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Jika vektor  dan  hasil ...

Pertanyaan

Jika vektor a with bar on top equals 4 i with bar on top minus 5 j with bar on top dan b with bar on top equals i with bar on top plus 6 j with bar on top hasil a with bar on top times b with bar on top equals...

  1. 34 

  2. negative 7 

  3. negative 34 

  4. 26 

  5. negative 26 

Pembahasan Soal:

Jika vektor a with bar on top equals 4 i with bar on top minus 5 j with bar on top dan b with bar on top equals i with bar on top plus 6 j with bar on top, berdasarkan operasi dot product maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with bar on top times b with bar on top end cell equals cell open parentheses 4 comma space minus 5 close parentheses times open parentheses 1 comma space 6 close parentheses end cell row blank equals cell open parentheses 4 times 1 close parentheses plus open parentheses negative 5 times 6 close parentheses end cell row blank equals cell 4 minus 30 end cell row blank equals cell negative 26 end cell end table 

Oleh karena itu jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Amamah

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika vektor  dan , hasil  adalah ....

Pembahasan Soal:

Perkalian titik dari dua vektor akan menghasilkan skalar. 

table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards harpoon with barb upwards on top times v with rightwards harpoon with barb upwards on top end cell equals cell open parentheses table row 2 row cell negative 1 end cell end table close parentheses open parentheses table row cell negative 3 end cell row cell negative 4 end cell end table close parentheses end cell row blank equals cell 2 times open parentheses negative 3 close parentheses plus open parentheses negative 1 close parentheses times open parentheses negative 4 close parentheses end cell row blank equals cell negative 6 plus 4 end cell row blank equals cell negative 2 end cell end table 

Jadi, hasil u with rightwards harpoon with barb upwards on top times v with rightwards harpoon with barb upwards on top equals negative 2 

Roboguru

Diketahui vektor  dan . Jika hasil , tentukan: a. nilai   b. hasil

Pembahasan Soal:

Nilai size 14px a adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top times v with rightwards arrow on top end cell equals cell negative 6 end cell row cell open parentheses table row cell negative 5 end cell row 7 end table close parentheses times open parentheses table row 4 row a end table close parentheses end cell equals cell negative 6 end cell row cell open parentheses negative 5 close parentheses times 4 plus 7 a end cell equals cell negative 6 end cell row cell negative 20 plus 7 a end cell equals cell negative 6 end cell row cell 7 a end cell equals cell negative 6 plus 20 end cell row a equals cell 14 over 7 end cell row a equals 2 end table end style 

Sehingga, hasil begin mathsize 14px style open parentheses 2 u with rightwards arrow on top plus 3 v with rightwards arrow on top close parentheses times open parentheses u with rightwards arrow on top minus 2 v with rightwards arrow on top close parentheses end style adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 2 u with rightwards arrow on top plus 3 v with rightwards arrow on top right parenthesis times left parenthesis u with rightwards arrow on top minus 2 v with rightwards arrow on top right parenthesis end cell equals cell open parentheses 2 open parentheses table row cell negative 5 end cell row 7 end table close parentheses plus 3 open parentheses table row 4 row 2 end table close parentheses close parentheses times open parentheses open parentheses table row cell negative 5 end cell row 7 end table close parentheses minus 2 open parentheses table row 4 row 2 end table close parentheses close parentheses end cell row blank equals cell open parentheses open parentheses table row cell negative 10 end cell row 14 end table close parentheses plus open parentheses table row 12 row 6 end table close parentheses close parentheses times open parentheses open parentheses table row cell negative 5 end cell row 7 end table close parentheses minus open parentheses table row 8 row 4 end table close parentheses close parentheses end cell row blank equals cell open parentheses table row 2 row 20 end table close parentheses times open parentheses table row cell negative 13 end cell row 3 end table close parentheses end cell row blank equals cell 2 times open parentheses negative 13 close parentheses plus 20 times 3 end cell row blank equals cell negative 26 plus 60 end cell row blank equals 34 end table end style 

Jadi, begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis 2 u with rightwards arrow on top plus 3 v with rightwards arrow on top right parenthesis times left parenthesis u with rightwards arrow on top minus 2 v with rightwards arrow on top right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 34 end table end style.

