Pertanyaan

Diketahui vektor a = 2 i + 3 j ​ + x k tegak lurus dengan vektor b = 3 i − 6 j ​ + 2 k . Jika c = x i − 5 j ​ + k maka a − b + c =

Diketahui vektor  tegak lurus dengan vektor . Jika  maka  

  1. 5 stack text i end text with rightwards arrow on top minus 4 stack text j end text with rightwards arrow on top plus 5 stack text k end text with rightwards arrow on top 

  2. 5 stack text i end text with rightwards arrow on top plus 4 stack text j end text with rightwards arrow on top plus 5 stack text k end text with rightwards arrow on top 

  3. 5 stack text i end text with rightwards arrow on top minus 4 stack text j end text with rightwards arrow on top minus 5 stack text k end text with rightwards arrow on top 

  4. 5 stack text i end text with rightwards arrow on top minus 8 stack text j end text with rightwards arrow on top plus 9 stack text k end text with rightwards arrow on top 

  5. 11 stack text i end text with rightwards arrow on top minus 8 stack text j end text with rightwards arrow on top plus 9 stack text k end text with rightwards arrow on top 

H. Eka

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

 jawaban yang tepat adalah B.

Pembahasan

Jika vektor dan , maka Jika vektor tegak lurus vektor , maka Berdasarkan konsep di atas, nilai dapat ditentukan sebagai berikut. Diperoleh vektor sehingga dan . Dapat ditentukan operasi penjumlahan dan pengurangan vektor berikut. Diperoleh Oleh karena itu,jawaban yang tepat adalah B.

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Jika vektor a with rightwards arrow on top tegak lurus vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px 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 space cos space alpha end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Berdasarkan konsep di atas, nilai x dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 row cell open parentheses table row 2 row 3 row x end table close parentheses open parentheses table row 3 row cell negative 6 end cell row 2 end table close parentheses end cell equals 0 row cell 2 times 3 plus 3 open parentheses negative 6 close parentheses plus 2 x end cell equals 0 row cell 6 minus 18 plus 2 x end cell equals 0 row cell negative 12 plus 2 x end cell equals 0 row cell 2 x end cell equals 12 row x equals 6 end table

Diperoleh vektor x equals 6 sehingga a with rightwards arrow on top equals 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top plus 6 k with rightwards arrow on top dan c with rightwards arrow on top equals 6 i with rightwards arrow on top minus 5 j with rightwards arrow on top plus k with rightwards arrow on top. Dapat ditentukan operasi penjumlahan dan pengurangan vektor berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top minus b with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell open parentheses table row 2 row 3 row 6 end table close parentheses minus open parentheses table row 3 row cell negative 6 end cell row 2 end table close parentheses plus open parentheses table row 6 row cell negative 5 end cell row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 2 minus 3 plus 6 end cell row cell 3 minus open parentheses negative 6 close parentheses plus open parentheses negative 5 close parentheses end cell row cell 6 minus 2 plus 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 5 row 4 row 5 end table close parentheses end cell end table

Diperoleh 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 minus 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 blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell c 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 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell i 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 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 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 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell k with rightwards arrow on top end cell end table

Oleh karena itu, jawaban yang tepat adalah B.

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