Roboguru

Diketahui  dan . Tentukan: c.

Pertanyaan

Diketahui a with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row 1 row 2 end table close parentheses dan b with rightwards arrow on top equals open parentheses table row 7 row 0 row cell negative 5 end cell end table close parentheses. Tentukan:
c. a with rightwards arrow on top times b with rightwards arrow on top 

Pembahasan Soal:

Operasi perkalian vektor atau dot product dapat dihitung seperti berikut,

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open parentheses table row cell negative 3 end cell row 1 row 2 end table close parentheses times open parentheses table row 7 row 0 row cell negative 5 end cell end table close parentheses end cell row blank equals cell open parentheses negative 3 close parentheses left parenthesis 7 right parenthesis plus left parenthesis 1 right parenthesis left parenthesis 0 right parenthesis plus left parenthesis 2 right parenthesis left parenthesis negative 5 right parenthesis end cell row blank equals cell negative 21 plus 0 minus 10 end cell row blank equals cell negative 31 end cell end table 

Jadi, hasil dari perkalian vektor di atas adalah negative 31 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Kumaralalita

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui ,  dan . Besar sudut antara  dan  adalah ....

Pembahasan Soal:

Ingat rumus untuk menentukan besar sudut antara dua vektor adalah sebagai berikut.

begin mathsize 14px style cos space alpha equals fraction numerator a with rightwards arrow on top times 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 style  

Berdasarkan rumus di atas, untuk menentukan besar sudut antara begin mathsize 14px style a with rightwards arrow on top end style dan begin mathsize 14px style b with rightwards arrow on top plus c with rightwards arrow on top end style dibutuhkan panjang dari vektor undefined, panjang dari vektor undefined dan hasil perkalian skalar vektor undefined dan vektor undefined.

Diketahui vektor begin mathsize 14px style a with rightwards arrow on top equals 6 i with rightwards arrow on top minus 3 j with rightwards arrow on top minus 3 k with rightwards arrow on top end style, maka panjang vektor undefined adalah.

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 close vertical bar end cell equals cell square root of open parentheses x subscript a close parentheses squared plus open parentheses y subscript a close parentheses squared plus open parentheses z subscript a close parentheses squared end root end cell row blank equals cell square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses squared end root end cell row blank equals cell square root of 36 plus 9 plus 9 end root end cell row blank equals cell square root of 54 end cell row blank equals cell 3 square root of 6 end cell end table end style 

Diketahui vektor begin mathsize 14px style b with rightwards arrow on top equals 2 i with rightwards arrow on top minus j with rightwards arrow on top plus 3 k with rightwards arrow on top end style dan vektor begin mathsize 14px style c with rightwards arrow on top equals negative 5 i with rightwards arrow on top minus 2 j with rightwards arrow on top plus 3 k with rightwards arrow on top end style, maka vektor undefined adalah.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell b with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell open parentheses table row 2 row cell negative 1 end cell row 3 end table close parentheses plus open parentheses table row cell negative 5 end cell row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell 2 plus open parentheses negative 5 close parentheses end cell row cell negative 1 plus open parentheses negative 2 close parentheses end cell row cell 3 plus 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 6 end table close parentheses end cell end table end style 

Panjang vektor undefined adalah.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top plus c with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses squared plus 6 squared end root end cell row blank equals cell square root of 9 plus 9 plus 36 end root end cell row blank equals cell square root of 54 end cell row blank equals cell 3 square root of 6 end cell end table end style 

Hasil perkalian skalar undefined dan vektor undefined adalah.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times open parentheses b with rightwards arrow on top plus c with rightwards arrow on top close parentheses end cell equals cell open parentheses table row 6 row cell negative 3 end cell row cell negative 3 end cell end table close parentheses times open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 6 end table close parentheses end cell row blank equals cell negative 18 plus 9 plus open parentheses negative 18 close parentheses end cell row blank equals cell negative 27 end cell end table end style 

Sehingga sudut antara vektor undefined dan vektor begin mathsize 14px style b with rightwards arrow on top plus c with rightwards arrow on top end style adalah. 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator vertical line a with rightwards arrow on top vertical line vertical line b with rightwards arrow on top vertical line end fraction end cell row blank equals cell fraction numerator negative 27 over denominator open parentheses 3 square root of 6 close parentheses open parentheses 3 square root of 6 close parentheses end fraction end cell row blank equals cell fraction numerator negative 27 over denominator 9 cross times 6 end fraction end cell row blank equals cell fraction numerator negative 27 over denominator 54 end fraction end cell row blank equals cell negative 1 half end cell row alpha equals cell cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses end cell row alpha equals cell 120 degree space text dan end text space 240 degree end cell end table end style 

Oleh karena itu, tidak ada jawaban yang benar.

0

Roboguru

Diberikan vektor-vektor ; dan . Nilai dari  sama dengan ...

