Roboguru

Diketahui  dengan koordinat  dan .  Tentukan:  Proyeksi vektor 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 vektor 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 

Maka proyeksi vektor ortogonal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight c with rightwards arrow on top end cell equals cell 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 squared end fraction times AC with rightwards arrow on top end cell row blank equals cell 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 open parentheses square root of 14 close parentheses squared end fraction times open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell fraction numerator open parentheses negative 9 close parentheses plus open parentheses 5 close parentheses plus open parentheses 2 close parentheses over denominator 14 end fraction times open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell fraction numerator negative 2 over denominator 14 end fraction times open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell negative 1 over 7 times open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses end cell end table 

Jadi,  proyeksi vektor ortogonal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses

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

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Pertanyaan yang serupa

Vektor yang dikenal dan . Proyeksi ortogonal dari vektor a ke vektor b adalah

Pembahasan Soal:

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Pembahasan Soal:

Ingat konsep proyeksi vektor orthognal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah 

Proyeksi space vektor space AC with rightwards arrow on top equals fraction numerator stack AB times with rightwards arrow on top AC with rightwards arrow on top over denominator open vertical bar AC with rightwards arrow on top close vertical bar squared end fraction stack times AC with rightwards arrow on top  

Diketahui straight A equals open parentheses 2 comma space minus 1 comma space minus 1 close parenthesesstraight 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 with rightwards arrow on top minus straight A with rightwards arrow on top 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 cell AC with rightwards arrow on top end cell equals cell straight C with rightwards arrow on top minus straight A with rightwards arrow on top 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

maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell stack AB times with rightwards arrow on top AC with rightwards arrow on top end cell equals cell 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 end cell row blank equals cell negative 9 plus 5 plus 2 end cell row blank equals cell negative 2 end cell row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of 3 squared plus 1 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 

sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell Proyeksi space vektor space AC with rightwards arrow on top end cell equals cell 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 squared end fraction times AC with rightwards arrow on top end cell row blank equals cell fraction numerator negative 2 over denominator open parentheses square root of 14 close parentheses squared end fraction open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell fraction numerator negative 2 over denominator 14 end fraction open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell negative 1 over 7 open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses end cell end table  

Dengan demikian proyeksi vektor orthognal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses end cell end table.

Roboguru

Diketahui vektor-vektor  dan . Sudut antara vektor  dan  adalah  dengan . Proyeksi  pada  adalah . Nilai dari b = ...

Pembahasan Soal:

Proyeksi vektor begin mathsize 12px style u with rightwards harpoon with barb upwards on top end style pada begin mathsize 12px style v with rightwards harpoon with barb upwards on top end style adalah vektor begin mathsize 12px style p with rightwards harpoon with barb upwards on top end style maka berlaku:

  • begin mathsize 12px style open vertical bar p with rightwards harpoon with barb upwards on top close vertical bar equals open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar cos theta end style
  • begin mathsize 12px style cos theta equals fraction numerator u with rightwards harpoon with barb upwards on top times v with rightwards harpoon with barb upwards on top over denominator open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar open vertical bar v with rightwards harpoon with barb upwards on top close vertical bar end fraction end style
  • Vektor begin mathsize 12px style p with rightwards harpoon with barb upwards on top end style sejajar dengan vektor begin mathsize 12px style v with rightwards harpoon with barb upwards on top end style maka begin mathsize 12px style p with rightwards harpoon with barb upwards on top equals n times v with rightwards harpoon with barb upwards on top end style

Diketahui:

  • begin mathsize 12px style u with rightwards harpoon with barb upwards on top equals b i with rightwards harpoon with barb upwards on top plus a j with rightwards harpoon with barb upwards on top plus 9 k with rightwards harpoon with barb upwards on top end style
  • begin mathsize 12px style v with rightwards harpoon with barb upwards on top equals a i with rightwards harpoon with barb upwards on top minus b j with rightwards harpoon with barb upwards on top plus a k with rightwards harpoon with barb upwards on top end style
  • begin mathsize 12px style p with rightwards harpoon with barb upwards on top equals 4 a with rightwards harpoon with barb upwards on top minus 2 j with rightwards harpoon with barb upwards on top plus 4 k with rightwards harpoon with barb upwards on top end style
  • begin mathsize 12px style cos theta equals 6 over 11 end style

