Roboguru

Diketahui ,  serta . Proyeksi pada  adalah...

Pertanyaan

Diketahui begin mathsize 14px style straight m with rightwards arrow on top equals straight i with rightwards arrow on top plus 2 straight j with rightwards arrow on top minus 3 straight k with rightwards arrow on top end stylebegin mathsize 14px style straight n with rightwards arrow on top equals 3 straight i with rightwards arrow on top minus straight j with rightwards arrow on top minus 3 straight k with rightwards arrow on top end style serta begin mathsize 14px style straight p with rightwards arrow on top equals 2 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus 5 straight k with rightwards arrow on top end style. Proyeksi begin mathsize 14px style straight m with rightwards arrow on top end style pada begin mathsize 14px style straight n with rightwards arrow on top plus straight p with rightwards arrow on top end style adalah...   

Pembahasan Soal:

Sebelum kita mencari hasil proyeksi dari Proyeksi begin mathsize 14px style straight m with rightwards arrow on top end style pada begin mathsize 14px style straight n with rightwards arrow on top plus straight p with rightwards arrow on top end style , terlebih dahulu kita mencari  begin mathsize 14px style straight n with rightwards arrow on top plus straight p with rightwards arrow on top end style dan  :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight n with rightwards arrow on top plus straight p with rightwards arrow on top end cell equals cell open parentheses table row 3 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses plus open parentheses table row 2 row cell negative 3 end cell row 5 end table close parentheses end cell row cell straight n with rightwards arrow on top plus straight p with rightwards arrow on top end cell equals cell open parentheses table row 5 row cell negative 4 end cell row 2 end table close parentheses end cell row cell straight n with rightwards arrow on top plus straight p with rightwards arrow on top end cell equals cell 5 straight i with rightwards arrow on top minus 4 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top end cell end table end style

Maka, hasil proyeksi dari Proyeksi begin mathsize 14px style straight m with rightwards arrow on top end style pada begin mathsize 14px style straight n with rightwards arrow on top plus straight p with rightwards arrow on top end style (z) adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight z equals cell open parentheses fraction numerator straight m with rightwards arrow on top times open parentheses straight n with rightwards arrow on top plus straight p with rightwards arrow on top close parentheses over denominator open vertical bar straight n with rightwards arrow on top plus straight p with rightwards arrow on top close vertical bar end fraction close parentheses open parentheses straight n with rightwards arrow on top plus straight p with rightwards arrow on top close parentheses end cell row blank equals cell open parentheses fraction numerator 1 times 5 plus 2 times open parentheses negative 4 close parentheses plus open parentheses negative 3 close parentheses 2 over denominator 5 squared plus open parentheses negative 4 close parentheses squared plus 2 squared end fraction close parentheses open parentheses table row 5 row cell negative 4 end cell row 2 end table close parentheses end cell row blank equals cell fraction numerator negative 9 over denominator 45 end fraction open parentheses table row 5 row cell negative 4 end cell row 2 end table close parentheses end cell row blank equals cell fraction numerator negative 1 over denominator 5 end fraction open parentheses table row 5 row cell negative 4 end cell row 2 end table close parentheses end cell row blank equals cell space open parentheses table row cell negative 1 end cell row cell 4 over 5 end cell row cell negative 2 over 5 end cell end table close parentheses end cell row blank equals cell negative i with rightwards arrow on top plus 4 over 5 j with rightwards arrow on top minus 2 over 5 k with rightwards arrow on top end cell end table end style

Jadi, hasil proyeksinya adalah begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell i with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 4 over 5 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell j with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 2 over 5 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell k with rightwards arrow on top end cell end table end style  

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Indah

Mahasiswa/Alumni Universitas Diponegoro

Terakhir diupdate 13 September 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 . Jika koordinat titik , koordinat titik Q adalah

Pembahasan Soal:

