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

Pertanyaan

Diberikan segitiga ABC dengan titik-titik sudut straight A open parentheses 4 comma negative 3 comma space 2 close parenthesesstraight B open parentheses 2 comma negative 2 comma space 6 close parentheses, dan straight C open parentheses 3 comma space 4 comma space 5 close parentheses.

b. Tunjukan bahwa proyeksi vektor ortogonal AC with rightwards arrow on top pada arah BC with rightwards arrow on top diwakili oleh vektor i with hat on top plus 6 j with hat on top minus k with hat on top.

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

terbukti bahwa proyeksi vektor ortogonal stack A C with rightwards arrow on top pada arah stack B C with rightwards arrow on top benar diwakili oleh vektor i with hat on top plus 6 j with hat on top minus k with hat on top.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah proyeksi vektor ortogonal stack bold italic A bold italic C with bold rightwards arrow on top pada arah stack bold italic B bold italic C with bold rightwards arrow on top benar diwakili oleh vektor bold i with bold hat on top bold plus bold 6 bold j with bold hat on top bold minus bold k with bold hat on top.

Ingat!

  • 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 sudut straight A open parentheses 4 comma negative 3 comma space 2 close parentheses

Titik sudut straight B open parentheses 2 comma negative 2 comma space 6 close parentheses

Titik sudut straight C open parentheses 3 comma space 4 comma space 5 close parentheses.

Ditanya:

Tunjukan bahwa proyeksi vektor ortogonal stack A C with rightwards arrow on top pada arah stack B C with rightwards arrow on top diwakili oleh vektor i with hat on top plus 6 j with hat on top minus k with hat on top.

Jawab: 

Ruas garis berarah stack A C with rightwards arrow on top adalah sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top end cell equals cell open parentheses table row cell 3 minus 4 end cell row cell 4 plus 3 end cell row cell 5 minus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table attributes columnalign center end attributes row cell negative 1 end cell row 7 row 3 end table close parentheses end cell end table

Ruas garis berarah stack B C with rightwards arrow on top adalah sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell stack B C with rightwards arrow on top end cell equals cell open parentheses table row cell 3 minus 2 end cell row cell 4 plus 2 end cell row cell 5 minus 6 end cell end table close parentheses end cell row blank equals cell open parentheses table attributes columnalign center end attributes row 1 row 6 row cell negative 1 end cell end table close parentheses end cell end table 

Jadi, proyeksi vektor ortogonal stack A C with rightwards arrow on top pada arah stack B C with rightwards arrow on top adalah 

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 stack A C with rightwards arrow on top space. space stack B C with rightwards arrow on top over denominator open vertical bar stack B C with rightwards arrow on top close vertical bar squared end fraction close parentheses stack B C with rightwards arrow on top end cell row cell i with hat on top plus 6 j with hat on top minus k with hat on top end cell equals cell open parentheses fraction numerator negative 1 cross times 1 plus 7 cross times 6 plus 3 cross times open parentheses negative 1 close parentheses over denominator open parentheses square root of 1 squared plus 6 squared plus open parentheses negative 1 close parentheses squared end root close parentheses squared end fraction close parentheses open parentheses table row 1 row 6 row cell negative 1 end cell end table close parentheses end cell row cell i with hat on top plus 6 j with hat on top minus k with hat on top end cell equals cell fraction numerator negative 1 plus 42 minus 3 over denominator open parentheses square root of 1 plus 36 plus 1 end root close parentheses squared end fraction open parentheses table row 1 row 6 row cell negative 1 end cell end table close parentheses end cell row cell i with hat on top plus 6 j with hat on top minus k with hat on top end cell equals cell 38 over 38 open parentheses table row 1 row 6 row cell negative 1 end cell end table close parentheses end cell row cell i with hat on top plus 6 j with hat on top minus k with hat on top end cell equals cell i with hat on top plus 6 j with hat on top minus k with hat on top end cell end table

Dengan demikian, terbukti bahwa proyeksi vektor ortogonal stack A C with rightwards arrow on top pada arah stack B C with rightwards arrow on top benar diwakili oleh vektor i with hat on top plus 6 j with hat on top minus k with hat on top.

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Jawaban terverifikasi

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