Diketahui vektor a=12i−8j​ dan vektor b=9i−15j​. Tentukan: e. 21​a+31​b

Pertanyaan

Diketahui vektor a with rightwards arrow on top equals 12 i with hat on top minus 8 j with hat on top dan vektor b with rightwards arrow on top equals 9 i with hat on top minus 15 j with hat on top. Tentukan:
e. 1 half a with rightwards arrow on top plus 1 third b with rightwards arrow on top

T. Prita

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jember

Jawaban terverifikasi

Jawaban

hasil perhitungan table attributes columnalign right center left columnspacing 0px end attributes row cell 1 half a with rightwards arrow on top plus 1 third b with rightwards arrow on top end cell equals cell 9 i with hat on top minus 9 j with rightwards arrow on top end cell end table.

Pembahasan

Jawaban yang benar dari pertanyaan tersebut adalah  table attributes columnalign right center left columnspacing 0px end attributes row cell bold 1 over bold 2 bold a with bold rightwards arrow on top bold plus bold 1 over bold 3 bold b with bold rightwards arrow on top end cell bold equals cell bold 9 bold i with bold hat on top bold minus bold 9 bold j with bold rightwards arrow on top end cell end table.

Penulisan vektor dalam bentuk r with rightwards arrow on top equals x i with hat on top plus y j with hat on top dapat dinyatakan dalam bentuk vektor baris sebagai r with rightwards arrow on top equals open parentheses table row x y end table close parentheses atau vektor kolom sebagai r with rightwards arrow on top equals open parentheses table row x row y end table close parentheses.

Ingat teori mengenai penjumlahan vektor. Penjumlahan vektor secara aljabar dapat dilakukan dengan cara menjumlahkan komponen yang seletak. Misalkan diketahui vektor a with rightwards arrow on top equals open parentheses table row cell x subscript a end cell row cell y subscript a end cell end table close parentheses dan vektor b with rightwards arrow on top equals open parentheses table row cell x subscript b end cell row cell y subscript b end cell end table close parentheses . Jika c with rightwards arrow on top equals a with rightwards arrow on top plus b with rightwards arrow on top, maka vektor c with rightwards arrow on top ditentukan oleh:

c with rightwards arrow on top equals open parentheses table row cell x subscript a end cell row cell y subscript a end cell end table close parentheses plus open parentheses table row cell x subscript b end cell row cell y subscript b end cell end table close parentheses equals open parentheses table row cell x subscript a plus x subscript b end cell row cell y subscript a plus y subscript b end cell end table close parentheses

Selanjutnya hasil skalar dengan vektor. Misalkan m adalah suatu skalar dan a with rightwards arrow on top adalah vektor dengan a with rightwards arrow on top equals open parentheses table row cell x subscript a end cell row cell y subscript a end cell end table close parentheses. Hasil kali skalar m dengan vektor a with rightwards arrow on top ditulis sebagai c with rightwards arrow on top equals m a with rightwards arrow on top, ditentukan oleh:

c with rightwards arrow on top equals m open parentheses table row cell x subscript a end cell row cell y subscript a end cell end table close parentheses equals open parentheses table row cell m x subscript a end cell row cell m y subscript a end cell end table close parentheses

Diketahui:
vektor a with rightwards arrow on top equals 12 i with hat on top minus 8 j with hat on top rightwards arrow a with rightwards arrow on top equals open parentheses table row 12 row cell negative 8 end cell end table close parentheses
vektor b with rightwards arrow on top equals 9 i with hat on top minus 15 j with hat on top rightwards arrow b with rightwards arrow on top equals open parentheses table row 9 row cell negative 15 end cell end table close parentheses

Hasil perhitungan 1 half a with rightwards arrow on top plus 1 third b with rightwards arrow on top:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 half a with rightwards arrow on top plus 1 third b with rightwards arrow on top end cell equals cell 1 half open parentheses table row 12 row cell negative 8 end cell end table close parentheses plus 1 third open parentheses table row 9 row cell negative 15 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell open parentheses 1 half close parentheses open parentheses 12 close parentheses end cell row cell open parentheses 1 half close parentheses open parentheses negative 8 close parentheses end cell end table close parentheses plus open parentheses table row cell open parentheses 1 third close parentheses open parentheses 9 close parentheses end cell row cell open parentheses 1 third close parentheses open parentheses negative 15 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row 6 row cell negative 4 end cell end table close parentheses plus open parentheses table row 3 row cell negative 5 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 6 plus 3 end cell row cell negative 4 plus open parentheses negative 5 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row 9 row cell negative 9 end cell end table close parentheses end cell row blank equals cell 9 i with hat on top minus 9 j with rightwards arrow on top end cell end table

Dengan demikian hasil perhitungan table attributes columnalign right center left columnspacing 0px end attributes row cell 1 half a with rightwards arrow on top plus 1 third b with rightwards arrow on top end cell equals cell 9 i with hat on top minus 9 j with rightwards arrow on top end cell end table.

26

5.0 (3 rating)

Pertanyaan serupa

Diketahui vektor a=4i−3j​, vektor b=−2i−j​, dan vektor c=−i+5j​. Tentukan vektor-vektor berikut (nyatakan hasilnya dalam vektor-vektor basis i dan j​ c. 3a−2b+3c

76

5.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia