Roboguru

, , , dan  dengan  dan  bilangan real tidak nol. J...

bold italic a equals negative i with italic rightwards arrow on top plus 4 straight j with rightwards arrow on topbold italic b equals 2 straight i with bold rightwards arrow on top plus straight j with rightwards arrow on topbold italic c equals 3 i with rightwards arrow on top minus 4 j with rightwards arrow on top, dan bold italic x equals p bold italic a plus q bold italic b dengan p dan q bilangan real tidak nol. Jika bold italic x sejajar bold italic c, maka p dan q memenuhi hubungan...

  1. 8 p minus 11 q equals 0 

  2. 8 p plus 11 q equals 0 

  3. 11 p minus 8 q equals 0 

  4. 11 p minus 9 q equals 0 

  5. 11 p plus 8 q equals 0 

Jawaban:

Akan ditentukan bold italic x 

table attributes columnalign right center left columnspacing 0px end attributes row bold italic x equals cell p bold italic a plus q bold italic b end cell row blank equals cell p left parenthesis negative i with rightwards arrow on top plus 4 j with rightwards arrow on top right parenthesis plus q left parenthesis 2 i with rightwards arrow on top plus j with rightwards arrow on top right parenthesis end cell row blank equals cell negative p i with rightwards arrow on top plus 4 p j with rightwards arrow on top plus 2 q i with rightwards arrow on top plus q j with rightwards arrow on top end cell row blank equals cell left parenthesis negative p plus 2 q right parenthesis i with rightwards arrow on top plus left parenthesis 4 p plus q right parenthesis j with rightwards arrow on top end cell end table 

Diketahui bold italic x sejajar bold italic c, maka berlaku

table attributes columnalign right center left columnspacing 0px end attributes row bold italic x equals cell k bold italic c end cell row cell left parenthesis negative p plus 2 q right parenthesis i with rightwards arrow on top plus left parenthesis 4 p plus q right parenthesis j with rightwards arrow on top end cell equals cell k left parenthesis 3 i with rightwards arrow on top minus 4 j with rightwards arrow on top right parenthesis end cell row cell left parenthesis negative p plus 2 q right parenthesis i with rightwards arrow on top plus left parenthesis 4 p plus q right parenthesis j with rightwards arrow on top end cell equals cell 3 k i with rightwards arrow on top minus 4 k j with rightwards arrow on top end cell end table 

Diperoleh

 table attributes columnalign right center left columnspacing 0px end attributes row cell negative p plus 2 q end cell equals cell 3 k end cell row cell fraction numerator negative p plus 2 q over denominator 3 end fraction end cell equals k end table 

dan

 table attributes columnalign right center left columnspacing 0px end attributes row cell 4 p plus q end cell equals cell negative 4 k end cell row cell fraction numerator 4 p plus q over denominator negative 4 end fraction end cell equals k end table

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row k equals k row cell fraction numerator negative p plus 2 q over denominator 3 end fraction end cell equals cell fraction numerator 4 p plus q over denominator negative 4 end fraction end cell row cell left parenthesis negative 4 right parenthesis left parenthesis negative p plus 2 q right parenthesis end cell equals cell 3 left parenthesis 4 p plus q right parenthesis end cell row cell 4 p minus 8 q end cell equals cell 12 p plus 3 q end cell row cell left parenthesis 4 p minus 8 q right parenthesis minus left parenthesis 12 p plus 3 q right parenthesis end cell equals 0 row cell 4 p minus 12 p minus 8 q minus 3 q end cell equals 0 row cell negative 8 p minus 11 q end cell equals 0 row cell 8 p plus 11 q end cell equals 0 end table 

Jadi, jawaban yang tepat adalah B.

0

Ruangguru

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved