Roboguru

Diketahui , dan . Jika  dan  memenuhi persamaan , berapa nilai dari  ?

Pertanyaan

Diketahui straight a with rightwards arrow on top equals 2 straight i plus straight j comma space straight b with rightwards arrow on top equals straight i plus 3 straight j, dan straight c with rightwards arrow on top equals straight i minus 7 straight j. Jika straight m dan straight n memenuhi persamaan straight c with rightwards arrow on top equals straight m straight a with rightwards arrow on top plus straight n straight b with rightwards arrow on top, berapa nilai dari open parentheses straight m minus straight n close parentheses ?

Pembahasan Soal:

Diketahui straight a with rightwards arrow on top equals 2 straight i plus straight j comma space straight b with rightwards arrow on top equals straight i plus 3 straight j, dan straight c with rightwards arrow on top equals straight i minus 7 straight j. Jika straight m dan straight n memenuhi persamaan straight c with rightwards arrow on top equals straight m straight a with rightwards arrow on top plus straight n straight b with rightwards arrow on top, maka:

space space space space space space space space straight c with rightwards arrow on top equals straight m straight a with rightwards arrow on top plus straight n straight b with rightwards arrow on top open parentheses table row 1 row cell negative 7 end cell end table close parentheses equals straight m open parentheses table row 2 row 1 end table close parentheses plus straight n open parentheses table row 1 row 3 end table close parentheses open parentheses table row 1 row cell negative 7 end cell end table close parentheses equals open parentheses table row cell 2 straight m end cell row straight m end table close parentheses plus open parentheses table row straight n row cell 3 straight n end cell end table close parentheses open parentheses table row 1 row cell negative 7 end cell end table close parentheses equals open parentheses table row cell 2 straight m plus straight n end cell row cell straight m plus 3 straight n end cell end table close parentheses space rightwards arrow table row cell 2 straight m plus straight n equals 1 space... space open parentheses 1 close parentheses end cell row cell straight m plus 3 straight n equals negative 7 space... space open parentheses 2 close parentheses end cell end table 

Eliminasi (1) dan (2)

Error converting from MathML to accessible text. 

Subtitusi nilai text m end text ke (1)

space space 2 straight m plus straight n equals 1 2 open parentheses 2 close parentheses plus straight n equals 1 space space space space space 4 plus straight n equals 1 space space space space space space space space space space straight n equals 1 minus 4 space space space space space space space space space space straight n equals negative 3 

Sehingga:

straight m minus straight n equals 2 minus open parentheses negative 3 close parentheses equals 2 plus 3 equals 5 

Jadi, nilai dari open parentheses straight m minus straight n close parentheses adalah 5.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

W. Lestari

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui ,  dan . Jika  maka nilai  sama dengan ...

Pembahasan Soal:

Diketahui a with bar on top equals open parentheses table row cell negative 3 end cell row 4 end table close parenthesesb with bar on top equals open parentheses table row 1 row cell negative 1 end cell end table close parentheses dan c with bar on top equals open parentheses table row cell negative 2 end cell row 1 end table close parentheses

Ingat!

  • Secara aljabar, perkalian vektor dengan skalar di R squared dapat dirumuskan dengan a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses maka k a with rightwards arrow on top equals k open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses equals open parentheses table row cell k a subscript 1 end cell row cell k a subscript 2 end cell end table close parentheses.
  • Pada vektor, jika a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses maka a with rightwards arrow on top plus b with rightwards arrow on top equals open parentheses table row cell a subscript 1 plus b subscript 1 end cell row cell a subscript 2 plus b subscript 2 end cell end table close parentheses.

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x a with bar on top plus y b with bar on top end cell equals cell c with bar on top end cell row cell x open parentheses table row cell negative 3 end cell row 4 end table close parentheses plus y open parentheses table row 1 row cell negative 1 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell row 1 end table close parentheses end cell row cell open parentheses table row cell negative 3 x end cell row cell 4 x end cell end table close parentheses plus open parentheses table row y row cell negative y end cell end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell row 1 end table close parentheses end cell row cell open parentheses table row cell negative 3 x plus y end cell row cell 4 x minus y end cell end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell row 1 end table close parentheses end cell end table 

Dari kesamaan vektor di atas didapat:

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 x plus y end cell equals cell negative 2 space... space left parenthesis 1 right parenthesis end cell row cell 4 x minus y end cell equals cell 1 space...... space left parenthesis 2 right parenthesis end cell end table 

Dengan menggunakan metode eliminasi, maka:

table row cell negative 3 x plus y equals negative 2 end cell row cell 4 x minus y equals 1 space space space space space plus end cell row cell x equals negative 1 end cell end table

Kita sustitusikan x=-1 ke salah satu persamaan yaitu 4x-y=1, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 x minus y end cell equals 1 row cell 4 left parenthesis negative 1 right parenthesis minus y end cell equals 1 row cell negative 4 minus y end cell equals 1 row cell negative y end cell equals cell 1 plus 4 end cell row cell negative y end cell equals 5 row y equals cell negative 5 end cell end table  

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell x y end cell equals cell left parenthesis negative 1 right parenthesis cross times left parenthesis negative 5 right parenthesis end cell row blank equals 5 end table 

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Diketahui vektor  dan . Tentukan .

