Roboguru
SD

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.

Pertanyaan

Diketahui kubus satuan ABCD.EFGH. Misalkan vektor-vektor AB with rightwards arrow on top equals bold italic i with hat on top equals left parenthesis 1 comma space 0 comma space 0 right parenthesisAD with rightwards arrow on top equals bold italic j with hat on top equals left parenthesis 0 comma space 1 comma space 0 right parenthesisAE with rightwards arrow on top equals bold italic k with hat on top equals left parenthesis 0 comma space 0 comma space 1 right parenthesis. Titik P adalah titik pusat sisi BCGF. Tentukan proyeksi FP with rightwards arrow on top pada vektor AC with rightwards arrow on top.

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Jawaban

proyeksi FP with rightwards arrow on top pada vektor AC with rightwards arrow on top adalah open parentheses 1 fourth comma space 1 fourth comma space 0 close parentheses.

Pembahasan

Ilustrasi gambar kubus yang dimaksud pada soal adalah sebagai berikut:

Ingat bahwa jika diketahui vektor u with rightwards arrow on top equals open parentheses u subscript 1 comma space u subscript 2 comma space u subscript 3 close parentheses dan v with rightwards arrow on top equals open parentheses v subscript 1 comma space v subscript 2 comma space v subscript 3 close parentheses, maka perkalian titik (dot product) antara u with rightwards arrow on top dan v with rightwards arrow on top mengikuti rumus berikut:

straight u with rightwards arrow on top times v with rightwards arrow on top equals u subscript 1 v subscript 1 plus u subscript 2 v subscript 2 plus u subscript 3 v subscript 3

Panjang vektor v with rightwards arrow on top dapat dicari dengan rumus berikut:

open vertical bar v with rightwards arrow on top close vertical bar equals square root of v subscript 1 squared plus v subscript 2 squared plus v subscript 3 squared end root

Proyeksi vektor u with rightwards arrow on top pada v with rightwards arrow on top dapat dicari dengan rumus berikut:

stack u subscript v with rightwards arrow on top equals fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar squared end fraction v with rightwards arrow on top

Menentukan vektor FP with rightwards arrow on top dan AC with rightwards arrow on top:

Dari ilustrasi, diketahui bahwa 

table attributes columnalign right center left columnspacing 0px end attributes row cell BC with rightwards arrow on top end cell equals cell AD with rightwards arrow on top end cell row blank equals cell open parentheses 0 comma space 1 comma space 0 close parentheses end cell end table

 table attributes columnalign right center left columnspacing 0px end attributes row cell FB with rightwards arrow on top end cell equals cell EA with rightwards arrow on top end cell row blank equals cell negative AE with rightwards arrow on top end cell row blank equals cell negative open parentheses 0 comma space 0 comma space 1 close parentheses end cell row blank equals cell open parentheses 0 comma space 0 comma space minus 1 close parentheses end cell end table

Maka vektor AC with rightwards arrow on top dapat diperoleh dari penjumlahan dua vektor secara aljabar sebagaimana perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell AC with rightwards arrow on top end cell equals cell AB with rightwards arrow on top plus BC with rightwards arrow on top end cell row blank equals cell open parentheses 1 comma space 0 comma space 0 close parentheses plus open parentheses 0 comma space 1 comma space 0 close parentheses end cell row blank equals cell open parentheses 1 comma space 1 comma space 0 close parentheses end cell end table

Sedangkan vektor FP with rightwards arrow on top dapat diperoleh dari vektor FC with rightwards arrow on top, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell FC with rightwards arrow on top end cell equals cell FB with rightwards arrow on top plus BC with rightwards arrow on top end cell row blank equals cell open parentheses 0 comma space 0 comma space minus 1 close parentheses plus open parentheses 0 comma space 1 comma space 0 close parentheses end cell row blank equals cell open parentheses 0 comma space 1 comma space minus 1 close parentheses end cell end table

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell FP with rightwards arrow on top end cell equals cell 1 half FC with rightwards arrow on top end cell row blank equals cell 1 half open parentheses 0 comma space 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses 0 comma space 1 half comma space minus 1 half close parentheses end cell end table

Proyeksi FP with rightwards arrow on top pada vektor AC with rightwards arrow on top, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell q with rightwards arrow on top end cell equals cell fraction numerator FP with rightwards arrow on top times AC with rightwards arrow on top over denominator open vertical bar AC with rightwards arrow on top close vertical bar squared end fraction AC with rightwards arrow on top end cell row blank equals cell fraction numerator open parentheses 0 comma space begin display style 1 half end style comma space minus begin display style 1 half end style close parentheses times open parentheses 1 comma space 1 comma space 0 close parentheses over denominator open vertical bar open parentheses 1 comma space 1 comma space 0 close parentheses close vertical bar squared end fraction open parentheses 1 comma space 1 comma space 0 close parentheses end cell row blank equals cell fraction numerator open parentheses 0 close parentheses open parentheses 1 close parentheses plus open parentheses begin display style 1 half end style close parentheses open parentheses 1 close parentheses plus open parentheses negative begin display style 1 half end style close parentheses open parentheses 0 close parentheses over denominator open parentheses square root of 1 squared plus 1 squared plus 0 squared end root close parentheses squared end fraction open parentheses 1 comma space 1 comma space 0 close parentheses end cell row blank equals cell fraction numerator begin display style 1 half end style over denominator 2 end fraction open parentheses 1 comma space 1 comma space 0 close parentheses end cell row blank equals cell 1 fourth open parentheses 1 comma space 1 comma space 0 close parentheses end cell row blank equals cell open parentheses 1 fourth comma space 1 fourth comma space 0 close parentheses end cell end table

Dengan demikian, proyeksi FP with rightwards arrow on top pada vektor AC with rightwards arrow on top adalah open parentheses 1 fourth comma space 1 fourth comma space 0 close parentheses.

128

5.0 (2 rating)

Pertanyaan serupa

Diketahui titik A(2, 3, −1), titik B(−2, −4, 3), dan vektor p​=4i−3j​+k.   b. Tentukan proyeksi vektor ortogonal vektor p​ pada arah AB.

47

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