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Pertanyaan

  1. Tentukan dua vektor satuan yang masing-masing tegak lurus pada vektor bold italic j with italic hat on top plus 4 bold italic k with bold hat on top dan 3 bold italic i with bold hat on top plus 2 bold italic j with italic hat on top plus 4 bold italic k with bold hat on top.
  2. Tentukan juga sudut antara masing-masing vektor satuan ini dengan vektor 1 half open parentheses bold italic i with bold hat on top minus bold italic j with italic hat on top minus bold italic k with bold hat on top close parentheses.

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Jawaban terverifikasi

Jawaban

 sudut antara masing-masing vektor satuan ini dengan vektor 1 half open parentheses bold italic i with bold hat on top minus bold italic j with italic hat on top minus bold italic k with bold hat on top close parentheses adalah theta subscript 1 equals 54 comma 74 degree dan theta subscript 2 equals 125 comma 26 degree.

Pembahasan

Jawaban yang benar untuk bagian a adalah bold italic v subscript bold 1 equals 4 over 13 bold italic i with bold hat on top minus 12 over 13 bold italic j with bold hat on top plus 3 over 13 bold italic k with bold hat on top dan bold italic v subscript bold 2 equals negative 4 over 13 bold italic i with bold hat on top plus 12 over 13 bold italic j with bold hat on top minus 3 over 13 bold italic k with bold hat on top.

Jawaban yang benar untuk bagian b adalah theta subscript 1 equals 54 comma 74 degree dan theta subscript 2 equals 125 comma 26 degree.

Bagian a:

Ingat bahwa hasil perkalian dot product antara dua vektor bold italic u equals u subscript 1 bold italic i with bold hat on top plus u subscript 2 bold italic j with bold hat on top plus u subscript 3 bold italic k with bold hat on top dan bold italic v equals v subscript 1 bold italic i with bold hat on top plus v subscript 2 bold italic j with bold hat on top plus v subscript 3 bold italic k with bold hat on top yang saling tegak lurus adalah 0. Sehingga, 

table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic u times bold italic v end cell equals 0 row cell u subscript 1 v subscript 1 plus u subscript 2 v subscript 2 plus u subscript 3 v subscript 3 end cell equals 0 end table

Panjang vektor tiga dimensi bold italic u equals u subscript 1 bold italic i with bold hat on top plus u subscript 2 bold italic j with bold hat on top plus u subscript 3 bold italic k with bold hat on top dapat dihitung menggunakan rumus berikut:

open vertical bar bold italic u close vertical bar equals square root of u subscript 1 squared plus u subscript 2 squared plus u subscript 3 squared end root

Misalkan vektor satuan yang tegak lurus terhadap vektor bold italic j with italic hat on top plus 4 bold italic k with bold hat on top dan 3 bold italic i with bold hat on top plus 2 bold italic j with italic hat on top plus 4 bold italic k with bold hat on top adalah vektor a bold italic i with bold hat on top plus b bold italic j with italic hat on top plus c bold italic k with bold hat on top. Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a bold italic i with bold hat on top plus b bold italic j with italic hat on top plus c bold italic k with bold hat on top close parentheses times open parentheses bold italic j with italic hat on top plus 4 bold italic k with bold hat on top close parentheses end cell equals 0 row cell open parentheses a close parentheses open parentheses 0 close parentheses plus open parentheses b close parentheses open parentheses 1 close parentheses plus open parentheses c close parentheses open parentheses 4 close parentheses end cell equals 0 row cell b plus 4 c end cell equals 0 row b equals cell negative 4 c end cell end table

dan

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a bold italic i with bold hat on top plus b bold italic j with italic hat on top plus c bold italic k with bold hat on top close parentheses times open parentheses 3 bold italic i with bold hat on top plus 2 bold italic j with italic hat on top plus 4 bold italic k with bold hat on top close parentheses end cell equals 0 row cell open parentheses a close parentheses open parentheses 3 close parentheses plus open parentheses b close parentheses open parentheses 2 close parentheses plus open parentheses c close parentheses open parentheses 4 close parentheses end cell equals 0 row cell 3 a plus 2 b plus 4 c end cell equals 0 end table

