Vektor x dengan panjang 5​ membuat sudut lancip dengan vektor y​=(3, 4). Jika vektor x diproyeksikan ke vektor y​, panjang proyeksinya 2. Vektor x tersebut adalah ....

Pertanyaan

Vektor x with rightwards arrow on top dengan panjang square root of 5 membuat sudut lancip dengan vektor y with rightwards arrow on top equals open parentheses 3 comma space 4 close parentheses. Jika vektor x with rightwards arrow on top diproyeksikan ke vektor y with rightwards arrow on top, panjang proyeksinya 2. Vektor x with rightwards arrow on top tersebut adalah ....

  1. open parentheses 1 comma space 2 close parentheses space atau space open parentheses 2 over 5 comma space 11 over 5 close parentheses 

  2. open parentheses 2 comma space 1 close parentheses space atau space open parentheses 2 over 5 comma space 11 over 5 close parentheses 

  3. open parentheses 1 comma space 2 close parentheses space atau space open parentheses 4 over 5 square root of 5 comma space 3 over 5 square root of 5 close parentheses 

  4. open parentheses 2 comma space 1 close parentheses space atau space open parentheses 3 over 5 square root of 5 comma space 4 over 5 square root of 5 close parentheses 

  5. open parentheses 2 over 5 comma space 11 over 5 close parentheses space atau space open parentheses 4 over 5 square root of 5 comma space 3 over 5 square root of 5 close parentheses 

P. Tessalonika

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Jika diketahui vektor a with rightwards arrow on top equals open parentheses x subscript 1 comma space y subscript 1 close parentheses dan vektor b with rightwards arrow on top equals open parentheses x subscript 2 comma space y subscript 2 close parentheses, maka panjang proyeksi a with rightwards arrow on top pada/terhadap b with rightwards arrow on top dapat dicari dengan cara berikut :

open vertical bar stack a subscript b with rightwards arrow on top close vertical bar equals fraction numerator a with rightwards arrow on top bullet b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar end fraction 

Dimana : 

 a with rightwards arrow on top bullet b with rightwards arrow on top equals open parentheses x subscript 1 times x subscript 2 close parentheses plus open parentheses y subscript 1 times y subscript 2 close parentheses 

open vertical bar b with rightwards arrow on top close vertical bar equals square root of open parentheses x subscript 2 close parentheses squared plus open parentheses y subscript 2 close parentheses squared end root 

Diketahui :

open vertical bar x with rightwards arrow on top close vertical bar equals square root of 5 

y with rightwards arrow on top equals open parentheses 3 comma space 4 close parentheses

open vertical bar stack x subscript y with rightwards arrow on top close vertical bar equals 2 

Misalkan x with rightwards arrow on top equals open parentheses a comma space b close parentheses maka diperoleh :

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 square root of a squared plus b squared end root end cell row cell square root of 5 end cell equals cell square root of a squared plus b squared end root end cell row 5 equals cell a squared plus b squared space horizontal ellipsis space open parentheses 1 close parentheses end cell end table 

Selanjutnya panjang proyeksi vektor x with rightwards arrow on top pada vektor y with rightwards arrow on top diperoleh dengan rumus :

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar stack x subscript y with rightwards arrow on top close vertical bar end cell equals 2 row cell fraction numerator x with rightwards arrow on top bullet y with rightwards arrow on top over denominator open vertical bar y with rightwards arrow on top close vertical bar end fraction end cell equals 2 row cell fraction numerator 3 a plus 4 b over denominator square root of 3 squared plus 4 squared end root end fraction end cell equals 2 row cell fraction numerator 3 a plus 4 b over denominator square root of 9 plus 16 end root end fraction end cell equals 2 row cell fraction numerator 3 a plus 4 b over denominator square root of 25 end fraction end cell equals 2 row cell fraction numerator 3 a plus 4 b over denominator 5 end fraction end cell equals 2 row cell 3 a plus 4 b end cell equals 10 row cell 3 a end cell equals cell 10 minus 4 b end cell row a equals cell fraction numerator 10 minus 4 b over denominator 3 end fraction end cell end table 

