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Diketahui a = 3 i − 4 j ​ + k , b = − 5 i + 6 j ​ + 2 x k dan c = 14 i − 18 j ​ + x k sebidang (koplanar) sehingga 3 a − b = c . Nilai x 2 + x − 2 = ...

Diketahui  dan  sebidang (koplanar) sehingga . Nilai  ...

  1. ...space 

  2. ...space 

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S. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Prof. DR. Hamka

Jawaban terverifikasi

Pembahasan

Tiga buah vektor ( , dan ) dikatakan koplanar (terletak dalam satu bidang) jika: dengan dan adalah suatu konstanta. Diketahui: , dan Karena vektor , dan koplanar maka berlaku: Diperoleh. Gunakan metode eliminasi persamaan dan seperti berikut: Substitusi ke persamaan seperti berikut: Kemudian substitusi dan ke persamaan seperti berikut: Maka, nilai .

Tiga buah vektor (begin mathsize 14px style a end stylebegin mathsize 14px style b end style dan begin mathsize 14px style c end style) dikatakan koplanar (terletak dalam satu bidang) jika:

begin mathsize 14px style m times a plus n times b equals c end style

dengan begin mathsize 14px style m end style dan begin mathsize 14px style n end style adalah suatu konstanta.

Diketahui:

begin mathsize 14px style a with rightwards arrow on top equals 3 i with rightwards arrow on top minus 4 j with rightwards arrow on top plus k with rightwards arrow on top end stylebegin mathsize 14px style b with rightwards arrow on top equals negative 5 i with rightwards arrow on top plus 6 j with rightwards arrow on top plus 2 x k with rightwards arrow on top end style dan begin mathsize 14px style c with rightwards arrow on top equals 14 i with rightwards arrow on top minus 18 j with rightwards arrow on top plus x k with rightwards arrow on top end style

 Karena  vektor begin mathsize 14px style a end stylebegin mathsize 14px style b end style dan begin mathsize 14px style c end style koplanar maka berlaku:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell m times a plus n times b end cell equals c row cell m open square brackets table row 3 row cell negative 4 end cell row 1 end table close square brackets plus n open square brackets table row cell negative 5 end cell row 6 row cell 2 x end cell end table close square brackets end cell equals cell open square brackets table row 14 row cell negative 18 end cell row x end table close square brackets end cell row cell open square brackets table row cell 3 m end cell row cell negative 4 m end cell row m end table close square brackets plus open square brackets table row cell negative 5 n end cell row cell 6 n end cell row cell 2 x n end cell end table close square brackets end cell equals cell open square brackets table row 14 row cell negative 18 end cell row x end table close square brackets end cell row cell open square brackets table row cell 3 m minus 5 n end cell row cell negative 4 m plus 6 n end cell row cell m plus 2 x n end cell end table close square brackets end cell equals cell open square brackets table row 14 row cell negative 18 end cell row x end table close square brackets end cell end table end style

Diperoleh.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 m minus 5 n end cell equals cell 14 space space space space space space space... open parentheses 1 close parentheses end cell row cell negative 4 m plus 6 n end cell equals cell negative 18 space space space... open parentheses 2 close parentheses end cell row cell m plus 2 x n end cell equals cell x space space space space space space space space... open parentheses 3 close parentheses end cell end table end style

Gunakan metode eliminasi persamaan begin mathsize 14px style open parentheses 1 close parentheses end style dan begin mathsize 14px style open parentheses 2 close parentheses end style seperti berikut:

begin mathsize 14px style table attributes columnalign right center left end attributes row cell 3 m minus 5 n end cell equals 14 row cell negative 4 m plus 6 n end cell equals cell negative 18 end cell end table space left enclose right enclose table row cell cross times 4 end cell row cell cross times 3 end cell end table end enclose end enclose  table attributes columnalign right center left end attributes row cell 12 m minus 20 n end cell equals 56 row cell negative 12 m plus 18 n end cell equals cell negative 54 space plus end cell row cell negative 2 n end cell equals 2 row n equals cell fraction numerator 2 over denominator negative 2 end fraction end cell row n equals cell negative 1 end cell end table end style

Substitusi begin mathsize 14px style n equals negative 1 end style ke persamaan begin mathsize 14px style open parentheses 1 close parentheses end style seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 m minus 5 n end cell equals 14 row cell 3 m minus 5 open parentheses negative 1 close parentheses end cell equals 14 row cell 3 m plus 5 end cell equals 14 row cell 3 m end cell equals cell 14 minus 5 end cell row cell 3 m end cell equals 9 row m equals cell 9 over 3 end cell row m equals 3 end table end style

Kemudian substitusi begin mathsize 14px style n equals negative 1 end style dan begin mathsize 14px style m equals 3 end style ke persamaan begin mathsize 14px style open parentheses 3 close parentheses end style seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell m plus 2 x n end cell equals x row cell 3 plus 2 x open parentheses negative 1 close parentheses end cell equals x row cell 3 plus open parentheses negative 2 x close parentheses end cell equals x row cell negative 2 x minus x end cell equals cell negative 3 end cell row cell negative 3 x end cell equals cell negative 3 end cell row x equals cell fraction numerator negative 3 over denominator negative 3 end fraction end cell row x equals 1 end table end style

Maka, nilai begin mathsize 14px style x squared plus x minus 2 equals 1 squared plus 1 minus 2 equals 1 plus 1 minus 2 equals 0 end style.

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