Diketahui vektor u=−6i+2j dan v=i+4j. Nyatakan komponen dari resultan kedua vektor tersebut dalam bentuk vektor baris! Tentukan hasil dari u⋅v dan ∣∣​u∣∣​⋅∣∣​v∣∣​! Jika sudut terkecil yang terbentuk antara kedua vektor adalah θ, maka tentukan cosinus !

Pertanyaan

Diketahui vektor u with rightwards arrow on top equals negative 6 i plus 2 j dan v with rightwards arrow on top equals i plus 4 j.

  1. Nyatakan komponen dari resultan kedua vektor tersebut dalam bentuk vektor baris!
  2. Tentukan hasil dari u with rightwards arrow on top times v with rightwards arrow on top dan open vertical bar u with rightwards arrow on top close vertical bar times open vertical bar v with rightwards arrow on top close vertical bar!
  3. Jika sudut terkecil yang terbentuk antara kedua vektor adalah begin mathsize 14px style theta end style, maka tentukan cosinus undefined!

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

diperoleh cos space theta equals 1 over 170 square root of 170.

Pembahasan

a. Menyatakan dalam vektor baris 

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell table row cell u with rightwards arrow on top equals negative 6 i plus 2 j end cell rightwards arrow cell u with rightwards arrow on top equals open parentheses table row cell negative 6 end cell 2 end table close parentheses end cell row cell v with rightwards arrow on top equals i plus 4 j end cell rightwards arrow cell v with rightwards arrow on top equals open parentheses table row 1 4 end table close parentheses end cell end table end cell end table 

b. Hasil dari u with rightwards arrow on top times v with rightwards arrow on top dan open vertical bar u with rightwards arrow on top close vertical bar times open vertical bar v with rightwards arrow on top close vertical bar

Error converting from MathML to accessible text.  

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar u with rightwards arrow on top close vertical bar times open vertical bar v with rightwards arrow on top close vertical bar end cell equals cell open parentheses square root of open parentheses negative 6 close parentheses squared plus 2 squared end root close parentheses times open parentheses square root of 1 squared plus 4 squared end root close parentheses end cell row blank equals cell left parenthesis square root of 36 plus 4 end root right parenthesis times left parenthesis square root of 1 plus 16 end root right parenthesis end cell row blank equals cell square root of 40 times square root of 17 end cell row blank equals cell square root of 680 end cell row blank equals cell 2 square root of 170 end cell end table 

Dengan demikian, hasil dari u with rightwards arrow on top times v with rightwards arrow on top equals 2 dan open vertical bar u with rightwards arrow on top close vertical bar times open vertical bar v with rightwards arrow on top close vertical bar equals 2 square root of 170.

c. Menenentukan cosinus undefined

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator vertical line u with rightwards arrow on top vertical line times vertical line v with rightwards arrow on top vertical line end fraction end cell row blank equals cell fraction numerator 2 over denominator 2 square root of 170 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 170 end fraction times fraction numerator square root of 170 over denominator square root of 170 end fraction end cell row blank equals cell 1 over 170 square root of 170 end cell end table   

Dengan demikian, diperoleh cos space theta equals 1 over 170 square root of 170.

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