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.

Pertanyaan

Diketahui titik straight A open parentheses 2 comma space 3 comma space minus 1 close parentheses, titik straight B open parentheses negative 2 comma space minus 4 comma space 3 close parentheses, dan vektor p with rightwards arrow on top equals 4 i with hat on top minus 3 j with hat on top plus k with hat on top.

  b. Tentukan proyeksi vektor ortogonal vektor p with rightwards arrow on top pada arah AB with rightwards arrow on top.

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

 proyeksi vektor ortogonal vektor p with rightwards arrow on top pada arah stack A B with rightwards arrow on top adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell negative 4 over 9 end cell row cell negative 7 over 9 end cell row cell 4 over 9 end cell end table close parentheses end cell end table.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell bold minus bold 4 over bold 9 end cell row cell bold minus bold 7 over bold 9 end cell row cell bold 4 over bold 9 end cell end table close parentheses end cell end table.

Ingat!

  • Jika diketahui vektor di bidang r with rightwards arrow on top equals x i with hat on top plus y j with hat on top maka dapat ditulis dalam bentuk vektor kolom sebagai berikut:

r with rightwards arrow on top equals open parentheses table row x row y end table close parentheses

  • Jika koordinat titik straight A open parentheses x subscript 1 comma space y subscript 1 close parentheses dan straight B open parentheses x subscript 2 comma space y subscript 2 close parentheses maka dapat ditetapkan:

stack A B with rightwards arrow on top equals open parentheses table row cell x subscript 2 minus x subscript 1 end cell row cell y subscript 2 minus y subscript 1 end cell end table close parentheses

  • Misalkan vektor a with rightwards arrow on top dan vektor b with rightwards arrow on top adalah vektor-vektor sembarang, dan vektor c with rightwards arrow on top adalah proyeksi vektor a with rightwards arrow on top pada arah vektor b with rightwards arrow on top maka proyeksi vektor ortogonal dari vektor a with rightwards arrow on top pada arah vektor b with rightwards arrow on top ditentukan oleh:

c with rightwards arrow on top equals open parentheses fraction numerator a with rightwards arrow on top space. space 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 b with rightwards arrow on top 

  • Rumus untuk menentukan panjang vektor r with rightwards arrow on top equals open parentheses table row x row y end table close parentheses adalah sebagai  berikut:

open vertical bar r with rightwards arrow on top close vertical bar equals square root of x squared plus y squared end root 

  • Rumus untuk menentukan hasil kali a with rightwards arrow on top space. space b with rightwards arrow on top jika diketahui vektor a with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses dan vektor b with rightwards arrow on top equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell end table close parentheses adalah sebagai berikut:

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

  • Rumus untuk perkalian skalar m dengan vektor a with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses adalah sebagai berikut:

m a with rightwards arrow on top equals m open parentheses table row cell x subscript 1 end cell row cell x subscript 2 end cell end table close parentheses equals open parentheses table row cell m x subscript 1 end cell row cell m x subscript 2 end cell end table close parentheses

Diketahui: 

Titik straight A open parentheses 2 comma space 3 comma space minus 1 close parentheses

Titik straight B open parentheses negative 2 comma space minus 4 comma space 3 close parentheses

Vektor p with rightwards arrow on top equals 4 i with hat on top minus 3 j with hat on top plus k with hat on top space rightwards arrow p with rightwards arrow on top equals open parentheses table row 4 row cell negative 3 end cell row 1 end table close parentheses 

Ditanya: proyeksi vektor ortogonal vektor p with rightwards arrow on top pada arah stack A B with rightwards arrow on top.

Jawab:

Dengan menggunakan rumus di atas, maka stack A B with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell open parentheses table row cell x subscript 2 minus x subscript 1 end cell row cell y subscript 2 minus y subscript 1 end cell row cell z subscript 2 minus z subscript 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 minus 2 end cell row cell negative 4 minus 3 end cell row cell 3 minus left parenthesis negative 1 right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 end cell row cell negative 7 end cell row 4 end table close parentheses end cell end table 

Jadi, proyeksi vektor ortogonal vektor p with rightwards arrow on top pada arah stack A B with rightwards arrow on top adalah sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top end cell equals cell open parentheses fraction numerator p with rightwards arrow on top space. space stack A B with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar squared end fraction close parentheses stack A B with rightwards arrow on top end cell row blank equals cell open parentheses fraction numerator 4 cross times open parentheses negative 4 close parentheses plus open parentheses negative 3 close parentheses cross times open parentheses negative 7 close parentheses plus 1 cross times 4 over denominator open parentheses square root of open parentheses negative 4 close parentheses squared plus open parentheses negative 7 close parentheses squared plus 4 squared end root close parentheses squared end fraction close parentheses open parentheses table row cell negative 4 end cell row cell negative 7 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses fraction numerator negative 16 plus 21 plus 4 over denominator 81 end fraction close parentheses open parentheses table row cell negative 4 end cell row cell negative 7 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses 9 over 81 close parentheses open parentheses table row cell negative 4 end cell row cell negative 7 end cell row 4 end table close parentheses end cell row blank equals cell 1 over 9 open parentheses table row cell negative 4 end cell row cell negative 7 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 over 9 end cell row cell negative 7 over 9 end cell row cell 4 over 9 end cell end table close parentheses end cell end table 

Dengan demikian, proyeksi vektor ortogonal vektor p with rightwards arrow on top pada arah stack A B with rightwards arrow on top adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row cell negative 4 over 9 end cell row cell negative 7 over 9 end cell row cell 4 over 9 end cell end table close parentheses end cell end table.

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Pertanyaan serupa

Diberikan segitiga ABC dengan titik-titik sudut A(4,−3, 2), B(2,−2, 6), dan C(3, 4, 5). Tunjukan bahwa proyeksi vektor ortogonal CA pada arah BA diwakili oleh vektor 2i−j​−4k.

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