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Proyeksi vektor p​=⎝⎛​2−51​⎠⎞​ dan q​=⎝⎛​x−24​⎠⎞​ adalah O (vektor nol). Tentukan nilai x.

Pertanyaan

Proyeksi vektor p with rightwards arrow on top equals open parentheses table row 2 row cell negative 5 end cell row 1 end table close parentheses dan q with rightwards arrow on top equals open parentheses table row x row cell negative 2 end cell row 4 end table close parentheses adalah O with rightwards arrow on top (vektor nol). Tentukan nilai x.

N. Puspita

Master Teacher

Jawaban terverifikasi

Pembahasan

Diketahui vektor p with rightwards arrow on top equals open parentheses table row 2 row cell negative 5 end cell row 1 end table close parenthesesq with rightwards arrow on top equals open parentheses table row x row cell negative 2 end cell row 4 end table close parentheses dan proyeksi vektor p pada q adala vektor c with rightwards arrow on top equals open parentheses table row 0 row 0 row 0 end table close parentheses, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top end cell equals cell fraction numerator p with rightwards arrow on top. q with rightwards arrow on top over denominator open vertical bar q with rightwards arrow on top close vertical bar squared end fraction cross times q with rightwards arrow on top end cell row cell open parentheses table row 0 row 0 row 0 end table close parentheses end cell equals cell fraction numerator left parenthesis 2 right parenthesis left parenthesis x right parenthesis plus left parenthesis negative 5 right parenthesis left parenthesis negative 2 right parenthesis plus left parenthesis 1 right parenthesis left parenthesis 4 right parenthesis over denominator open parentheses square root of x squared plus left parenthesis negative 2 right parenthesis squared plus 4 squared end root close parentheses squared end fraction cross times open parentheses table row x row cell negative 2 end cell row 4 end table close parentheses end cell row cell open parentheses table row 0 row 0 row 0 end table close parentheses end cell equals cell fraction numerator 2 x plus 10 plus 4 over denominator x squared plus 4 plus 16 end fraction cross times open parentheses table row x row cell negative 2 end cell row 4 end table close parentheses end cell row cell open parentheses table row 0 row 0 row 0 end table close parentheses end cell equals cell fraction numerator 2 x plus 14 over denominator x squared plus 20 end fraction cross times open parentheses table row x row cell negative 2 end cell row 4 end table close parentheses end cell row cell open parentheses table row 0 row 0 row 0 end table close parentheses end cell equals cell open parentheses table row cell fraction numerator 2 x squared plus 14 x over denominator x squared plus 20 end fraction end cell row cell fraction numerator negative 4 x minus 28 over denominator x squared plus 20 end fraction end cell row cell fraction numerator 8 x plus 56 over denominator x squared plus 20 end fraction end cell end table close parentheses end cell end table

Dari kesamaan vektor di atas, didapatkan table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x squared plus 14 x over denominator x squared plus 20 end fraction end cell equals 0 end table, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x squared plus 14 x over denominator x squared plus 20 end fraction end cell equals 0 row cell 2 x squared plus 14 x end cell equals 0 row cell x squared plus 7 x end cell equals 0 row cell x left parenthesis x plus 7 right parenthesis end cell equals 0 end table 

x equals 0 space atau space x equals negative 7

Karena yang memenuhi hanyalah x equals negative 7, jadi nilai x adalah negative sign 7

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