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Diketahui vektor a = 4 i − 2 3 ​ j ​ + 2 k ; b = 2 i + 5 j ​ − 3 k dan c = 2 j ​ − 3 k . Hasil dari a − 2 1 ​ b + 2 1 ​ c adalah ...

Diketahui vektor ; dan . Hasil dari adalah ...

  1. 3 straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top

  2. 3 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top

  3. 5 straight i with rightwards arrow on top minus 5 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top

  4. 5 straight i with rightwards arrow on top minus 5 straight j with rightwards arrow on top plus 3 straight k with rightwards arrow on top

  5. 5 straight i with rightwards arrow on top minus 5 straight j with rightwards arrow on top plus 5 straight k with rightwards arrow on top

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D. Nuryani

Master Teacher

Mahasiswa/Alumni Universitas Padjadjaran

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

jawaban yang benar adalah B.

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Pembahasan

Misalkan terdapat vektor , dan adalah skalar. maka berlaku: Berdasarkan sifat-sifat operasi vektor tersebut, maka: Dengan demikian, hasil operasi vektor pada soal adalah . Oleh karena itu, jawaban yang benar adalah B.

Misalkan terdapat vektor straight a with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 end cell row cell straight a subscript 2 end cell row cell straight a subscript 3 end cell end table close parentheses equals straight a subscript 1 straight i with rightwards arrow on top plus straight a subscript 2 straight j with rightwards arrow on top plus straight a subscript 3 straight k with rightwards arrow on top, straight b with rightwards arrow on top equals open parentheses table row cell straight b subscript 1 end cell row cell straight b subscript 2 end cell row cell straight b subscript 3 end cell end table close parentheses equals straight b subscript 1 straight i with rightwards arrow on top plus straight b subscript 2 straight j with rightwards arrow on top plus straight b subscript 3 straight k with rightwards arrow on top dan straight z adalah skalar. maka berlaku:

open parentheses straight i close parentheses space straight a with rightwards arrow on top plus-or-minus straight b with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 plus-or-minus straight b subscript 1 end cell row cell straight a subscript 2 plus-or-minus straight b subscript 2 straight a subscript 3 plus-or-minus straight b subscript 3 end cell end table close parentheses equals open parentheses straight a subscript 1 plus-or-minus straight b subscript 1 close parentheses straight i with rightwards arrow on top plus open parentheses straight a subscript 2 plus-or-minus straight b subscript 2 close parentheses straight j with rightwards arrow on top plus open parentheses straight a subscript 3 plus-or-minus straight b subscript 3 close parentheses straight k with rightwards arrow on top left parenthesis ii right parenthesis space straight z straight a with rightwards arrow on top equals open parentheses table row cell za subscript 1 end cell row cell za subscript 2 end cell row cell za subscript 3 end cell end table close parentheses equals za subscript 1 straight i with rightwards arrow on top plus za subscript 2 straight j with rightwards arrow on top plus za subscript 3 straight k with rightwards arrow on top

Berdasarkan sifat-sifat operasi vektor tersebut, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with rightwards arrow on top minus 1 half straight b with rightwards arrow on top plus 1 half straight c with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative begin inline style 3 over 2 end style end cell row 2 end table close parentheses minus 1 half open parentheses table row 2 row 5 row cell negative 3 end cell end table close parentheses plus 1 half open parentheses table row 0 row 2 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row cell negative begin inline style 3 over 2 end style end cell row 2 end table close parentheses minus open parentheses table row 1 row cell begin inline style 5 over 2 end style end cell row cell negative begin inline style 3 over 2 end style end cell end table close parentheses plus open parentheses table row 0 row 1 row cell negative begin inline style 3 over 2 end style end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 4 minus 1 end cell row cell negative begin inline style 3 over 2 end style minus begin inline style 5 over 2 end style plus 1 end cell row cell 2 plus begin inline style 3 over 2 end style minus begin inline style 3 over 2 end style end cell end table close parentheses end cell row blank equals cell open parentheses table row 3 row cell negative 3 end cell row 2 end table close parentheses end cell row blank equals cell 3 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top end cell end table 

Dengan demikian, hasil operasi vektor pada soal adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight i with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight j with rightwards arrow 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 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight k with rightwards arrow on top end cell end table.

Oleh karena itu, jawaban yang benar adalah B.

Latihan Bab

Pengertian dan Operasi Vektor I

Operasi Vektor II

Kedudukan Vektor

Aljabar Vektor I

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20_A1_Muhammad Rizky

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