Diketahui ∣a∣=6​×(a+b)×(a−b)=0 dan a×(a−b)=3. Besar sudut antara vektor a  dan b adalah ...

Pertanyaan

Diketahui open vertical bar a close vertical bar equals square root of 6 cross times open parentheses top enclose a plus top enclose b close parentheses cross times open parentheses top enclose a minus top enclose b close parentheses equals 0 space dan space top enclose a cross times open parentheses top enclose a minus top enclose b close parentheses equals 3. Besar sudut antara vektor top enclose a space dan top enclose b adalah ...

  1. 30 degree 

  2. 45 degree  

  3. 60 degree  

  4. 90 degree  

  5. 180 degree 

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Pembahasan

Terdapat kesalahan pada soal sehingga revisi soal yang benar adalah sebagai berikut :

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a close vertical bar end cell equals cell square root of 6 end cell row cell open parentheses top enclose a plus top enclose b close parentheses open parentheses top enclose a minus top enclose b close parentheses end cell equals cell 0 space dan space end cell row cell top enclose a open parentheses top enclose a minus top enclose b close parentheses end cell equals 3 end table

Ingat konsep :

a with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar squared 

Ingat rumus aljabar :

x squared minus y squared equals open parentheses x minus y close parentheses open parentheses x plus y close parentheses   

Ingat konsep perkalian titik dua vektor jika diketahui sudutnya :

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top. b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar. cos space theta end cell end table   

Berdasarkan konsep di atas diperoleh perhitungan sebagai berikut.  

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a with rightwards arrow on top minus b with rightwards arrow on top close parentheses open parentheses a with rightwards arrow on top plus b with rightwards arrow on top close parentheses end cell equals 0 row cell a with rightwards arrow on top squared minus b with rightwards arrow on top squared end cell equals 0 row cell open vertical bar a with rightwards arrow on top close vertical bar squared minus open vertical bar b with rightwards arrow on top close vertical bar squared end cell equals 0 row cell open parentheses square root of 6 close parentheses squared minus open vertical bar b with rightwards arrow on top close vertical bar squared end cell equals 0 row cell open parentheses square root of 6 minus open vertical bar b with rightwards arrow on top close vertical bar close parentheses open parentheses square root of 6 plus open vertical bar b with rightwards arrow on top close vertical bar close parentheses end cell equals 0 row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of 6 space left parenthesis panjang space vektor space pasti space positif right parenthesis end cell end table 

Selanjutnya :

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top open parentheses a with rightwards arrow on top minus b with rightwards arrow on top close parentheses end cell equals 3 row cell a with rightwards arrow on top. a with rightwards arrow on top minus a with rightwards arrow on top. b with rightwards arrow on top end cell equals 3 row cell open vertical bar a with rightwards arrow on top close vertical bar squared minus a with rightwards arrow on top. b with rightwards arrow on top end cell equals 3 row cell 6 minus 3 end cell equals cell a with rightwards arrow on top b with rightwards arrow on top end cell end table 

Selanjutnya

:table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top. b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos space theta end cell row cell cos space theta end cell equals cell fraction numerator a with rightwards arrow on top. b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator 3 over denominator square root of 6 square root of 6 end fraction end cell row blank equals cell 3 over 6 end cell row cell cos space theta end cell equals cell 1 half end cell row theta equals cell 60 degree end cell end table 

Dengan demikian diperoleh besar sudut antara vektor top enclose a dan top enclose b adalah 60 degree.

Oleh karena itu jawaban yang tepat adalah C.

 

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