Pembahasan Soal:

Ingatlah rumus dot product antara dua vektor straight a with rightwards arrow on top equals straight a subscript 1 straight i with rightwards arrow on top plus straight a subscript 2 straight j with rightwards arrow on top plus straight a subscript 3 straight k with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 end cell row cell straight a subscript 2 end cell row cell straight a subscript 3 end cell end table close parentheses dan straight b with rightwards arrow on top equals straight b subscript 1 straight i with rightwards arrow on top plus straight b subscript 2 straight j with rightwards arrow on top plus straight b subscript 3 straight k with rightwards arrow on top equals open parentheses table row cell straight b subscript 1 end cell row cell straight b subscript 2 end cell row cell straight b subscript 3 end cell end table close parentheses yaitu:

straight a with rightwards arrow on top times straight b with rightwards arrow on top equals straight a subscript 1 straight b subscript 1 plus straight a subscript 2 straight b subscript 2 plus straight a subscript 3 straight b subscript 3

Pada soal, diketahui straight a with rightwards arrow on top equals 3 straight i with rightwards arrow on top minus 2 straight j with rightwards arrow on top minus straight k with rightwards arrow on top; straight b with rightwards arrow on top equals straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top minus 3 straight k with rightwards arrow on top dan straight c with rightwards arrow on top equals 2 straight i with rightwards arrow on top plus straight j with rightwards arrow on top minus 2 straight k with rightwards arrow on top, maka:

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

Dengan demikian, hasil dari straight a with rightwards arrow on top times open parentheses straight b with rightwards arrow on top plus straight c with rightwards arrow on top close parentheses adalah 6.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Jika  dan , maka besar sudut yang dibentuk vektor  dan  sama dengan...

Pembahasan Soal:

Ingat kembali beberapa aturan berikut.

Diketahui vektor top enclose a equals left parenthesis a subscript 1 comma space a subscript 2 comma space a subscript 3 right parenthesis space dan space top enclose b equals left parenthesis b subscript 1 comma space b subscript 2 comma space b subscript 3 right parenthesis, maka

  • a bullet b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3 

 

  • vertical line top enclose a vertical line equals square root of a subscript 1 squared plus a subscript 2 squared plus a subscript 3 squared end root 

 

  • Misalkan alpha adalah sudut yang dibentuk oleh vektor top enclose a dan top enclose b, maka

 cos space alpha equals fraction numerator a bullet b over denominator vertical line top enclose a vertical line vertical line top enclose b vertical line end fraction 

  • Merasionalkan bentuk akar

fraction numerator square root of a over denominator square root of b end fraction times fraction numerator square root of b over denominator square root of b end fraction 

 

  • table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 135 degree end cell equals cell cos space left parenthesis 180 degree minus 45 degree right parenthesis end cell row blank equals cell negative cos space 45 degree end cell row blank equals cell negative 1 half square root of 2 end cell end table 

 

Dari rumus di atas, diperoleh perhitungan sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator a bullet b over denominator vertical line top enclose a vertical line vertical line top enclose b vertical line end fraction end cell row blank equals cell fraction numerator left parenthesis negative 1 right parenthesis cross times 1 plus 1 cross times left parenthesis negative 2 right parenthesis plus 0 cross times 2 over denominator square root of left parenthesis negative 1 right parenthesis squared plus 1 squared plus 0 squared space end root space cross times square root of 1 squared plus left parenthesis negative 2 right parenthesis squared plus 2 squared end root end fraction end cell row blank equals cell fraction numerator negative 1 minus 2 plus 0 over denominator square root of 1 plus 1 plus 0 end root cross times square root of 1 plus 4 plus 4 end root end fraction end cell row blank equals cell fraction numerator negative 3 over denominator square root of 2 cross times square root of 9 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator square root of 2 cross times 3 end fraction end cell row blank equals cell fraction numerator negative 1 over denominator square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator negative 1 over denominator 2 end fraction square root of 2 end cell end table 

Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row alpha equals cell cos to the power of negative 1 end exponent left parenthesis negative 1 half square root of 2 right parenthesis end cell row blank equals cell 135 degree end cell end table

Jadi besar sudut yang dibentuk vektor top enclose a dan top enclose b adalah 135 degree.

Oleh karena itu, jawaban yang benar adalah E.

 

0

Roboguru

Misal vektor  dan  membentuk sudut sebesar , maka

Pembahasan Soal:

Untuk mengetahui sudut antar vektor dapat dilakukan sebagai berikut :

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with bar on top times straight b with bar on top end cell equals cell open vertical bar straight a with bar on top close vertical bar open vertical bar straight b with bar on top close vertical bar cos space theta end cell row blank equals cell open vertical bar straight a with bar on top close vertical bar open vertical bar straight b with bar on top close vertical bar cos space 60 degree end cell row blank equals cell open vertical bar straight a with bar on top close vertical bar open vertical bar straight b with bar on top close vertical bar cross times 1 half end cell row blank equals cell 1 half open vertical bar straight a with bar on top close vertical bar open vertical bar straight b with bar on top close vertical bar end cell end table  

Maka, table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with bar on top times straight b with bar on top end cell equals cell 1 half open vertical bar straight a with bar on top close vertical bar open vertical bar straight b with bar on top close vertical bar end cell end table.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Diketahui vektor  dan , dan  maka nilai dari a−b adalah ...

Pembahasan Soal:

Rumus perkalian titik (dot product) antara dua vektor satuan dalam sistem koordinat tiga dimensi (x,y,z) adalah sebagai berikut:

ab=axbx+ayby+azbz 

Sehingga diperoleh:

ab213m212m+2+32m2mm======3333521 

Kita substitusi nilai m pada vektor b, maka nilai ab diperoleh:

ab==213121112 

Dengan demikian, nilai dari begin mathsize 14px style a with rightwards arrow on top minus b with rightwards arrow on top end style adalah (ij2k) .

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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