Substitusi b equals negative 4 c ke persamaan 3 a plus b plus 4 c equals 0:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 a plus 2 b plus 4 c end cell equals 0 row cell 3 a plus 2 open parentheses negative 4 c close parentheses plus 4 c end cell equals 0 row cell 3 a minus 8 c plus 4 c end cell equals 0 row cell 3 a minus 4 c end cell equals 0 row cell 3 a end cell equals cell 4 c end cell row a equals cell 4 over 3 c end cell end table

Karena vektor a bold italic i with bold hat on top plus b bold italic j with italic hat on top plus c bold italic k with bold hat on top merupakan vektor satuan, maka panjang vektornya adalah 1. Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a bold i with bold hat on top plus b bold j with bold hat on top plus c bold k with bold hat on top close vertical bar end cell equals 1 row cell square root of a squared plus b squared plus c squared end root end cell equals 1 row cell a squared plus b squared plus c squared end cell equals 1 end table

Substitusi b equals negative 4 c dan a equals 4 over 3 c ke persamaan a squared plus b squared plus c squared equals 1:

table attributes columnalign right center left columnspacing 0px end attributes row cell a squared plus b squared plus c squared end cell equals 1 row cell open parentheses 4 over 3 c close parentheses squared plus open parentheses negative 4 c close parentheses squared plus c squared end cell equals 1 row cell 16 over 9 c squared plus 16 c squared plus c squared end cell equals 1 row cell fraction numerator 16 c squared plus 144 c squared plus 9 c squared over denominator 9 end fraction end cell equals 1 row cell 169 c squared end cell equals 9 row cell c squared end cell equals cell 9 over 169 end cell row c equals cell plus-or-minus square root of 9 over 169 end root end cell row blank equals cell plus-or-minus 3 over 13 end cell end table

Dari perhitungan di atas, diperoleh c subscript 1 equals 3 over 13 dan c subscript 2 equals negative 3 over 13. Untuk c subscript 1 equals 3 over 13, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell b subscript 1 end cell equals cell negative 4 c subscript 1 end cell row blank equals cell negative 4 open parentheses 3 over 13 close parentheses end cell row blank equals cell negative 12 over 13 end cell row blank blank blank row cell a subscript 1 end cell equals cell 4 over 3 c subscript 1 end cell row blank equals cell 4 over 3 open parentheses 3 over 13 close parentheses end cell row blank equals cell 4 over 13 end cell end table

Untuk c subscript 2 equals negative 3 over 13, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell b subscript 2 end cell equals cell negative 4 c subscript 2 end cell row blank equals cell negative 4 open parentheses negative 3 over 13 close parentheses end cell row blank equals cell 12 over 13 end cell row blank blank blank row cell a subscript 1 end cell equals cell 4 over 3 c subscript 1 end cell row blank equals cell 4 over 3 open parentheses negative 3 over 13 close parentheses end cell row blank equals cell negative 4 over 13 end cell end table

Sehingga vektor yang tegak lurus terhadap vektor bold italic j with italic hat on top plus 4 bold italic k with bold hat on top dan 3 bold italic i with bold hat on top plus 2 bold italic j with italic hat on top plus 4 bold italic k with bold hat on top, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic v subscript bold 1 end cell equals cell 4 over 13 bold italic i with bold hat on top minus 12 over 13 bold italic j with bold hat on top plus 3 over 13 bold italic k with bold hat on top end cell row blank blank blank row cell bold italic v subscript bold 2 end cell equals cell negative 4 over 13 bold italic i with bold hat on top plus 12 over 13 bold italic j with bold hat on top minus 3 over 13 bold italic k with bold hat on top end cell end table

Dengan demikian, dua vektor yang tegak lurus terhadap vektor bold italic j with italic hat on top plus 4 bold italic k with bold hat on top dan 3 bold italic i with bold hat on top plus 2 bold italic j with italic hat on top plus 4 bold italic k with bold hat on top adalah  bold italic v subscript bold 1 equals 4 over 13 bold italic i with bold hat on top minus 12 over 13 bold italic j with bold hat on top plus 3 over 13 bold italic k with bold hat on top dan bold italic v subscript bold 2 equals negative 4 over 13 bold italic i with bold hat on top plus 12 over 13 bold italic j with bold hat on top minus 3 over 13 bold italic k with bold hat on top.