Karena begin mathsize 12px style p with rightwards harpoon with barb upwards on top end style sejajar dengan vektor begin mathsize 12px style v with rightwards harpoon with barb upwards on top end style maka: 

begin mathsize 12px style n times v with rightwards harpoon with barb upwards on top equals p with rightwards harpoon with barb upwards on top  n equals open parentheses table row a row cell negative b end cell row a end table close parentheses equals open parentheses table row 4 row cell negative 2 end cell row 4 end table close parentheses end style

Dari persamaan di atas diperoleh:

na = 4begin mathsize 12px style left right double arrow end stylen = begin mathsize 12px style 4 over a end style    ...  (i)

-bn = -2begin mathsize 12px style left right double arrow end stylen = begin mathsize 12px style 2 over b end style  ...  (ii)

Dari persamaan (i) dan (ii) diperoleh:

begin mathsize 12px style 4 over a end stylebegin mathsize 12px style 2 over b end stylebegin mathsize 12px style left right double arrow end stylea = 2b ... (iii)

 

Sudut antara kedua vektor adalah begin mathsize 12px style theta end style sehingga berlaku:

begin mathsize 12px style cos theta equals fraction numerator u with rightwards harpoon with barb upwards on top times v with rightwards harpoon with barb upwards on top over denominator open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar open vertical bar v with rightwards harpoon with barb upwards on top close vertical bar end fraction space  left right double arrow 6 over 11 equals fraction numerator a b minus a b plus 9 a over denominator square root of b squared plus a squared plus 81 end root times square root of a squared plus b squared plus a squared end root end fraction  left right double arrow 6 over 11 equals fraction numerator 9 a over denominator square root of b squared plus a squared plus 81 end root times square root of 2 a squared plus b squared end root end fraction space space... space left parenthesis i v right parenthesis end style

Substitusikan persamaan (iii) ke persamaan (iv), diiperoleh:

begin mathsize 12px style 6 over 11 equals fraction numerator 18 b over denominator square root of b squared plus 4 b squared plus 81 end root begin display style times end style begin display style square root of 8 b squared plus b squared end root end style end fraction  left right double arrow 6 over 11 equals fraction numerator 18 b over denominator square root of 5 b squared plus 81 end root begin display style times end style begin display style 3 end style begin display style b end style end fraction  left right double arrow 6 over 11 equals fraction numerator 6 over denominator square root of 5 b squared plus 81 end root end fraction  left right double arrow square root of 5 b squared plus 81 end root equals 11  left right double arrow 5 b squared plus 81 equals 121  left right double arrow 5 b squared equals 40  left right double arrow b squared equals 8  left right double arrow b equals plus-or-minus 2 square root of 2 end style

Jadi, nilai b = 2begin mathsize 12px style square root of 2 end style

Roboguru

Vektor yang merupakan proyeksi vektor  pada vektor  adalah ...

Pembahasan Soal:

Panjang vektor a with rightwards arrow on top equals x i with rightwards arrow on top plus y j with rightwards arrow on top plus z k with rightwards arrow on top, yaitu open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Proyeksi vektor orthogonal vektor a with rightwards arrow on top pada b with rightwards arrow on top dapat ditentukan menggunakan rumus berikut.

c with rightwards arrow on top equals 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

