Roboguru

Diketahui  dengan koordinat  dan .  Tentukan:  Proyeksi skalar ortogonal  pada

Pertanyaan

Diketahui triangle ABC dengan koordinat straight A open parentheses 2 comma space minus 1 comma space minus 1 close parentheses comma space straight B open parentheses negative 1 comma space 4 comma space minus 2 close parentheses dan straight C open parentheses 5 comma space 0 comma space minus 3 close parentheses
Tentukan: 

  • Proyeksi skalar ortogonal AB with rightwards arrow on top pada AC with rightwards arrow on top   

Pembahasan Soal:

Diketahui triangle ABC dengan koordinat straight A open parentheses 2 comma space minus 1 comma space minus 1 close parentheses comma space straight B open parentheses negative 1 comma space 4 comma space minus 2 close parentheses dan straight C open parentheses 5 comma space 0 comma space minus 3 close parentheses

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell straight B minus straight A end cell row blank equals cell open parentheses negative 1 comma space 4 comma space minus 2 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses negative 3 comma space 5 comma space minus 1 close parentheses end cell row blank blank blank row cell AC with rightwards arrow on top end cell equals cell straight C minus straight A end cell row blank equals cell open parentheses 5 comma space 0 comma space minus 3 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell end table 

Sehingga 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 3 close parentheses squared plus open parentheses 1 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 9 plus 1 plus 4 end root end cell row blank equals cell square root of 14 end cell end table 

Proyeksi skalar ortogonal AB with rightwards arrow on top pada AC with rightwards arrow on top yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell straight c with rightwards arrow on top end cell equals cell open vertical bar fraction numerator AB 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 end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator open parentheses negative 3 comma space 5 comma space minus 1 close parentheses times open parentheses 3 comma space 1 comma space minus 2 close parentheses over denominator square root of 14 end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator open parentheses negative 9 close parentheses plus 5 plus 2 over denominator square root of 14 end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator negative 2 over denominator square root of 14 end fraction close vertical bar end cell row blank equals cell fraction numerator 2 over denominator square root of 14 end fraction end cell end table 

Jadi, proyeksi skalar ortogonal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah fraction numerator 2 over denominator square root of 14 end fraction

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

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 vektor  dan  panjang proyeksi vektor skalar  pada vektor  adalah ...

Pembahasan Soal:

Diketahui vektor straight u with rightwards arrow on top equals open parentheses 2 comma space minus 1 comma space 3 close parentheses dan straight v with rightwards arrow on top equals open parentheses negative 3 comma space 2 comma space 6 close parentheses, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top end cell equals cell 3 open parentheses 2 comma space minus 1 comma space 3 close parentheses plus 2 open parentheses negative 3 comma space 2 comma space 6 close parentheses end cell row blank equals cell open parentheses 6 comma space minus 3 comma space 9 close parentheses plus open parentheses negative 6 comma space 4 comma space 12 close parentheses end cell row blank equals cell open parentheses 0 comma space 1 comma space 21 close parentheses end cell end table 

Hasil kali titik 3 straight u with rightwards arrow on top plus 2 straight v with rightwards arrow on top dengan straight v with rightwards arrow on top adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 u with rightwards harpoon with barb upwards on top plus 2 v with rightwards arrow on top close parentheses times open parentheses v with rightwards arrow on top close parentheses end cell equals cell open parentheses 0 comma space 1 comma space 21 close parentheses times open parentheses negative 3 comma space 2 comma space 6 close parentheses end cell row blank equals cell open parentheses 0 plus 2 plus 126 close parentheses end cell row blank equals 128 end table 

Panjang vektor v with rightwards arrow on top adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar v with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 3 close parentheses squared plus 2 squared plus 6 squared end root end cell row blank equals cell square root of 9 plus 4 plus 36 end root end cell row blank equals cell square root of 49 end cell row blank equals 7 end table  

Sehingga, panjang proyeksi vektor skalar 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top pada vektor v with rightwards arrow on top adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar c with rightwards arrow on top close vertical bar end cell equals cell fraction numerator open parentheses 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top close parentheses times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell 128 over 7 end cell end table 

Dengan demikian, panjang proyeksi vektor skalar 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top pada vektor v with rightwards arrow on top adalah 128 over 7 satuan panjang.  

Roboguru

Diketahui , dan proyeksi skalar vektor  pada  adalah . Nilai x = ....

Pembahasan Soal:

begin mathsize 14px style Jika space straight c with rightwards arrow on top space merupakan space proyeksi space straight q with rightwards arrow on top space pada space space straight p with rightwards arrow on top space maka  open vertical bar straight c with rightwards arrow on top close vertical bar equals fraction numerator straight p with rightwards arrow on top straight space. straight space straight q with rightwards arrow on top over denominator open vertical bar straight p with rightwards arrow on top close vertical bar end fraction  8 over 7 equals fraction numerator open parentheses table row 2 row 3 row 6 end table close parentheses. open parentheses table row 1 row cell 2 straight x end cell row 2 end table close parentheses over denominator square root of 2 squared plus 3 squared plus 6 squared end root end fraction  8 over 7 equals fraction numerator 2 open parentheses 1 close parentheses plus 3 open parentheses 2 straight x close parentheses plus 6 open parentheses 2 close parentheses over denominator square root of 49 end fraction    Maka  8 equals 2 plus 6 straight x plus 12  6 straight x equals 8 minus 2 minus 12  6 straight x equals negative 6  straight x equals negative 1 end style

Roboguru

Vektor  dengan panjang membentuk sudut lancip dengan vektor . Jika panjang proyeksi orthogonal  pada  adalah 2, tentukan vektor a.

