet open parentheses a with bar on top plus b with bar on top close parentheses end cell equals cell a with bar on top bullet a with bar on top plus a with bar on top bullet b with bar on top plus b with bar on top bullet a with bar on top plus b with bar on top bullet b with bar on top end cell row blank equals cell open vertical bar a with bar on top close vertical bar squared plus 2 open parentheses a with bar on top bullet b with bar on top close parentheses plus open vertical bar b with bar on top close vertical bar squared end cell row blank equals cell open vertical bar a with bar on top close vertical bar squared plus 2 times open vertical bar a with bar on top close vertical bar times open vertical bar b with bar on top close vertical bar times cos space alpha plus open vertical bar b with bar on top close vertical bar squared end cell row blank equals cell open vertical bar a with bar on top close vertical bar squared plus open vertical bar b with bar on top close vertical bar squared plus 2 times stack open vertical bar a close vertical bar with bar on top times open vertical bar b with bar on top close vertical bar times cos space alpha end cell row blank equals cell open vertical bar a with bar on top plus b with bar on top close vertical bar squared end cell end table   

Dengan demikian table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a with bar on top plus b with bar on top close parentheses bullet open parentheses a with bar on top plus b with bar on top close parentheses end cell equals cell open vertical bar a with bar on top plus b with bar on top close vertical bar squared end cell end table

Jadi, jawaban yang benar adalah A.

0

Roboguru

Jika ∣a∣=1dan∣∣​b∣∣​=4. Jika a∙(a+b)=3 maka besar sudut yang dibentuk oleh vektor adanb sama dengan...

Pembahasan Soal:

Ingat kembali  besar sudut pada vektor dan sifat dot product berikut.

  • Jika theta adalah sudut antara vektor top enclose a space dan space top enclose b, maka cos space theta equals fraction numerator a bullet b over denominator open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar end fraction  sehingga a bullet b equals open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar times cos space theta

 

  • Sudut antara vektor top enclose a space dan space top enclose a adalah 0 degree.

 

  • top enclose a left parenthesis top enclose a plus top enclose b right parenthesis equals top enclose a bullet top enclose a plus top enclose a bullet top enclose b 

 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a bullet left parenthesis top enclose a plus top enclose b right parenthesis end cell equals 3 row cell top enclose a bullet top enclose a plus top enclose a bullet top enclose b end cell equals 3 row cell open vertical bar top enclose a close vertical bar times open vertical bar top enclose a close vertical bar times cos space 0 degree plus open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar times cos space theta end cell equals 3 row cell 1 times 1 times 1 plus 1 times 4 times cos space theta end cell equals 3 row cell 1 plus 4 space cos space theta end cell equals 3 row cell 4 space cos space theta end cell equals cell 3 minus 1 end cell row cell 4 space cos space theta end cell equals 2 row cell cos space theta end cell equals cell 2 over 4 end cell row cell cos space theta end cell equals cell 1 half end cell row cell cos space theta end cell equals cell cos space 60 degree end cell row theta equals cell 60 degree end cell row blank blank blank row blank blank blank end table  

Jadi, besar sudut yang dibentuk oleh vektor top enclose a space dan space top enclose b adalah 60 degree.

Oleh karena itu, jawaban yang benar adalah C.

 

0

Roboguru

Jika ∣a∣=6; ∣∣​b∣∣​=5, ∣∣​a+b∣∣​=9 maka a∙(a+b)=…

Pembahasan Soal:

Ingat kembali rumus berikut.

  • table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar p with bar on top plus q with bar on top close vertical bar squared end cell equals cell open vertical bar p with bar on top close vertical bar squared plus 2 open parentheses p with bar on top bullet q with bar on top close parentheses plus open vertical bar q with bar on top close vertical bar squared end cell end table 
  • top enclose p bullet top enclose p equals open vertical bar top enclose p close vertical bar squared
  • top enclose a bullet left parenthesis top enclose a plus top enclose b right parenthesis equals top enclose a bullet top enclose a plus top enclose a bullet top enclose b 

Dari rumus diatas diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with bar on top plus b with bar on top close vertical bar squared end cell equals cell open vertical bar a with bar on top close vertical bar squared plus 2 open parentheses a with bar on top bullet b with bar on top close parentheses plus open vertical bar b with bar on top close vertical bar squared end cell row cell 9 squared end cell equals cell 6 squared plus 2 open parentheses a with bar on top bullet b with bar on top close parentheses plus 5 squared end cell row 81 equals cell 36 plus 2 open parentheses a with bar on top bullet b with bar on top close parentheses plus 25 end cell row 81 equals cell 36 plus 25 plus 2 open parentheses a with bar on top bullet b with bar on top close parentheses end cell row 81 equals cell 61 plus 2 open parentheses a with bar on top bullet b with bar on top close parentheses end cell row cell 81 minus 61 end cell equals cell 2 open parentheses a with bar on top bullet b with bar on top close parentheses end cell row 20 equals cell 2 open parentheses a with bar on top bullet b with bar on top close parentheses end cell row cell 20 over 2 end cell equals cell open parentheses a with bar on top bullet b with bar on top close parentheses end cell row 10 equals cell open parentheses a with bar on top bullet b with bar on top close parentheses end cell end table 

Karena open parentheses a with bar on top bullet b with bar on top close parentheses equals 10 maka

 table attributes columnalign right center left columnspacing 0px end attributes row cell a with bar on top bullet open parentheses a with bar on top plus b with bar on top close parentheses end cell equals cell a with bar on top bullet a with bar on top plus a with bar on top bullet b with bar on top end cell row blank equals cell open vertical bar a with bar on top close vertical bar squared plus 10 end cell row blank equals cell 6 squared plus 10 end cell row blank equals cell 36 plus 10 end cell row blank equals 46 end table

Dengan demikian a with bar on top bullet open parentheses a with bar on top plus b with bar on top close parentheses equals 46

Jadi, jawaban yang benar adalah B.

0

Roboguru

Diketahui vektor a=i+2j​−2k, b=3i−2j​+k dan c=2i+j​+2k. Jika a tegak lurus c maka (a+b)⋅(a−c) adalah ...

Pembahasan Soal:

Ingat!

Sifat-sifat perkalian dot product pada vektor:

  • a(b±c)=ab±ac(sifatdistributif) 
  • aa=a2 
  • Jika a=0b=0 jika ab, maka ab=0 

Sehingga:

(a+b).(ac)====a.aa.c+b.ab.c(12+22+(2)2)20+(3×1+(2)×2+1×(2))+(3×2+(2)×1+1×2)90+(3)+612  

Dengan demikian, nilai dari (a+b).(ac) adalah 12.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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