Diketahui PQ with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row cell negative 1 end cell end table close parentheses dan koordinat titik straight P open parentheses 4 comma negative 3 close parentheses, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell PQ with rightwards arrow on top end cell equals cell open parentheses table row cell negative 2 end cell row cell negative 1 end cell end table close parentheses end cell row cell straight Q minus straight P end cell equals cell open parentheses table row cell negative 2 end cell row cell negative 1 end cell end table close parentheses end cell row cell straight Q minus open parentheses table row 4 row cell negative 3 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 2 end cell row cell negative 1 end cell end table close parentheses end cell row straight Q equals cell open parentheses table row cell negative 2 end cell row cell negative 1 end cell end table close parentheses plus open parentheses table row 4 row cell negative 3 end cell end table close parentheses end cell row straight Q equals cell open parentheses table row 2 row cell negative 4 end cell end table close parentheses end cell end table

Jadi, koordinat titik Q adalah straight Q open parentheses 2 comma negative 4 close parentheses.

Roboguru

Diketahui a=3i−j​; b=2i+13j​ dan c=−2i−8j​. Tentukan: a. a+b dan ∣∣​a+b∣∣​

Pembahasan Soal:

Ingat!

Pada vektor, jika p=ai+bj dan q=ci+dj maka:

  • p+q=(ab)+(cd)=(a+cb+d) 
  • p=a2+b2 

Maka:

a+b====(31)+(213)(3+21+13)(512)5i+12j 

Sehingga:

a+b====52+12225+14416913satuanpanjang 

Dengan demikian, a+b=5i+12j dan a+b=13satuanpanjang.

Roboguru

Diketahui koordinat-koordinat ,  dan . Jika , , . Tunjukkan bahwa !

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell A B end cell equals cell B minus A end cell row blank equals cell open parentheses table row cell negative 2 end cell row 5 row 3 end table close parentheses minus open parentheses table row 3 row 4 row 7 end table close parentheses end cell row blank equals cell open parentheses negative 5 comma space 1 comma space minus 4 close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell B C end cell equals cell C minus B end cell row blank equals cell open parentheses table row 6 row 3 row 5 end table close parentheses minus open parentheses table row cell negative 2 end cell row 5 row 3 end table close parentheses end cell row blank equals cell open parentheses 8 comma space minus 2 comma space 2 close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell A C end cell equals cell C minus A end cell row blank equals cell open parentheses table row 6 row 3 row 5 end table close parentheses minus open parentheses table row 3 row 4 row 7 end table close parentheses end cell row blank equals cell open parentheses 3 comma space minus 1 comma space minus 2 close parentheses end cell end table

Dengan penjumlahan vektor secara aljabar:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row 3 row cell negative 1 end cell row cell negative 2 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 5 end cell row 1 row cell negative 4 end cell end table close parentheses plus open parentheses table row 8 row cell negative 2 end cell row 2 end table close parentheses end cell row cell A C end cell equals cell A B plus B C end cell row r equals cell p plus q space end cell end table

jadi r equals p plus q.

Roboguru

Diketahui  dan . Tentukan panjang proyeksi dan vektor proyeksi a terhadap vektor b.

Pembahasan Soal:

Ingat bahwa, rumus panjang proyeksi dan vektor preyeksi u terhadap v berturut-turut adalah

open vertical bar x with rightwards arrow on top close vertical bar 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 space dan space x 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 times v with rightwards arrow on top

maka, panjang proyeksi a terhadap vektor b adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x with rightwards arrow on top close vertical bar end cell equals cell open vertical bar fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b 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 table row 2 row 2 row cell negative 1 end cell end table close parentheses times open parentheses table row 6 row cell negative 3 end cell row 2 end table close parentheses over denominator square root of 6 squared plus left parenthesis negative 3 right parenthesis squared plus 2 to the power of 2 end exponent end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 12 minus 6 minus 2 over denominator square root of 36 plus 9 plus 4 end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 4 over denominator square root of 49 end fraction close vertical bar end cell row blank equals cell open vertical bar 4 over 7 close vertical bar end cell row blank equals cell 4 over 7 end cell end table