Pembahasan Soal:

Misalkan vekor bold italic a equals open parentheses x comma space y close parentheses, dengan rumus panjang proyeksi orthogonal  bold italic a pada bold italic b, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar c close vertical bar end cell equals cell fraction numerator a times b over denominator open vertical bar b close vertical bar end fraction end cell row 2 equals cell fraction numerator open parentheses table row x row y end table close parentheses open parentheses table row 6 row 8 end table close parentheses over denominator square root of 6 squared plus 8 squared end root end fraction end cell row 2 equals cell fraction numerator 6 x plus 8 y over denominator 10 end fraction end cell row 20 equals cell 6 x plus 8 y end cell row 10 equals cell 3 x plus 4 y... left parenthesis 1 right parenthesis end cell end table

Selanjutnya diketahui vektor bold italic a dengan panjang square root of 5 space, maka

open vertical bar a close vertical bar equals square root of 5 square root of x squared plus y squared end root equals square root of 5 x squared plus y squared equals 5... left parenthesis 2 right parenthesis

Substitusi persamaan (1) ke persamaan (2),

3 x plus 4 y equals 10 rightwards arrow x equals fraction numerator 10 minus 4 y over denominator 3 end fraction x squared plus y squared equals 5 open parentheses fraction numerator 10 minus 4 y over denominator 3 end fraction close parentheses squared plus y squared equals 5 fraction numerator 100 minus 80 y plus 16 y squared over denominator 9 end fraction plus y squared equals 5 space left parenthesis k a l i space 9 right parenthesis 16 y squared minus 80 y plus 100 plus 9 y squared equals 45 25 y squared minus 80 y plus 55 equals 0 5 y squared minus 16 y plus 11 equals 0 left parenthesis 5 y minus 11 right parenthesis left parenthesis y minus 1 right parenthesis equals 0 y subscript 1 equals 11 over 5 space atau space y subscript 2 equals 1 x subscript 1 equals fraction numerator 10 minus 4 open parentheses begin display style 11 over 5 end style close parentheses over denominator 3 end fraction equals 2 over 5 x subscript 2 equals fraction numerator 10 minus 4 left parenthesis 1 right parenthesis over denominator 3 end fraction equals 2

Jadi, diperoleh vektor bold italic a equals left parenthesis 2 comma space 1 right parenthesis space atau space bold italic a equals open parentheses 2 over 5 comma space 11 over 5 close parentheses.

Roboguru

Diketahui vektor dan . Tentukan b. Proyeksi skalar ortogonal   pada

Pembahasan Soal:

Pertama kita tentukan vektor u with rightwards arrow on top plus v with rightwards arrow on top :

table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top plus v with rightwards arrow on top end cell equals cell open square brackets table row 5 row 2 row cell negative 8 end cell end table close square brackets plus open square brackets table row 3 row cell negative 2 end cell row 6 end table close square brackets end cell row blank equals cell open square brackets table row 8 row 0 row cell negative 2 end cell end table close square brackets end cell end table 

Iingat rumus Proyeksi skalar ortogonal u pada v

equals open vertical bar 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 end fraction close vertical bar 

Sehingga,  proyeksi skalar ortogonal 2 u with rightwards arrow on top pada v with rightwards arrow on top

table attributes columnalign right center left columnspacing 0px end attributes row blank equals cell open vertical bar fraction numerator open parentheses u with rightwards arrow on top plus v with rightwards arrow on top close parentheses times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator open square brackets table row 8 row 0 row cell negative 2 end cell end table close square brackets times open square brackets table row 3 row cell negative 2 end cell row 6 end table close square brackets over denominator square root of 3 squared plus open parentheses negative 2 close parentheses squared plus open parentheses 6 close parentheses squared end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 24 plus 0 minus 12 over denominator square root of 9 plus 4 plus 36 end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 12 over denominator square root of 49 ‬ end root end fraction close vertical bar end cell row blank equals cell 12 over 7 end cell end table  

Jadi, Proyeksi skalar ortogonal u with rightwards arrow on top plus v with rightwards arrow on top pada v with rightwards arrow on topadalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 12 over 7 end cell end table 

Roboguru

Diketahui  dan . Panjang vektor proyeksi ortogonal dari  pada  adalah ....

Pembahasan Soal:

Misalkan vektor proyeksi ortogonal dari undefined pada undefined adalah begin mathsize 14px style italic p with rightwards arrow on top end style. Sehingga panjang proyeksi ortogonalnya adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar italic p with rightwards arrow on top close vertical bar end cell equals cell fraction numerator italic a with rightwards arrow on top ⋅ b with rightwards arrow on top over denominator open vertical bar italic a with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell open vertical bar fraction numerator open parentheses table row cell negative sign 1 end cell row 2 row 4 end table close parentheses ⋅ open parentheses table row 8 row cell negative sign 3 end cell row 2 end table close parentheses over denominator square root of open parentheses negative sign 1 close parentheses squared plus 2 squared plus 4 squared end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator open parentheses negative sign 1 close parentheses ⋅ 8 plus 2 ⋅ open parentheses negative sign 3 close parentheses plus 4 ⋅ 2 over denominator square root of 1 plus 4 plus 16 end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator negative sign 8 minus sign 6 plus 8 over denominator square root of 21 end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator negative sign 6 over denominator square root of 21 end fraction close vertical bar end cell row blank equals cell fraction numerator 6 over denominator square root of 21 end fraction end cell row blank equals cell fraction numerator 6 over denominator square root of 21 end fraction ⋅ fraction numerator square root of 21 over denominator square root of 21 end fraction end cell row blank equals cell fraction numerator 6 square root of 21 over denominator 21 end fraction end cell row blank equals cell fraction numerator 2 square root of 21 over denominator 7 end fraction end cell end table     

Jadi, jawaban yang tepat adalah A. 

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