Roboguru

Panjang vektor dan vektor masing-masing 12 satuan dan 18 satuan. Besar sudut antara vektor dan vektor adalah , maka hasil kali skalar antara vektor dan vektor adalah...

Pertanyaan

Panjang vektor begin mathsize 14px style u with rightwards arrow on top end style dan vektor begin mathsize 14px style v with rightwards arrow on top end style masing-masing 12 satuan dan 18 satuan. Besar sudut antara vektor begin mathsize 14px style u with rightwards arrow on top end style dan vektor begin mathsize 14px style v with rightwards arrow on top end style adalah begin mathsize 14px style 60 degree end style, maka hasil kali skalar antara vektor begin mathsize 14px style u with rightwards arrow on top end style dan vektor begin mathsize 14px style v with rightwards arrow on top end style adalah...

Pembahasan Soal:

Diketahui:

begin mathsize 14px style open vertical bar u with rightwards arrow on top close vertical bar equals 12 end style

begin mathsize 14px style open vertical bar v with rightwards arrow on top close vertical bar equals 18 end style

begin mathsize 14px style theta equals 60 degree end style 

Ditanya:

hasil kali skalar vektor begin mathsize 14px style u with rightwards arrow on top end style dan vektor begin mathsize 14px style v with rightwards arrow on top end style adalah....

Penyelesaian:

Untuk menentukan hasil kali skalar vektor begin mathsize 14px style u with rightwards arrow on top end style dan vektor begin mathsize 14px style v with rightwards arrow on top end style dapat menggunakan rumus:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top times v with rightwards arrow on top end cell equals cell open vertical bar u with rightwards arrow on top close vertical bar times open vertical bar u with rightwards arrow on top close vertical bar space cos space theta end cell row blank equals cell 12 cross times 18 cross times cos space 60 degree end cell row blank equals cell 216 cross times 1 half end cell row blank equals 108 end table end style

Jadi, hasil kali skalar vektor begin mathsize 14px style u with rightwards arrow on top end style dan vektor begin mathsize 14px style v with rightwards arrow on top end style adalah 108

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Nasrullah

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 31 Maret 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Diketahui . Jika  adalah sudut antara  dan . Nilai dari

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards harpoon with barb upwards on top end cell equals cell open parentheses table row 6 row cell negative 3 end cell row cell negative 2 end cell end table close parentheses comma space q with rightwards harpoon with barb upwards on top equals open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses end cell row cell p with rightwards harpoon with barb upwards on top times q with rightwards harpoon with barb upwards on top end cell equals cell open parentheses table row 6 row cell negative 3 end cell row cell negative 2 end cell end table close parentheses times open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses 6 cross times 3 close parentheses plus open parentheses negative 3 cross times 2 close parentheses plus open parentheses negative 2 cross times open parentheses negative 1 close parentheses close parentheses end cell row blank equals cell 18 minus 6 plus 2 end cell row blank equals 14 row cell open vertical bar p with rightwards harpoon with barb upwards on top close vertical bar end cell equals cell square root of 6 squared plus open parentheses negative 3 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 36 plus 9 plus 4 end root end cell row blank equals cell square root of 49 end cell row blank equals 7 row cell open vertical bar q with rightwards harpoon with barb upwards on top close vertical bar end cell equals cell square root of 3 squared plus 2 squared plus open parentheses negative 1 close parentheses squared end root end cell row blank equals cell square root of 9 plus 4 plus 1 end root end cell row blank equals cell square root of 14 end cell row cell cos space gamma end cell equals cell fraction numerator p with rightwards harpoon with barb upwards on top times q with rightwards harpoon with barb upwards on top over denominator open vertical bar p with rightwards harpoon with barb upwards on top close vertical bar times open vertical bar q with rightwards harpoon with barb upwards on top close vertical bar end fraction end cell row blank equals cell fraction numerator 14 over denominator 7 square root of 14 end fraction end cell row blank equals cell fraction numerator 14 over denominator 7 square root of 14 end fraction cross times fraction numerator square root of 14 over denominator square root of 14 end fraction end cell row blank equals cell fraction numerator 14 over denominator 7 cross times 14 end fraction square root of 14 end cell row blank equals cell 1 over 7 square root of 14 end cell row gamma equals cell a r c cos open parentheses 1 over 7 square root of 14 close parentheses end cell row blank equals cell 57 comma 6884668 degree end cell row cell tan space gamma end cell equals cell tan space open parentheses 57 comma 6884668 degree close parentheses end cell row blank equals cell 1 comma 58113883 end cell row blank equals cell 1 half square root of 10 end cell end table 