Roboguru

Perhatikan gambar berikut!   Dengan aturan cosinus yaitu:   tunjukkan bahwa !

Pembahasan Soal:

Jika terdapat titik straight A open parentheses a subscript 1 comma space a subscript 2 close parentheses dan titik straight B open parentheses b subscript 1 comma space b subscript 2 close parentheses, maka berlaku:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell open parentheses a subscript 1 comma space a subscript 2 close parentheses comma space b with rightwards arrow on top equals open parentheses b subscript 1 comma space b subscript 2 close parentheses end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open parentheses a subscript 1 comma space a subscript 2 close parentheses times open parentheses b subscript 1 comma space b subscript 2 close parentheses end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 end cell row cell AB with rightwards arrow on top end cell equals cell b with rightwards arrow on top minus a with rightwards arrow on top end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of a subscript 1 squared plus a subscript 2 squared end root end cell end table  

Keterangan:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell perkalian space titik space a with rightwards arrow on top space dan space b with rightwards arrow on top end cell row cell AB with rightwards arrow on top end cell equals cell vektor space posisi space dari space titik space straight B space ke space titik space straight A end cell row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell panjang space a with rightwards arrow on top end cell end table 

Berdasarkan gambar, diketahui:

straight O open parentheses 0 comma space 0 close parentheses comma space straight P open parentheses a subscript 1 comma space a subscript 2 close parentheses comma space straight Q open parentheses b subscript 1 comma space b subscript 2 close parentheses space stack O P with rightwards arrow on top equals a with rightwards arrow on top stack O Q with rightwards arrow on top equals b with rightwards arrow on top  

Bukti:

begin mathsize 11px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar PQ with rightwards arrow on top close vertical bar squared end cell equals cell open vertical bar OP with rightwards arrow on top close vertical bar squared plus open vertical bar OQ with rightwards arrow on top close vertical bar squared minus 2 open vertical bar OP with rightwards arrow on top close vertical bar open vertical bar OQ with rightwards arrow on top close vertical bar cos space theta end cell row cell open parentheses square root of open parentheses b subscript 1 minus a subscript 1 close parentheses squared plus open parentheses b subscript 2 minus a subscript 2 close parentheses squared end root close parentheses squared end cell equals cell open parentheses square root of a subscript 1 squared plus a subscript 2 squared end root close parentheses squared plus open parentheses square root of b subscript 1 squared plus b subscript 2 squared end root close parentheses squared minus 2 times square root of a subscript 1 squared plus a subscript 2 squared end root times square root of b subscript 1 squared plus b subscript 2 squared end root times cos space theta end cell row cell open parentheses b subscript 1 minus a subscript 1 close parentheses squared plus open parentheses b subscript 2 minus a subscript 2 close parentheses squared end cell equals cell a subscript 1 squared plus a subscript 2 squared plus b subscript 1 squared plus b subscript 2 squared minus 2 times square root of a subscript 1 squared plus a subscript 2 squared end root times square root of b subscript 1 squared plus b subscript 2 squared end root times cos space theta end cell row cell b subscript 1 squared minus 2 a subscript 1 b subscript 1 plus a subscript 1 squared plus b subscript 2 squared minus 2 a subscript 2 b subscript 2 plus a subscript 2 squared end cell equals cell a subscript 1 squared plus a subscript 2 squared plus b subscript 1 squared plus b subscript 2 squared minus 2 times square root of a subscript 1 squared plus a subscript 2 squared end root times square root of b subscript 1 squared plus b subscript 2 squared end root times cos space theta end cell row cell negative 2 a subscript 1 b subscript 1 minus 2 a subscript 2 b subscript 2 end cell equals cell negative 2 times square root of a subscript 1 squared plus a subscript 2 squared end root times square root of b subscript 1 squared plus b subscript 2 squared end root times cos space theta end cell row cell negative 2 open parentheses a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 close parentheses end cell equals cell negative 2 times square root of a subscript 1 squared plus a subscript 2 squared end root times square root of b subscript 1 squared plus b subscript 2 squared end root times cos space theta end cell row cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 end cell equals cell square root of a subscript 1 squared plus a subscript 2 squared end root times square root of b subscript 1 squared plus b subscript 2 squared end root times cos space theta end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos space theta end cell end table end style 