Pembahasan Soal:

Apabila diketahui

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell row cell a subscript 3 end cell end table close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table dan table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight b with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell row cell b subscript 3 end cell end table close parentheses end cell end table

maka, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight b with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell a subscript 1 plus b subscript 1 end cell row cell a subscript 2 plus b subscript 2 end cell row cell a subscript 3 plus b subscript 3 end cell end table close parentheses end cell end table

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top plus v with rightwards arrow on top end cell equals cell open parentheses table row cell negative 4 end cell row 8 row 2 end table close parentheses plus open parentheses table row 5 row 1 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 plus 5 end cell row cell 8 plus 1 end cell row cell 2 plus left parenthesis negative 4 right parenthesis end cell end table close parentheses end cell row cell u with rightwards arrow on top plus v with rightwards arrow on top end cell equals cell open parentheses table row 1 row 9 row cell negative 2 end cell end table close parentheses end cell end table 
  

0

Roboguru

Diketahui koordinat titik  dan . Jika  berturut-turut vektor posisi titik A, B, dan C, hasil  adalah....

Pembahasan Soal:

Diketahui titik-titik koordinat straight A left parenthesis 4 comma space minus 3 right parenthesis comma space straight B left parenthesis negative 1 comma space minus 5 right parenthesis dan straight C open parentheses negative 2 comma space 3 close parentheses. karena a with rightwards arrow on top comma space b with rightwards arrow on top space dan space c with rightwards arrow on top berturut-turut vektor posisi titik A, B, dan C, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative 3 end cell end table close parentheses semicolon space b with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row cell negative 5 end cell end table close parentheses semicolon space c with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank blank blank row blank rightwards arrow cell 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top plus c with rightwards arrow on top equals 2 open parentheses table row 4 row cell negative 3 end cell end table close parentheses minus 3 open parentheses table row cell negative 1 end cell row cell negative 5 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell 2 cross times 4 end cell row cell 2 cross times open parentheses negative 3 close parentheses end cell end table close parentheses minus open parentheses table row cell 3 cross times open parentheses negative 1 close parentheses end cell row cell 3 cross times open parentheses negative 5 close parentheses end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 8 row cell negative 6 end cell end table close parentheses minus open parentheses table row cell negative 3 end cell row cell negative 15 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 9 row 12 end table close parentheses end cell end table    

Jadi, hasil table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell open parentheses table row 9 row 12 end table close parentheses end cell end table.

0

Roboguru

Jika  dan , hitunglah nilai dari: a.

Pembahasan Soal:

Dengan menerapkan operasi penjumlahan dua vektor dan perkalian skalar dengan vektor, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top plus 4 b with rightwards arrow on top end cell equals cell open parentheses table row 12 row 3 end table close parentheses plus 4 open parentheses table row cell negative 3 end cell row 9 end table close parentheses end cell row blank equals cell open parentheses table row 12 row 3 end table close parentheses plus open parentheses table row cell negative 12 end cell row 36 end table close parentheses end cell row blank equals cell open parentheses table row 0 row 39 end table close parentheses end cell end table end style 

Dengan menerapkan rumus panjang vektor pada dua dimensi, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top plus 4 b with rightwards arrow on top close vertical bar end cell equals cell square root of 0 squared plus 39 squared end root end cell row blank equals cell square root of 0 plus 1.521 end root end cell row blank equals cell square root of 1.521 end root end cell row blank equals 39 end table end style 

Dengan demikian, panjang vektor begin mathsize 14px style open vertical bar a with rightwards arrow on top plus stack 4 b with rightwards arrow on top close vertical bar end style adalah 39 satuan.

0

Roboguru

Diketahui vektor , , dan . Tentukan

Pembahasan Soal:

Penjumlahan vektor dapat dilakukan secara aljabar sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 over 3 a with rightwards arrow on top plus 1 half b with rightwards arrow on top end cell equals cell 2 over 3 open parentheses table row 2 row 3 end table close parentheses plus 1 half open parentheses table row 1 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 4 over 3 end cell row 2 end table close parentheses plus open parentheses table row cell 1 half end cell row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator 8 plus 3 over denominator 6 end fraction end cell row 0 end table close parentheses end cell row blank equals cell open parentheses table row cell 11 over 6 end cell row 0 end table close parentheses end cell row blank equals cell 11 over 6 i with hat on top end cell end table end style 

Dengan demikian, begin mathsize 14px style 2 over 3 a with rightwards arrow on top plus 1 half b with rightwards arrow on top end style adalah 611i.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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