Jadi, nilai tan space gamma equals 1 half square root of 10.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Diketahui , , dan  sudut antara vektor  dan . Nilai  adalah ...

Pembahasan Soal:

Berdasarkan konsep sudut antara dua vektor maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator p with bar on top times q with bar on top over denominator open vertical bar p with bar on top close vertical bar times open vertical bar q with bar on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 1 row 2 end table close parentheses times open parentheses table row 4 row 2 end table close parentheses over denominator square root of 1 squared plus 2 squared end root times square root of 4 squared plus 2 squared end root end fraction end cell row blank equals cell fraction numerator 4 plus 4 over denominator square root of 5 times square root of 20 end fraction end cell row blank equals cell fraction numerator 8 over denominator square root of 100 end fraction end cell row blank equals cell 8 over 10 end cell row blank equals cell 4 over 5 end cell end table end style 

Oleh karena definisi cos adalah 

begin mathsize 14px style cos space straight theta equals samping over miring equals 4 over 5 end style  

maka berdasarkan teorema Pythagoras diperoleh sisi depan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space depan end cell equals cell square root of 5 squared minus 4 squared end root end cell row blank equals cell square root of 25 minus 16 end root end cell row blank equals cell square root of 9 end cell row blank equals 3 end table end style 

sehingga diperoleh nilai:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight theta end cell equals cell depan over miring end cell row blank equals cell 3 over 5 end cell end table end style 

Oleh karena itu, jawaban yang benar adalah A.

1

Roboguru

Nilai kosinus sudut antara vektor  dan  adalah... .

Pembahasan Soal:

Diketahui vektor u with rightwards arrow on top equals open parentheses table row 3 row cell negative 6 end cell end table close parentheses dan v with rightwards arrow on top equals open parentheses table row 4 row cell negative 3 end cell end table close parentheses. Misal theta adalah sudut diantara dua vektor tersebut maka nilai kosinus sudut tersebut sebagai berikut.

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 open vertical bar u with rightwards arrow on top close vertical bar open vertical bar v with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 3 row cell negative 6 end cell end table close parentheses times open parentheses table row 4 row cell negative 3 end cell end table close parentheses over denominator open square brackets square root of 3 squared plus open parentheses negative 6 close parentheses squared end root close square brackets space open square brackets square root of 4 squared plus open parentheses negative 3 close parentheses squared end root close square brackets end fraction end cell row blank equals cell fraction numerator 12 plus 18 over denominator open parentheses square root of 9 plus 36 end root close parentheses open parentheses square root of 16 plus 9 end root close parentheses end fraction end cell row blank equals cell fraction numerator 30 over denominator open parentheses square root of 45 close parentheses open parentheses square root of 25 close parentheses end fraction end cell row blank equals cell fraction numerator 30 over denominator open parentheses square root of 9 times 5 end root close parentheses times 5 end fraction end cell row blank equals cell fraction numerator 30 over denominator 3 times 5 square root of 5 end fraction end cell row blank equals cell fraction numerator 30 over denominator 15 square root of 5 end fraction times fraction numerator square root of 5 over denominator square root of 5 end fraction end cell row blank equals cell fraction numerator 2 square root of 5 over denominator 5 end fraction end cell row blank equals cell 2 over 5 square root of 5 end cell end table 
 

Jadi, nilai kosinus sudut antara vektor u with rightwards arrow on top equals open parentheses table row 3 row cell negative 6 end cell end table close parentheses dan v with rightwards arrow on top equals open parentheses table row 4 row cell negative 3 end cell end table close parentheses adalah 2 over 5 square root of 5.

