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Misalkan diketahui vektor a=⎝⎛​1−23​−213​​⎠⎞​ dan vektor b=⎝⎛​−123​21​​⎠⎞​.  a. tentukan vektor (a+b) dan vektor (a−b).  b. Jika θ adalah sudut antara vektor (a+b) dan vektor (a−b), hitunglah nilai kosinus sudut θ.

Pertanyaan

Misalkan diketahui vektor a with rightwards arrow on top equals open parentheses table row 1 row cell negative 3 over 2 end cell row cell negative 13 over 2 end cell end table close parentheses dan vektor b with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row cell 3 over 2 end cell row cell 1 half end cell end table close parentheses

a. tentukan vektor left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top right parenthesis dan vektor left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis

b. Jika theta adalah sudut antara vektor left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top right parenthesis dan vektor left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis, hitunglah nilai kosinus sudut theta

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

jawaban yang benar untuk pertanyaan tersebut adalah seperti tersebut diatas.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah bold left parenthesis bold italic a with bold rightwards arrow on top bold plus bold italic b with bold rightwards arrow on top bold right parenthesis bold equals bold left parenthesis bold 0 bold comma bold 0 bold comma bold minus bold 6 bold right parenthesis,  bold left parenthesis bold italic a with bold rightwards arrow on top bold minus bold italic b with bold rightwards arrow on top bold right parenthesis bold equals bold left parenthesis bold 2 bold comma bold minus bold 3 bold comma bold minus bold 7 bold right parenthesis, dan nilai bold cos bold italic theta bold equals bold 7 over bold 30 square root of bold 15.

Ingat! 

Jika diketahui dua 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 row cell z subscript 1 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell row cell z subscript 2 end cell end table close parentheses, maka :

  • Penjumlahan vektor a with rightwards arrow on top plus b with rightwards arrow on top equals open parentheses table row cell x subscript 1 plus x subscript 2 end cell row cell y subscript 1 plus y subscript 2 end cell row cell z subscript 1 plus z subscript 2 end cell end table close parentheses 
  • Pengurangan vektor a with rightwards arrow on top minus b with rightwards arrow on top equals open parentheses table row cell x subscript 1 minus x subscript 2 end cell row cell y subscript 1 minus y subscript 2 end cell row cell z subscript 1 minus z subscript 2 end cell end table close parentheses 
  • Hasil kali a with rightwards arrow on top times b with rightwards arrow on top equals x subscript 1 times end subscript x subscript 2 plus y subscript 1 times y subscript 2 plus z subscript 1 times z subscript 2 
  • Panjang vektor open vertical bar a with rightwards arrow on top close vertical bar equals square root of open parentheses x subscript 1 close parentheses squared plus open parentheses y subscript 1 close parentheses squared plus open parentheses z subscript 1 close parentheses squared end root 
  • Nilai cos theta equals fraction numerator a with rightwards arrow on top times 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 

a. Tentukan vektor left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top right parenthesis dan vektor left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis

  • Nilai left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top right parenthesis

open parentheses a with rightwards arrow on top plus b with rightwards arrow on top close parentheses equals open parentheses table row cell 1 plus left parenthesis negative 1 right parenthesis end cell row cell negative 3 over 2 plus 3 over 2 end cell row cell negative 13 over 2 plus 1 half end cell end table close parentheses equals open parentheses table row 0 row 0 row cell negative 12 over 2 end cell end table close parentheses equals open parentheses table row 0 row 0 row cell negative 6 end cell end table close parentheses 

  • Nilai left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis

left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis equals open parentheses table row cell 1 minus left parenthesis negative 1 right parenthesis end cell row cell negative 3 over 2 minus 3 over 2 end cell row cell negative 13 over 2 minus 1 half end cell end table close parentheses equals open parentheses table row cell 1 plus 1 end cell row cell negative 6 over 2 end cell row cell negative 14 over 2 end cell end table close parentheses equals open parentheses table row 2 row cell negative 3 end cell row cell negative 7 end cell end table close parentheses 

b. Jika theta adalah sudut antara vektor left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top right parenthesis dan vektor left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis, hitunglah nilai kosinus sudut theta

table attributes columnalign right center left columnspacing 0px end attributes row cell cos theta end cell equals cell fraction numerator open parentheses a with rightwards arrow on top plus b with rightwards arrow on top close parentheses open parentheses a with rightwards arrow on top minus b with rightwards arrow on top close parentheses over denominator open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top minus b with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 0 row 0 row cell negative 6 end cell end table close parentheses open parentheses table row 2 row cell negative 3 end cell row cell negative 7 end cell end table close parentheses over denominator open parentheses square root of 0 squared plus 0 squared plus open parentheses negative 6 close parentheses squared end root close parentheses open parentheses square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus left parenthesis negative 7 right parenthesis squared end root close parentheses end fraction end cell row blank equals cell fraction numerator 0 plus 0 plus 42 over denominator open parentheses square root of 0 plus 0 plus 36 end root close parentheses open parentheses square root of 2 plus 9 plus 49 end root close parentheses end fraction end cell row blank equals cell fraction numerator 42 over denominator square root of 36 square root of 60 end fraction end cell row blank equals cell fraction numerator 42 over denominator 6 square root of 4 times 15 end root end fraction end cell row blank equals cell fraction numerator 7 over denominator 2 square root of 15 end fraction cross times fraction numerator square root of 15 over denominator square root of 15 end fraction end cell row blank equals cell fraction numerator 7 over denominator 2 times 15 end fraction square root of 15 end cell row blank equals cell 7 over 30 square root of 15 end cell end table 

Dengan demikian, jawaban yang benar untuk pertanyaan tersebut adalah seperti tersebut diatas.

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