Roboguru

Diketahui vektor . Tentukan lima vektor  yang memenuhi .

Pertanyaan

Diketahui vektor p with rightwards arrow on top equals open parentheses table row 5 row 2 end table close parentheses. Tentukan lima vektor q with rightwards arrow on top yang memenuhi p with rightwards arrow on top times q with rightwards arrow on top equals negative 8.

Pembahasan Soal:

Diketahui vektor p with rightwards arrow on top equals open parentheses table row 5 row 2 end table close parentheses. Misalkan vektor q with rightwards arrow on top equals open parentheses table row a row b end table close parentheses. Akan ditentukan lima vektor q with rightwards arrow on top yang memenuhi p with rightwards arrow on top times q with rightwards arrow on top equals negative 8, yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top times q with rightwards arrow on top end cell equals cell negative 8 end cell row cell open parentheses table row 5 row 2 end table close parentheses times open parentheses table row a row b end table close parentheses end cell equals cell negative 8 end cell row cell 5 a plus 2 b end cell equals cell negative 8 end cell row a equals cell fraction numerator negative 8 minus 2 b over denominator 5 end fraction end cell end table

Jadi, vektor q with rightwards arrow on top equals open parentheses table row cell table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator negative 8 minus 2 b over denominator 5 end fraction end cell end table end cell row b end table close parentheses.

Untuk b equals 0, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell q with rightwards arrow on top end cell equals cell open parentheses table row cell fraction numerator negative 8 minus 2 open parentheses 0 close parentheses over denominator 5 end fraction end cell row 0 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 8 over 5 end cell row 0 end table close parentheses end cell end table

Untuk b equals 1, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell q with rightwards arrow on top end cell equals cell open parentheses table row cell fraction numerator negative 8 minus 2 open parentheses 1 close parentheses over denominator 5 end fraction end cell row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator negative 8 minus 2 over denominator 5 end fraction end cell row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator negative 10 over denominator 5 end fraction end cell row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 1 end table close parentheses end cell end table

Untuk b equals 2, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell q with rightwards arrow on top end cell equals cell open parentheses table row cell fraction numerator negative 8 minus 2 open parentheses 2 close parentheses over denominator 5 end fraction end cell row 2 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 12 over 5 end cell row 2 end table close parentheses end cell end table

Untuk b equals 3, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell q with rightwards arrow on top end cell equals cell open parentheses table row cell fraction numerator negative 8 minus 2 open parentheses 3 close parentheses over denominator 5 end fraction end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 14 over 5 end cell row 3 end table close parentheses end cell end table

Untuk b equals 4, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell q with rightwards arrow on top end cell equals cell open parentheses table row cell fraction numerator negative 8 minus 2 open parentheses 4 close parentheses over denominator 5 end fraction end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 16 over 5 end cell row 4 end table close parentheses end cell end table

Jadi,  lima vektor q with rightwards arrow on top yang memenuhi p with rightwards arrow on top times q with rightwards arrow on top equals negative 8 adalah

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell negative 8 over 5 end cell row 0 end table close parentheses end cell end table comma table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell negative 2 end cell row 1 end table close parentheses end cell end table comma table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell negative 12 over 5 end cell row 2 end table close parentheses end cell end table comma table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell negative 14 over 5 end cell row 3 end table close parentheses comma open parentheses table row cell negative 16 over 5 end cell row 4 end table close parentheses end cell end table

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Umi

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 05 Juni 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 koordinat titik , , dan . Hasil perkalian  adalah ....

Pembahasan Soal:

Gunakan konsep dot produk pada vektor.

Diketahui koordinat titik straight P open parentheses 3 comma space minus 1 comma space 4 close parenthesesstraight Q open parentheses negative 2 comma space 1 comma space minus 5 close parentheses, dan straight R open parentheses 4 comma space 3 comma space minus 2 close parentheses. Akan ditentukan hasil perkalian PQ with rightwards arrow on top times PR with rightwards arrow on top.

*Menentukan PQ with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell PQ with rightwards arrow on top end cell equals cell straight Q minus straight P end cell row blank equals cell open parentheses table row cell negative 2 end cell row 1 row cell negative 5 end cell end table close parentheses minus open parentheses table row 3 row cell negative 1 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 minus 3 end cell row cell 1 minus open parentheses negative 1 close parentheses end cell row cell negative 5 minus 4 end cell end table close parentheses end cell row cell PQ with rightwards arrow on top end cell equals cell open parentheses table row cell negative 5 end cell row 2 row cell negative 9 end cell end table close parentheses end cell end table