 

Bagian b:

Ingat bahwa sudut antara dua vektor bold italic u equals u subscript 1 bold italic i with bold hat on top plus u subscript 2 bold italic j with bold hat on top plus u subscript 3 bold italic k with bold hat on top dan bold italic v equals v subscript 1 bold italic i with bold hat on top plus v subscript 2 bold italic j with bold hat on top plus v subscript 3 bold italic k with bold hat on top dapat dihitung menggunakan rumus berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator bold italic u times bold italic v over denominator open vertical bar bold italic u close vertical bar open vertical bar bold italic v close vertical bar end fraction end cell row blank equals cell fraction numerator u subscript 1 v subscript 1 plus u subscript 2 v subscript 2 plus u subscript 3 v subscript 3 over denominator square root of open parentheses u subscript 1 squared plus u subscript 2 squared plus u subscript 3 squared close parentheses open parentheses v subscript 1 squared plus v subscript 2 squared plus v subscript 3 squared close parentheses end root end fraction end cell end table

Sudut antara bold italic v subscript bold 1 equals 4 over 13 bold italic i with bold hat on top minus 12 over 13 bold italic j with bold hat on top plus 3 over 13 bold italic k with bold hat on top dan 1 half open parentheses bold italic i with bold hat on top minus bold italic j with italic hat on top minus bold italic k with bold hat on top close parentheses, yaitu:

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Sudut antara bold italic v subscript bold 2 equals negative 4 over 13 bold italic i with bold hat on top plus 12 over 13 bold italic j with bold hat on top minus 3 over 13 bold italic k with bold hat on top dan 1 half open parentheses bold italic i with bold hat on top minus bold italic j with italic hat on top minus bold italic k with bold hat on top close parentheses, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta subscript 2 end cell equals cell fraction numerator open parentheses negative begin display style 4 over 13 end style bold italic i with bold hat on top plus begin display style 12 over 13 end style bold italic j with bold hat on top minus begin display style 3 over 13 end style bold italic k with bold hat on top close parentheses times open parentheses begin display style 1 half bold italic i with bold hat on top minus 1 half bold italic j with bold hat on top minus 1 half bold italic k with bold hat on top end style close parentheses over denominator open vertical bar begin display style negative 4 over 13 bold italic i with bold hat on top plus 12 over 13 bold italic j with bold hat on top minus 3 over 13 bold italic k with bold hat on top end style close vertical bar open vertical bar begin bold style begin display style straight 1 over straight 2 end style bold italic i with bold italic hat on top straight minus begin display style straight 1 over straight 2 end style bold italic j with bold italic hat on top straight minus begin display style straight 1 over straight 2 end style bold italic k with bold italic hat on top end style close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses begin display style negative 4 over 13 end style close parentheses open parentheses begin display style 1 half end style close parentheses plus open parentheses begin display style 12 over 13 end style close parentheses open parentheses negative begin display style 1 half end style close parentheses plus open parentheses begin display style negative 3 over 13 end style close parentheses open parentheses negative begin display style 1 half end style close parentheses over denominator square root of open parentheses 1 close parentheses open parentheses open parentheses begin display style 1 half end style close parentheses squared plus open parentheses begin display style 1 half end style close parentheses squared plus open parentheses begin display style 1 half end style close parentheses squared close parentheses end root end fraction end cell row blank equals cell fraction numerator negative begin display style 4 over 26 end style minus begin display style 12 over 26 end style plus begin display style 3 over 26 end style over denominator square root of 3 cross times open parentheses begin display style 1 half end style close parentheses squared end root end fraction end cell row blank equals cell fraction numerator begin display style negative 13 over 26 end style over denominator begin display style 1 half end style square root of 2 end fraction end cell row blank equals cell fraction numerator negative up diagonal strike begin display style 1 half end style end strike over denominator up diagonal strike begin display style 1 half end style end strike square root of 3 end fraction end cell row blank equals cell negative fraction numerator 1 over denominator square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell negative 1 third square root of 3 end cell row cell theta subscript 2 end cell equals cell cos to the power of negative 1 end exponent space open parentheses negative 1 third square root of 3 close parentheses end cell row blank equals cell 125 comma 26 degree end cell end table 

Dengan demikian, sudut antara masing-masing vektor satuan ini dengan vektor 1 half open parentheses bold italic i with bold hat on top minus bold italic j with italic hat on top minus bold italic k with bold hat on top close parentheses adalah theta subscript 1 equals 54 comma 74 degree dan theta subscript 2 equals 125 comma 26 degree.

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