RoboguruRoboguru
SD

Diberikan dua vektor sebagai berikut. Tentukan proyeksi a pada b dan juga b pada a.

Pertanyaan

Diberikan dua vektor sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row bold italic a equals cell 2 bold italic i with bold hat on top minus 3 bold italic j with hat on top plus bold italic k with hat on top semicolon space end cell row blank blank blank row bold italic b equals cell negative bold italic i with bold hat on top plus 2 bold italic j with hat on top minus bold italic k with hat on top end cell end table

Tentukan proyeksi bold italic a pada bold italic b dan juga bold italic b pada bold italic a.

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Jawaban

proyeksi bold italic a pada bold italic b adalah 9 over 6 bold italic i with hat on top minus 9 over 3 bold italic j with hat on top plus 9 over 6 bold italic k with hat on top dan proyeksi bold italic b pada bold italic a adalah negative 9 over 7 bold italic i with hat on top plus 27 over 14 bold italic j with hat on top minus 9 over 14 bold italic k with hat on top.

Pembahasan

Asumsikan bahwa proyeksi yang dimaksud pada soal adalah proyeksi vektor. Ingat bahwa rumus proyeksi vektor bold italic u pada bold italic v, yaitu:

bold italic u subscript v equals fraction numerator bold italic u bold times bold italic v over denominator open vertical bar bold italic v close vertical bar squared end fraction bold italic v

Perkalian titik (dot product) dari dua vektor dimensi tiga mengikuti rumus berikut:

bold italic u times bold italic v bold equals u subscript 1 times v subscript 1 plus u subscript 2 times v subscript 2 plus u subscript 3 times v subscript 3

Sedangkan rumus panjang vektor pada dimensi tiga, yaitu:

open vertical bar bold italic v close vertical bar equals square root of v subscript 1 squared plus v subscript 2 squared plus v subscript 3 squared end root

Sehingga proyeksi bold italic a pada bold italic b, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic a subscript b end cell equals cell fraction numerator bold italic a bold times bold italic b over denominator vertical line bold italic b vertical line squared end fraction bold italic b end cell row blank equals cell fraction numerator open parentheses 2 bold italic i with hat on top minus 3 bold italic j with hat on top plus bold italic k with hat on top close parentheses times open parentheses negative bold italic i with hat on top plus 2 bold italic j with hat on top minus bold italic k with hat on top close parentheses over denominator open parentheses square root of open parentheses negative 1 close parentheses squared plus open parentheses 2 close parentheses squared plus open parentheses negative 1 close parentheses squared end root close parentheses squared end fraction open parentheses negative bold italic i with hat on top plus 2 bold italic j with hat on top minus bold italic k with hat on top close parentheses end cell row blank equals cell fraction numerator 2 open parentheses negative 1 close parentheses plus open parentheses negative 3 close parentheses 2 plus 1 open parentheses negative 1 close parentheses over denominator 1 plus 4 plus 1 end fraction open parentheses negative bold italic i with hat on top plus 2 bold italic j with hat on top minus bold italic k with hat on top close parentheses end cell row blank equals cell fraction numerator negative 2 minus 6 minus 1 over denominator 6 end fraction open parentheses negative bold italic i with hat on top plus 2 bold italic j with hat on top minus bold italic k with hat on top close parentheses end cell row blank equals cell negative 9 over 6 open parentheses negative bold italic i with hat on top plus 2 bold italic j with hat on top minus bold italic k with hat on top close parentheses end cell row blank equals cell 9 over 6 bold italic i with hat on top minus 9 over 3 bold italic j with hat on top plus 9 over 6 bold italic k with hat on top end cell end table

Sedangkan proyeksi bold italic b pada bold italic a, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic b subscript a end cell equals cell fraction numerator bold italic b bold times bold italic a over denominator vertical line bold italic a vertical line squared end fraction bold italic a end cell row blank equals cell fraction numerator open parentheses negative bold italic i with hat on top plus 2 bold italic j with hat on top minus bold italic k with hat on top close parentheses times open parentheses 2 bold italic i with hat on top minus 3 bold italic j with hat on top plus bold italic k with hat on top close parentheses over denominator open parentheses square root of open parentheses 2 close parentheses squared plus open parentheses negative 3 close parentheses squared plus open parentheses 1 close parentheses squared end root close parentheses squared end fraction open parentheses 2 bold italic i with hat on top minus 3 bold italic j with hat on top plus bold italic k with hat on top close parentheses end cell row blank equals cell fraction numerator open parentheses negative 1 close parentheses 2 plus 2 open parentheses negative 3 close parentheses plus open parentheses negative 1 close parentheses 1 over denominator 4 plus 9 plus 1 end fraction open parentheses 2 bold italic i with hat on top minus 3 bold italic j with hat on top plus bold italic k with hat on top close parentheses end cell row blank equals cell fraction numerator negative 2 minus 6 minus 1 over denominator 14 end fraction open parentheses 2 bold italic i with hat on top minus 3 bold italic j with hat on top plus bold italic k with hat on top close parentheses end cell row blank equals cell negative 9 over 14 open parentheses 2 bold italic i with hat on top minus 3 bold italic j with hat on top plus bold italic k with hat on top close parentheses end cell row blank equals cell negative 9 over 7 bold italic i with hat on top plus 27 over 14 bold italic j with hat on top minus 9 over 14 bold italic k with hat on top end cell end table

Dengan demikian, proyeksi bold italic a pada bold italic b adalah 9 over 6 bold italic i with hat on top minus 9 over 3 bold italic j with hat on top plus 9 over 6 bold italic k with hat on top dan proyeksi bold italic b pada bold italic a adalah negative 9 over 7 bold italic i with hat on top plus 27 over 14 bold italic j with hat on top minus 9 over 14 bold italic k with hat on top.

27

5.0 (1 rating)

Pertanyaan serupa

Diketahui kubus satuan ABCD.EFGH. Misalkan vektor-vektor AB=i=(1, 0, 0); AD=j​=(0, 1, 0); AE=k=(0, 0, 1). Titik P adalah titik pusat sisi BCGF. Tentukan proyeksi FP pada vektor AC.

132

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