dan proyeksi vektor a terhadap vektor b adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell x with rightwards arrow on top end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar squared end fraction times b with rightwards arrow on top end cell row blank equals cell fraction numerator open parentheses table row 2 row 2 row cell negative 1 end cell end table close parentheses times open parentheses table row 6 row cell negative 3 end cell row 2 end table close parentheses over denominator open square brackets square root of 6 squared plus left parenthesis negative 3 right parenthesis squared plus 2 squared end root close square brackets squared end fraction times open parentheses table row 6 row cell negative 3 end cell row 2 end table close parentheses end cell row blank equals cell 4 over 49 open parentheses table row 6 row cell negative 3 end cell row 2 end table close parentheses end cell row blank equals cell open parentheses table row cell 24 over 49 end cell row cell negative 12 over 49 end cell row cell 8 over 49 end cell end table close parentheses end cell end table

Oleh karena itu, panjang proyeksi dan vektor preyeksi a terhadap b berturut-turut adalah 4 over 7 space dan space 24 over 49 straight i minus 12 over 49 straight j plus 8 over 49 straight k.

Roboguru

Diketahui titik , , dan . Jika  mewakili  dan  mewakili , maka proyeksi ortogonal vektor  pada  adalah ....

Pembahasan Soal:

Gunakan konsep proyeksi orthogonal vektor u with rightwards arrow on top pada v with rightwards arrow on top.

Proyeksi space ortogonal equals open parentheses 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 close parentheses v with rightwards arrow on top

Diketahui:
Titik straight A open parentheses 2 comma space 7 comma space 8 close parenthesesstraight B open parentheses negative 1 comma space 1 comma space minus 1 close parentheses, dan straight C open parentheses 0 comma space 3 comma space 2 close parentheses. Jika u with rightwards arrow on top mewakili AB with rightwards arrow on top dan v with rightwards arrow on top mewakili BC with rightwards arrow on top.
Akan ditentukan proyeksi ortogonal vektor u with rightwards arrow on top pada v with rightwards arrow on top.

Tentukan vektor u with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell u with rightwards arrow on top end cell equals cell AB with rightwards arrow on top end cell row blank equals cell straight B minus straight A end cell row blank equals cell open parentheses negative 1 minus 2 comma space 1 minus 7 comma space minus 1 minus 8 close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses negative 3 comma space minus 6 comma space minus 9 close parentheses end cell end table

Tentukan vektor v with rightwards arrow on top.

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

Sehingga proyeksi ortogonal vektor u with rightwards arrow on top pada v with rightwards arrow on top dapat diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row Proyeksi equals cell open parentheses 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 close parentheses times v with rightwards arrow on top end cell row blank equals cell open parentheses fraction numerator open parentheses negative 3 close parentheses times 1 plus open parentheses negative 6 close parentheses times 2 plus open parentheses negative 9 close parentheses times 3 over denominator open parentheses square root of 1 squared plus 2 squared plus 3 squared end root close parentheses squared end fraction close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses fraction numerator negative 3 minus 12 minus 27 over denominator open parentheses square root of 1 plus 4 plus 9 end root close parentheses squared end fraction close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 42 over open parentheses square root of 14 close parentheses squared close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 42 over 14 close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell 3 cross times open parentheses 1 comma space 2 comma space 3 close parentheses end cell row Proyeksi equals cell open parentheses 3 comma space 6 comma space 9 close parentheses end cell end table

Sehingga diperoleh proyeksi ortogonalnya adalah open parentheses 3 comma space 6 comma space 9 close parentheses atau 3 i with rightwards arrow on top plus 6 j with rightwards arrow on top plus 9 k with rightwards arrow on top.

Jadi, jawaban yang tepat adalah E.

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