*Menentukan PR with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell PR with rightwards arrow on top end cell equals cell straight R minus straight P end cell row blank equals cell open parentheses table row 4 row 3 row cell negative 2 end cell end table close parentheses minus open parentheses table row 3 row cell negative 1 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell 4 minus 3 end cell row cell 3 minus open parentheses negative 1 close parentheses end cell row cell negative 2 minus 4 end cell end table close parentheses end cell row cell PR with rightwards arrow on top end cell equals cell open parentheses table row 1 row 4 row cell negative 6 end cell end table close parentheses end cell end table

Sehingga nilai PQ with rightwards arrow on top times PR with rightwards arrow on top dapat dihitung sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell PQ with rightwards arrow on top times PR with rightwards arrow on top end cell equals cell open parentheses table row cell negative 5 end cell row 2 row cell negative 9 end cell end table close parentheses times open parentheses table row 1 row 4 row cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses negative 5 close parentheses open parentheses 1 close parentheses plus open parentheses 2 close parentheses open parentheses 4 close parentheses plus open parentheses negative 9 close parentheses open parentheses negative 6 close parentheses end cell row blank equals cell negative 5 plus 8 plus 54 end cell row cell PQ with rightwards arrow on top times PR with rightwards arrow on top end cell equals 57 end table

Diperoleh hasil perkalian PQ with rightwards arrow on top times PR with rightwards arrow on top adalah 57.

Jadi, tidak ada jawaban yang tepat.

0

Roboguru

Diketahui vektor , , dan . Tentukan hasil dari: c.

Pembahasan Soal:

Dengan menerapkan konsep perkalian vektor dot product, diperoleh perhitungan sebagai berikut.

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

Selanjutnya, dengan menerapkan perkalian skalar dengan vektor diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top open parentheses q with rightwards arrow on top times r with rightwards arrow on top close parentheses end cell equals cell p with rightwards arrow on top times open parentheses 1 close parentheses end cell row blank equals cell p with rightwards arrow on top end cell row blank equals cell negative 4 i with rightwards arrow on top minus 7 j with rightwards arrow on top end cell end table end style 

Dengan demikian, hasil dari begin mathsize 14px style p with rightwards arrow on top open parentheses q with rightwards arrow on top times r with rightwards arrow on top close parentheses end style adalah begin mathsize 14px style negative 4 i with rightwards arrow on top minus 7 j with rightwards arrow on top end style.

0

Roboguru

Diketahui    Tentukan

Pembahasan Soal:

Perkalian titik (dot product) pada vektor dapat diselesaikan sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell straight p times straight q end cell equals cell open vertical bar straight p close vertical bar open vertical bar straight q close vertical bar cos angle 135 degree end cell row blank equals cell 6 times 4 times open parentheses cos open parentheses 180 degree minus 45 degree close parentheses close parentheses end cell row blank equals cell 24 times left parenthesis negative cos space 45 degree right parenthesis end cell row blank equals cell 24 times open parentheses negative 1 half square root of 2 close parentheses end cell row blank equals cell negative 12 square root of 2 end cell end table 

Jadi, hasil perkalian titik kedua vektor tersebut adalah negative 12 square root of 2.

0

Roboguru

Jika  dan , sudut antara vektor  adalah . Hasil kali skalar antara vektor   adalah ....

Pembahasan Soal:

Hasil kali skalar dua vektor dapat ditentukan dengan rumus 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 times open vertical bar b with rightwards arrow on top close vertical bar times cos space alpha 

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 5 row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals 12 row alpha equals cell 60 degree end cell end table 

Sehingga diperoleh hasil kali skalar dari vektor a with rightwards arrow on top space dan space b with rightwards arrow on top sebagai 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 vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar times cos space alpha end cell row blank equals cell 5 times 12 times cos space 60 degree end cell row blank equals cell 60 times 1 half end cell row blank equals 30 end table 

Jadi, jawaban yang benar adalah D.

0

Roboguru

Jika  dan sudut antara  adalah , tentukanlah !

Pembahasan Soal:

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar u with rightwards arrow on top close vertical bar end cell equals 6 row cell open vertical bar v with rightwards arrow on top close vertical bar end cell equals 4 row theta equals cell 60 degree end cell end table 

maka dapat ditentukan:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses u with rightwards arrow on top plus v with rightwards arrow on top close parentheses times v with rightwards arrow on top end cell equals cell u with rightwards arrow on top times v with rightwards arrow on top plus v with rightwards arrow on top times v with rightwards arrow on top end cell row blank equals cell open vertical bar u with rightwards arrow on top close vertical bar open vertical bar v with rightwards arrow on top close vertical bar cos space theta plus open vertical bar v with rightwards arrow on top close vertical bar squared end cell row blank equals cell 6 times 4 times cos space 60 degree plus 4 squared end cell row blank equals cell 24 times 1 half plus 16 end cell row blank equals cell 12 plus 16 end cell row blank equals 28 end table 

Dengan demikian, diperoleh table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses u with rightwards arrow on top plus v with rightwards arrow on top close parentheses 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 v 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 28 end table.

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