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Pertanyaan

Tentukan suatu vektor yang besarnya 1 dan tegak lurus terhadap kedua vektor bold italic a equals open parentheses 2 comma space minus 2 comma space 1 close parentheses dan bold italic b equals open parentheses 2 comma space 3 comma space minus 4 close parentheses 

K. Prameswari

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Jawaban

diperoleh bold italic c equals open parentheses 1 third comma space 2 over 3 comma space 2 over 3 close parentheses atau bold italic c equals open parentheses negative 1 third comma space minus 2 over 3 comma space minus 2 over 3 close parentheses 

Pembahasan

Ingat bahwa dua buah vektor bold italic p equals open parentheses p subscript 1 comma space p subscript 2 comma space p subscript 3 close parentheses dan bold italic q equals open parentheses q subscript 1 comma space q subscript 2 comma space q subscript 3 close parentheses dikatakan tegak lurus apabila

table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic p times bold italic q end cell equals 0 row cell open parentheses table row cell p subscript 1 end cell row cell p subscript 2 end cell row cell p subscript 3 end cell end table close parentheses times open parentheses table row cell q subscript 1 end cell row cell q subscript 2 end cell row cell q subscript 3 end cell end table close parentheses end cell equals 0 row cell p subscript 1 q subscript 1 plus p subscript 2 q subscript 2 plus p subscript 3 q subscript 3 end cell equals 0 end table  

Namakan vektor yang akan dicari dengan vektor bold italic c equals open parentheses c subscript 1 comma space c subscript 2 comma space c subscript 3 close parentheses. Diketahui besar vektor bold italic c adalah 1, berarti

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar bold italic c close vertical bar end cell equals 1 row cell square root of c subscript 1 squared plus c subscript 2 squared plus c subscript 3 squared end root end cell equals cell 1 space........... left parenthesis straight i right parenthesis end cell end table  

Selanjutnya, diketahui bahwa 

1. Vektor bold italic c tegak lurus terhadap vektor bold italic a equals open parentheses 2 comma space minus 2 comma space 1 close parentheses, berarti

table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic c times bold italic a end cell equals 0 row cell open parentheses table row cell c subscript 1 end cell row cell c subscript 2 end cell row cell c subscript 3 end cell end table close parentheses times open parentheses table row 2 row cell negative 2 end cell row 1 end table close parentheses end cell equals 0 row cell open parentheses c subscript 1 cross times 2 close parentheses plus open parentheses c subscript 2 cross times open parentheses negative 2 close parentheses close parentheses plus open parentheses c subscript 3 cross times 1 close parentheses end cell equals 0 row cell 2 c subscript 1 minus 2 c subscript 2 plus c subscript 3 end cell equals cell 0 space................ left parenthesis ii right parenthesis end cell end table  

2. Vektor bold italic c tegak lurus terhadap vektor bold italic b equals open parentheses 2 comma space 3 comma space minus 4 close parentheses, berarti

table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic c times bold italic b end cell equals 0 row cell open parentheses table row cell c subscript 1 end cell row cell c subscript 2 end cell row cell c subscript 3 end cell end table close parentheses times open parentheses table row 2 row 3 row cell negative 4 end cell end table close parentheses end cell equals 0 row cell open parentheses c subscript 1 cross times 2 close parentheses plus open parentheses c subscript 2 cross times 3 close parentheses plus open parentheses c subscript 3 cross times open parentheses negative 4 close parentheses close parentheses end cell equals 0 row cell 2 c subscript 1 plus 3 c subscript 2 minus 4 c subscript 3 end cell equals cell 0 space................ left parenthesis iii right parenthesis end cell end table 

Kemudian, akan dilakukan eliminasi antara persamaan (ii) dan (iii) sebagai berikut

table row cell 2 c subscript 1 minus 2 c subscript 2 plus c subscript 3 equals 0 space space space end cell row cell bottom enclose 2 c subscript 1 plus 3 c subscript 2 minus 4 c subscript 3 equals 0 space space minus end enclose end cell row cell negative 5 c subscript 2 plus 5 c subscript 3 equals 0 end cell row cell space space space space space space space space space space space space space space space space 5 c subscript 3 equals 5 c subscript 2 end cell row cell space space space space space space space space space space space space space space space c subscript 3 equals c subscript 2 end cell end table

Subtitusi ke persamaan (ii) diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 c subscript 1 minus 2 c subscript 2 plus c subscript 3 end cell equals 0 row cell 2 c subscript 1 minus 2 c subscript 2 plus c subscript 2 end cell equals 0 row cell 2 c subscript 1 minus c subscript 2 end cell equals 0 row cell 2 c subscript 1 end cell equals cell c subscript 2 end cell end table

Akibatnya c subscript 3 equals c subscript 2 equals 2 c subscript 1. Selanjutnya, subtitusi ke persamaan (i) diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of c subscript 1 squared plus c subscript 2 squared plus c subscript 3 squared end root end cell equals 1 row cell open parentheses square root of c subscript 1 squared plus c subscript 2 squared plus c subscript 3 squared end root close parentheses squared end cell equals cell 1 squared end cell row cell c subscript 1 squared plus c subscript 2 squared plus c subscript 3 squared end cell equals 1 row cell c subscript 1 squared plus open parentheses 2 c subscript 1 close parentheses squared plus open parentheses 2 c subscript 1 close parentheses squared end cell equals 1 row cell c subscript 1 squared plus 4 c subscript 1 squared plus 4 c subscript 1 squared end cell equals 1 row cell 9 c subscript 1 squared end cell equals 1 row cell c subscript 1 squared end cell equals cell 1 over 9 end cell row cell c subscript 1 end cell equals cell plus-or-minus square root of 1 over 9 end root end cell row cell c subscript 1 end cell equals cell plus-or-minus 1 third end cell row blank blank blank end table

  • Untuk c subscript 1 equals 1 third diperoleh

c subscript 2 equals c subscript 3 equals 2 c subscript 1 equals 2 open parentheses 1 third close parentheses equals 2 over 3

Berarti bold italic c equals open parentheses 1 third comma space 2 over 3 comma space 2 over 3 close parentheses

  • Untuk c subscript 1 equals negative 1 third diperoleh

c subscript 2 equals c subscript 3 equals 2 c subscript 1 equals 2 open parentheses negative 1 third close parentheses equals negative 2 over 3

Berarti bold italic c equals open parentheses negative 1 third comma space minus 2 over 3 comma space minus 2 over 3 close parentheses

Dengan demikian, diperoleh bold italic c equals open parentheses 1 third comma space 2 over 3 comma space 2 over 3 close parentheses atau bold italic c equals open parentheses negative 1 third comma space minus 2 over 3 comma space minus 2 over 3 close parentheses 

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