 

0

Roboguru

Diketahui vector  dan vector  membentuk sudut . Hitunglah !

Pembahasan Soal:

Jika terdapat vektor a with rightwards arrow on top equals open parentheses a subscript 1 comma space a subscript 2 comma space a subscript 3 close parentheses dan vektor b with rightwards arrow on top equals open parentheses b subscript 1 comma space b subscript 2 comma space b subscript 3 close parentheses di mana kedua vektor membentuk sudut sebesar theta, maka berlaku rumus perkalian dot sebagai berikut.

a with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space theta

Perkalian dot secara aljabar, yaitu

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

Pada soal di atas, panjang vektor a with rightwards arrow on top dan vektor b with rightwards arrow on top ditentukan sebagai berikut.

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 x squared plus y squared plus z squared end root end cell row blank equals cell square root of open parentheses negative 2 close parentheses squared plus 2 squared plus 1 squared end root end cell row blank equals cell square root of 9 end cell row blank equals 3 end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of x squared plus y squared plus z squared end root end cell row blank equals cell square root of 4 squared plus open parentheses negative 2 close parentheses squared plus 4 squared end root end cell row blank equals cell square root of 36 end cell row blank equals 6 end table

Perkalian dot secara aljabar, yaitu

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 square brackets table row cell negative 2 end cell row 2 row 1 end table close square brackets open square brackets table row 4 row cell negative 2 end cell row 4 end table close square brackets end cell row blank equals cell open parentheses negative 2 close parentheses times 4 plus 2 open parentheses negative 2 close parentheses plus 1 times 4 end cell row blank equals cell negative 8 minus 4 plus 4 end cell row blank equals cell negative 8 end cell end table

Nilai cos space beta dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space beta end cell equals cell 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 cell row blank equals cell fraction numerator negative 8 over denominator 3 times 6 end fraction end cell row blank equals cell fraction numerator negative 8 over denominator 18 end fraction end cell row blank equals cell negative 4 over 9 end cell end table

Dengan demikian, cos space beta equals negative 4 over 9 

0

Roboguru

Diketahui titik-titik  dan .  wakil dari u dan  wakil dari  v. Kosinus sudut yang dibentuk oleh vektor u dan v adalah

Pembahasan Soal:

  • Menentukan vektor u dan v:

uequals stack A B with rightwards arrow on top equals B minus A equals open parentheses 5 comma 1 comma 5 close parentheses minus open parentheses 3 comma 2 comma 4 close parentheses equals open parentheses 2 comma negative 1 comma 1 close parentheses

v equals stack A C with rightwards arrow on top equals C minus A equals open parentheses 4 comma 3 comma 6 close parentheses minus open parentheses 3 comma 2 comma 4 close parentheses equals open parentheses 1 comma 1 comma 2 close parentheses

  • Menentukan sudut kosinus yang dibentuk u dan v:

cos space alpha equals fraction numerator u. v over denominator open vertical bar u close vertical bar open vertical bar v close vertical bar end fraction cos space alpha equals fraction numerator open parentheses 2 close parentheses open parentheses 1 close parentheses plus open parentheses negative 1 close parentheses open parentheses 1 close parentheses plus open parentheses 1 close parentheses open parentheses 2 close parentheses over denominator square root of 2 squared plus open parentheses negative 1 close parentheses squared plus 1 squared end root square root of 1 squared plus 1 squared plus 2 squared end root end fraction cos space alpha equals fraction numerator 3 over denominator square root of 6 square root of 6 end fraction cos space alpha equals 3 over 6 cos space alpha equals 1 half 

Jadi, kosinus sudut yang dibentuk adalah 1 half.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved