Roboguru

Vektor  mempunyai panjang . Jika  dan , tentukan .

Pertanyaan

Vektor straight a mempunyai panjang 2 square root of 3. Jika straight a with rightwards arrow on top open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 15 dan angle open parentheses straight a with rightwards arrow on top comma space straight b with rightwards arrow on top close parentheses equals straight pi over 6, tentukan open vertical bar straight b with rightwards arrow on top close vertical bar.

Pembahasan Soal:

Ingat kembali:

cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator stack open vertical bar straight a close vertical bar with rightwards arrow on top times space open vertical bar straight b with rightwards arrow on top close vertical bar end fraction straight a with rightwards arrow on top times straight b with rightwards arrow on top equals cos space straight theta times stack open vertical bar straight a close vertical bar with rightwards arrow on top times space open vertical bar straight b with rightwards arrow on top close vertical bar 

Dari soal dapat diketahui:

open vertical bar straight a with rightwards arrow on top close vertical bar equals 2 square root of 3 straight a with rightwards arrow on top open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 15 angle open parentheses straight a with rightwards arrow on top comma space straight b with rightwards arrow on top close parentheses equals straight pi over 6 rightwards arrow straight theta equals straight pi over 6 equals 30 degree 

Maka:

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight a with rightwards arrow on top open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 15 space space space space space space space space space space space space space space space space space space space space space space space space space space space straight a with rightwards arrow on top times straight a with rightwards arrow on top plus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 15 space space space space space space space space space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus cos space straight theta times stack open vertical bar straight a close vertical bar with rightwards arrow on top times space open vertical bar straight b with rightwards arrow on top close vertical bar equals 15 open parentheses 2 square root of 3 close parentheses squared plus cos space 30 degree times 2 square root of 3 times space open vertical bar straight b with rightwards arrow on top close vertical bar equals 15 space space space space space space space space space space 4 times 3 plus 1 half square root of 3 times 2 square root of 3 times open vertical bar straight b with rightwards arrow on top close vertical bar equals 15 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 12 plus 1 times open vertical bar straight b with rightwards arrow on top close vertical bar equals 15 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 12 plus open vertical bar straight b with rightwards arrow on top close vertical bar equals 15 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar equals 15 minus 12 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar equals 3 

Jadi, open vertical bar straight b with rightwards arrow on top close vertical bar adlaah 3.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

W. Lestari

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 30 April 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

Jika vektor  dan vektor  membentuk sudut , panjang vektor  dan panjang vektor , maka  sama dengan ...

Pembahasan Soal:

Perhatikan gambar berikut.



 

Perkalian titik (dot product) antara a with rightwards arrow on top dan b with rightwards arrow on top dapat dirumuskan 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 

dimana

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 panjang space vektor space a with rightwards arrow on top end cell row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell panjang space vektor space b with rightwards arrow on top end cell end table  

Diketahui:

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 4 row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals 6 row theta equals cell 120 degree end cell end table 

Karena a with rightwards arrow on top times a with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar squared, maka nilai a with rightwards arrow on top times open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses adalah:

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 a with rightwards arrow on top close parentheses end cell equals cell a with rightwards arrow on top times b with rightwards arrow on top plus a with rightwards arrow on top times a with rightwards arrow on top end cell row blank 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 space cos space theta plus open vertical bar a with rightwards arrow on top close vertical bar squared end cell row blank equals cell 4 times 6 times cos space 120 degree plus 4 squared end cell row blank equals cell 24 times cos space left parenthesis 180 degree minus 120 degree right parenthesis plus 16 end cell row blank equals cell 24 times open parentheses negative cos space 60 degree close parentheses plus 16 end cell row blank equals cell 24 times open parentheses negative 1 half close parentheses plus 16 end cell row blank equals cell negative 12 plus 16 end cell row blank equals 4 end table 

Sehingga, 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 blank times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses 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 4 end table.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

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

Pembahasan Soal:

Diketahui open vertical bar straight a with rightwards arrow on top close vertical bar equals square root of 6 comma space open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 0, dan straight a with rightwards arrow on top times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals 3. Maka:

- Menentukan panjang straight b with rightwards arrow on top 

space space space space space space space space space space space space space space open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 0 straight a with rightwards arrow on top times straight a with rightwards arrow on top plus straight a with rightwards arrow on top times straight b with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top minus straight b with rightwards arrow on top times straight b with rightwards arrow on top equals 0 space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space open parentheses square root of 6 close parentheses squared minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space 6 minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals negative 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar equals plus-or-minus square root of 6 ,

- Menentukan straight a with rightwards arrow on top times straight b with rightwards arrow on top 

space space space space space space straight a with rightwards arrow on top times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals 3 space space space space straight a with rightwards arrow on top times straight a with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 open parentheses square root of 6 close parentheses squared minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 space space space space space space space space space space minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 minus 6 space space space space space space space space space space minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals negative 3 space space space space space space space space space space space space space straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

Sehingga besar sudut antara vektor straight a with rightwards arrow on top dan straight b with rightwards arrow on top adalah:

- Untuk open vertical bar straight b with rightwards arrow on top close vertical bar equals square root of 6 dan straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar times open vertical bar straight b with rightwards arrow on top close vertical bar end fraction cos space straight theta equals fraction numerator 3 over denominator square root of 6 times square root of 6 end fraction cos space straight theta equals 3 over 6 cos space straight theta equals 1 half cos space straight theta equals cos space 30 degree space space space space space space space straight theta equals 30 degree space space space space space space space straight theta equals 30 degree cross times fraction numerator straight pi over denominator 180 degree end fraction space space space space space space space space straight theta equals straight pi over 6 

- Untuk open vertical bar straight b with rightwards arrow on top close vertical bar equals negative square root of 6 dan straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar times open vertical bar straight b with rightwards arrow on top close vertical bar end fraction cos space straight theta equals fraction numerator 3 over denominator square root of 6 times negative square root of 6 end fraction cos space straight theta equals fraction numerator 3 over denominator negative 6 end fraction cos space straight theta equals negative 1 half cos space straight theta equals cos space 120 degree space space space space space space space straight theta equals 120 degree space space space space space space space straight theta equals 120 degree cross times fraction numerator straight pi over denominator 180 degree end fraction space space space space space space space space straight theta equals 2 over 3 straight pi 

Jadi, jawaban yang benar adalah A dan E.