Berdasarkan konsep tersebut dapat ditentukan penyelesaian soal sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top end cell equals cell fraction numerator x with rightwards arrow on top times y with rightwards arrow on top over denominator open vertical bar y with rightwards arrow on top close vertical bar squared end fraction times y with rightwards arrow on top end cell row blank equals cell fraction numerator open parentheses table row 3 row 1 row cell negative 1 end cell end table close parentheses open parentheses table row 2 row 5 row 1 end table close parentheses over denominator open parentheses square root of 2 squared plus 5 squared plus 1 end root close parentheses squared end fraction times open parentheses table row 2 row 5 row 1 end table close parentheses end cell row blank equals cell fraction numerator 3 times 2 plus 1 times 5 plus open parentheses negative 1 close parentheses times 1 over denominator open parentheses square root of 30 close parentheses squared end fraction times open parentheses table row 2 row 5 row 1 end table close parentheses end cell row blank equals cell 10 over 30 times open parentheses table row 2 row 5 row 1 end table close parentheses end cell row blank equals cell 1 third times open parentheses table row 2 row 5 row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 2 over 3 end cell row cell 5 over 3 end cell row cell 1 third end cell end table close parentheses end cell end table

Vektor proyeksi vektor x with rightwards arrow on top pada y with rightwards arrow on top adalah 2 over 3 i with rightwards arrow on top plus 5 over 3 j with rightwards arrow on top plus 1 third k with rightwards arrow on top 

Oleh karena itu, jawaban yang tepat adalah B.

Roboguru

Perhatikan gambar berikut.     Proyeksi vektor  pada  adalah ....

Pembahasan Soal:

Proyeksi ortogonal vektor a with rightwards arrow on top pada vektor b with rightwards arrow on top.

Proyeksi space vektor equals open parentheses 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 close parentheses times b with rightwards arrow on top

Akan ditentukan proyeksi vektor AB with bar on top pada AC with bar on top, yaitu

Proyeksi space vektor equals open parentheses 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 squared end fraction close parentheses times AC with rightwards arrow on top

*Menentukan vektor AB with bar on top pada AC with bar on top berdasarkan gambar diperoleh

AB with rightwards arrow on top equals open parentheses table row 4 row 0 end table close parentheses space dan space AC with rightwards arrow on top equals open parentheses table row 3 row 3 end table close parentheses

*Menentukan AB with rightwards arrow on top times AC with rightwards arrow on top

AB with rightwards arrow on top times AC with rightwards arrow on top equals open parentheses table row 4 row 0 end table close parentheses times open parentheses table row 3 row 3 end table close parentheses equals 4 times 3 plus 0 times 3 equals 12

*Menentukan open vertical bar AC with rightwards arrow on top close vertical bar

open vertical bar AC with rightwards arrow on top close vertical bar equals square root of 3 squared plus 3 squared end root equals square root of 9 plus 9 end root equals square root of 18 equals 3 square root of 2

*Menentukan proyeksi vektor AB with bar on top pada AC with bar on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell Proyeksi space vektor end cell equals cell open parentheses 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 squared end fraction close parentheses times AC with rightwards arrow on top end cell row blank equals cell open parentheses 12 over open parentheses 3 square root of 2 close parentheses squared close parentheses times open parentheses table row 3 row 3 end table close parentheses end cell row blank equals cell open parentheses 12 over 18 close parentheses times open parentheses table row 3 row 3 end table close parentheses end cell row blank equals cell 2 over 3 times open parentheses table row 3 row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator 2 over denominator up diagonal strike 3 end fraction cross times up diagonal strike 3 end cell row cell fraction numerator 2 over denominator up diagonal strike 3 end fraction cross times up diagonal strike 3 end cell end table close parentheses end cell row cell Proyeksi space vektor end cell equals cell open parentheses table row 2 row 2 end table close parentheses end cell end table

Diperoleh proyeksi vektornya adalah open parentheses table row 2 row 2 end table close parentheses, berdasarkan gambar diperoleh bahwa AE with rightwards arrow on top equals open parentheses table row 2 row 2 end table close parentheses, sehingga proyeksi vektor AB with bar on top pada AC with bar on top adalah AE with bar on top.

Jadi, jawaban yang tepat adalah B.

Roboguru

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