Substitusi nilai a pada persamaan (1) diperoleh :

table attributes columnalign right center left columnspacing 0px end attributes row 5 equals cell a squared plus b squared end cell row 5 equals cell open parentheses fraction numerator 10 minus 4 b over denominator 3 end fraction close parentheses squared plus b squared end cell row 5 equals cell fraction numerator 100 minus 80 b plus 16 b squared over denominator 9 end fraction plus b squared end cell row 45 equals cell 100 minus 80 b plus 16 b squared plus 9 b squared end cell row 0 equals cell 25 b squared minus 80 b plus 100 minus 45 end cell row 0 equals cell 25 b squared minus 80 b plus 55 end cell row 0 equals cell 5 b squared minus 16 b plus 11 end cell row 0 equals cell open parentheses 5 b minus 11 close parentheses open parentheses b minus 1 close parentheses end cell end table 

b equals 11 over 5 space atau space b equals 1

Menentukan nilai a yaitu :

table attributes columnalign right center left columnspacing 0px end attributes row a equals cell fraction numerator 10 minus 4 b over denominator 3 end fraction end cell row blank equals cell fraction numerator 10 minus 4 open parentheses begin display style 11 over 5 end style close parentheses over denominator 3 end fraction end cell row blank equals cell fraction numerator 10 minus begin display style 44 over 5 end style over denominator 3 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 50 minus 44 over denominator 5 end fraction end style over denominator 3 end fraction end cell row blank equals cell 6 over 15 end cell row blank equals cell 2 over 5 end cell end table  atau  table attributes columnalign right center left columnspacing 0px end attributes row a equals cell fraction numerator 10 minus 4 b over denominator 3 end fraction end cell row blank equals cell fraction numerator 10 minus 4 open parentheses 1 close parentheses over denominator 3 end fraction end cell row blank equals cell fraction numerator 10 minus 4 over denominator 3 end fraction end cell row blank equals cell 6 over 3 end cell row blank equals 2 end table 

Dikatakan bahwa vektor x with rightwards arrow on top membuat sudut lancip dengan vektor y with rightwards arrow on top maka berlaku 0 less than cos space theta less than 1. Kita akan cek apakah berlaku untuk open parentheses 2 over 5 comma space 11 over 5 close parentheses atau open parentheses 2 comma space 1 close parentheses.

  • Untuk vektor x with rightwards arrow on top equals open parentheses 2 over 5 comma space 11 over 5 close parentheses dan vektor y with rightwards arrow on top equals open parentheses 3 comma space 4 close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator x with rightwards arrow on top bullet y with rightwards arrow on top over denominator open vertical bar x with rightwards arrow on top close vertical bar open vertical bar y with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 2 over 5 end style cross times 3 close parentheses plus open parentheses begin display style 11 over 5 end style cross times 4 close parentheses over denominator open parentheses square root of 5 close parentheses open parentheses 5 close parentheses end fraction end cell row blank equals cell fraction numerator begin display style 6 over 5 end style plus begin display style 44 over 5 end style over denominator 5 square root of 5 end fraction end cell row blank equals cell fraction numerator begin display style 50 over 5 end style over denominator 5 square root of 5 end fraction end cell row blank equals cell fraction numerator 10 over denominator 5 square root of 5 end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 5 end fraction end cell row blank equals cell fraction numerator 2 over denominator 2 comma 236 end fraction end cell row blank equals cell 0 comma 89 space open parentheses memenuhi close parentheses end cell end table 

  • Untuk vektor x with rightwards arrow on top equals open parentheses 2 comma space 1 close parentheses dan vektor y with rightwards arrow on top equals open parentheses 3 comma space 4 close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator x with rightwards arrow on top bullet y with rightwards arrow on top over denominator open vertical bar x with rightwards arrow on top close vertical bar open vertical bar y with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 2 cross times 3 end style close parentheses plus open parentheses begin display style 1 cross times 4 end style close parentheses over denominator open parentheses square root of 5 close parentheses open parentheses 5 close parentheses end fraction end cell row blank equals cell fraction numerator 6 plus 4 over denominator 5 square root of 5 end fraction end cell row blank equals cell fraction numerator 10 over denominator 5 square root of 5 end fraction end cell row blank equals cell 0 comma 89 space open parentheses memenuhi close parentheses end cell end table 

Dengan demikian, vektor x with rightwards arrow on top tersebut adalah open parentheses 2 comma space 1 close parentheses atau open parentheses 2 over 5 comma space 11 over 5 close parentheses.

Oleh karena itu, jawaban yang tepat adalah B.

162

5.0 (8 rating)

Britania AR

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