0

Roboguru

Diketahui vektor dan . Besar sudut antara  adalah

Pembahasan Soal:

Perkalian titik atau dot product dari begin mathsize 14px style a with rightwards arrow on top space dan space b with rightwards arrow on top end style adalah begin mathsize 14px style 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 end style. Kita cari begin mathsize 14px style a with rightwards arrow on top times b with rightwards arrow on top end style terlebih dahulu.

begin mathsize 14px style 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 2 times 3 close parentheses plus open parentheses negative 3 times negative 2 close parentheses plus open parentheses 3 times negative 4 close parentheses end cell row blank equals cell 6 plus 6 minus 12 end cell row blank equals 0 end table end style

Kemudian kita cari begin mathsize 14px style open vertical bar a with rightwards arrow on top close vertical bar space dan space open vertical bar b with rightwards arrow on top close vertical bar end style.

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 2 squared plus open parentheses negative 3 close parentheses squared plus 3 cubed end root end cell row blank equals cell square root of 4 plus 9 plus 9 end root end cell row blank equals cell square root of 22 end cell row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of 3 squared plus open parentheses negative 2 close parentheses squared plus open parentheses negative 4 close parentheses squared end root end cell row blank equals cell square root of 9 plus 4 plus 16 end root end cell row blank equals cell square root of 29 end cell end table end style

Lalu kita cari sudutnya menggunakan perkalian titik.

begin mathsize 14px style 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 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 end cell row 0 equals cell square root of 22 times square root of 29 times cos space theta end cell row cell cos space theta end cell equals cell fraction numerator 0 over denominator square root of 22 times square root of 29 end fraction end cell row blank equals 0 row theta equals cell 90 degree end cell end table end style

Jadi, jawaban yang tepat adalah C.

1

Roboguru

Jika , dan, maka tentukanlah nilai a yang memenuhi ..

Pembahasan Soal:

Diketahui :

Jika u with rightwards arrow on top equals open parentheses table row 1 row cell square root of 2 end cell row a end table close parentheses semicolon space v with rightwards arrow on top equals open parentheses table row 1 row cell negative square root of 2 end cell row a end table close parentheses,

danangle open parentheses u with rightwards arrow on top comma space v with rightwards arrow on top close parentheses equals straight pi over 3,

Ditanya :

maka tentukanlah nilai a yang memenuhi ..

Penyelesaian :

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator u with rightwards arrow on top. space top enclose v over denominator open vertical bar u close vertical bar open vertical bar v close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 1 row cell square root of 2 end cell row a end table close parentheses open parentheses table row 1 row cell negative square root of 2 end cell row a end table close parentheses over denominator square root of 1 squared plus open parentheses square root of 2 close parentheses squared plus a squared end root square root of 1 squared plus open parentheses negative square root of 2 close parentheses squared plus a squared end root end fraction end cell row blank equals cell fraction numerator 1 minus 2 plus a squared over denominator 3 plus a squared end fraction end cell row blank equals cell fraction numerator negative 1 plus a squared over denominator 3 plus a squared end fraction end cell row cell cos space 180 over 30 end cell equals cell cos space 60 end cell row blank equals cell 1 half end cell row cell 1 half end cell equals cell fraction numerator negative 1 plus a squared over denominator 3 plus a squared end fraction end cell row cell 3 plus a squared end cell equals cell negative 2 plus 2 a squared end cell row 5 equals cell a squared end cell row cell plus-or-minus square root of 5 end cell equals a end table

Jadi, nilai a adalah plus-or-minus square root of 5.

0

Roboguru

Pembahasan Soal:

a space ⃗ equals open parentheses table row 2 row 1 row 3 end table close parentheses comma b space ⃗ equals open parentheses table row cell negative 1 end cell row 2 row 2 end table close parentheses  S u d u t space y a n g space d i b e n t u k space k e d u a space v e k t o r space t e r s e b u t space m e m e n u h i  cos invisible function application theta equals fraction numerator a space ⃗ space space. b space ⃗ over denominator vertical line a space ⃗ space vertical line space space vertical line b space ⃗ space vertical line end fraction  cos invisible function application theta equals fraction numerator 2 left parenthesis negative 1 right parenthesis plus 1 left parenthesis 2 right parenthesis plus 3 left parenthesis 2 right parenthesis over denominator square root of 2 squared plus 1 squared plus 3 squared end root. square root of left parenthesis negative 1 right parenthesis squared plus 2 squared plus 2 squared end root end fraction  cos invisible function application theta equals fraction numerator 6 over denominator 3 square root of 14 end fraction  cos invisible function application theta equals fraction numerator 2 over denominator square root of 14 end fraction  G a m b a r space s e g i t i g a space s i k u minus s i k u space b a n t u a n comma

sin invisible function application theta equals fraction numerator square root of left parenthesis square root of 14 right parenthesis squared minus 2 squared right parenthesis end root over denominator square root of 14 end fraction  sin invisible function application theta equals fraction numerator square root of 10 over denominator square root of 14 end fraction  sin invisible function application theta equals 1 over 7 square root of 35

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