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Diketahui p ​ = ( t 5 ​ ) dan q ​ = ( 1 − 6 − t ​ ) .Tentukan nilai t jika vektor p ​ sejajar vektor q ​ adalah ...

Diketahui  dan . Tentukan nilai  jika vektor  sejajar vektor  adalah ...space 

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A. Armanda

Master Teacher

Mahasiswa/Alumni Universitas Indraprasta PGRI

Jawaban terverifikasi

Jawaban

nilai adalah dan .

nilai t adalah negative 1 dan negative 5.

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Pembahasan

Jika vektor sejajar vektor , maka berlaku: dengan merupakan panjang vektor dan adalah panjang vektor . Sehingga, faktorkan persamaan di atas. Jadi, nilai adalah dan .

Jika vektor p with rightwards arrow on top sejajar vektor q with rightwards arrow on top, maka berlaku:

p with rightwards arrow on top times q with rightwards arrow on top equals open vertical bar p with rightwards arrow on top close vertical bar open vertical bar q with rightwards arrow on top close vertical bar 

dengan open vertical bar p with rightwards arrow on top close vertical bar merupakan panjang vektor p with rightwards arrow on top dan open vertical bar q with rightwards arrow on top close vertical bar adalah panjang vektor q with rightwards arrow on top.

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top times q with rightwards arrow on top end cell equals cell open vertical bar p with rightwards arrow on top close vertical bar open vertical bar q with rightwards arrow on top close vertical bar end cell row cell open parentheses table row t row 5 end table close parentheses times open parentheses table row 1 row cell negative 6 minus t end cell end table close parentheses end cell equals cell square root of t squared plus 5 squared end root times square root of 1 squared plus open parentheses negative 6 minus t close parentheses squared end root end cell row cell t times 1 plus 5 open parentheses negative 6 minus t close parentheses end cell equals cell square root of t squared plus 5 squared end root times square root of 1 squared plus open parentheses negative 6 minus t close parentheses squared end root end cell row cell t minus 30 minus 5 t end cell equals cell square root of t squared plus 5 squared end root times square root of 1 squared plus open parentheses negative 6 minus t close parentheses squared end root end cell row cell negative 4 t minus 30 end cell equals cell square root of t squared plus 5 squared end root times square root of 1 squared plus open parentheses negative 6 minus t close parentheses squared end root space open parentheses kedua space ruas space dikuadratkan close parentheses end cell row cell open parentheses negative 4 t minus 30 close parentheses squared end cell equals cell open parentheses square root of t squared plus 5 squared end root times square root of 1 squared plus open parentheses negative 6 minus t close parentheses squared end root close parentheses squared end cell row cell 16 t squared plus 240 t plus 900 end cell equals cell open parentheses t squared plus 25 close parentheses times open parentheses 1 plus 36 plus 12 t plus t squared close parentheses end cell row cell 16 t squared plus 240 t plus 900 end cell equals cell open parentheses t squared plus 25 close parentheses times open parentheses 37 plus 12 t plus t squared close parentheses end cell row cell 16 t squared plus 240 t plus 900 end cell equals cell t to the power of 4 plus 12 t cubed plus 62 t squared plus 300 t plus 925 end cell row cell t to the power of 4 plus 12 t cubed plus 46 t squared plus 60 t plus 25 end cell equals 0 end table 

faktorkan persamaan di atas.

table attributes columnalign right center left columnspacing 0px end attributes row cell t to the power of 4 plus 12 t cubed plus 46 squared plus 60 t plus 25 end cell equals 0 row cell open parentheses t plus 1 close parentheses squared open parentheses t plus 5 close parentheses squared end cell equals 0 row cell therefore space t end cell equals cell negative 1 logical or t equals negative 5 end cell end table  

Jadi, nilai t adalah negative 1 dan negative 5.

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