Jadi, terbukti bahwa a with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space theta.

Roboguru

Tentukan  jika diketahui:

Pembahasan Soal:

Dengan memperhatikan konsep dot product dan sifat-sifatnya, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell a times b end cell equals cell open parentheses table row 3 row cell negative 4 end cell end table close parentheses times open parentheses table row 3 row 1 end table close parentheses end cell row blank equals cell open parentheses table row 9 row cell negative 4 end cell end table close parentheses end cell row blank equals cell 9 i minus 4 j end cell end table

Oleh karena itu, table attributes columnalign right center left columnspacing 0px end attributes row cell a times b end cell equals cell 9 i minus 4 j end cell end table.

Roboguru

Diketahui koordinat titik , , dan . Hasil perkalian  adalah ....

Pembahasan Soal:

Gunakan konsep dot produk pada vektor.

Diketahui koordinat titik straight P open parentheses 3 comma space minus 1 comma space 4 close parenthesesstraight Q open parentheses negative 2 comma space 1 comma space minus 5 close parentheses, dan straight R open parentheses 4 comma space 3 comma space minus 2 close parentheses. Akan ditentukan hasil perkalian PQ with rightwards arrow on top times PR with rightwards arrow on top.

*Menentukan PQ with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell PQ with rightwards arrow on top end cell equals cell straight Q minus straight P end cell row blank equals cell open parentheses table row cell negative 2 end cell row 1 row cell negative 5 end cell end table close parentheses minus open parentheses table row 3 row cell negative 1 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 minus 3 end cell row cell 1 minus open parentheses negative 1 close parentheses end cell row cell negative 5 minus 4 end cell end table close parentheses end cell row cell PQ with rightwards arrow on top end cell equals cell open parentheses table row cell negative 5 end cell row 2 row cell negative 9 end cell end table close parentheses end cell end table

*Menentukan PR with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell PR with rightwards arrow on top end cell equals cell straight R minus straight P end cell row blank equals cell open parentheses table row 4 row 3 row cell negative 2 end cell end table close parentheses minus open parentheses table row 3 row cell negative 1 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell 4 minus 3 end cell row cell 3 minus open parentheses negative 1 close parentheses end cell row cell negative 2 minus 4 end cell end table close parentheses end cell row cell PR with rightwards arrow on top end cell equals cell open parentheses table row 1 row 4 row cell negative 6 end cell end table close parentheses end cell end table

Sehingga nilai PQ with rightwards arrow on top times PR with rightwards arrow on top dapat dihitung sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell PQ with rightwards arrow on top times PR with rightwards arrow on top end cell equals cell open parentheses table row cell negative 5 end cell row 2 row cell negative 9 end cell end table close parentheses times open parentheses table row 1 row 4 row cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses negative 5 close parentheses open parentheses 1 close parentheses plus open parentheses 2 close parentheses open parentheses 4 close parentheses plus open parentheses negative 9 close parentheses open parentheses negative 6 close parentheses end cell row blank equals cell negative 5 plus 8 plus 54 end cell row cell PQ with rightwards arrow on top times PR with rightwards arrow on top end cell equals 57 end table

Diperoleh hasil perkalian PQ with rightwards arrow on top times PR with rightwards arrow on top adalah 57.

Jadi, tidak ada jawaban yang tepat.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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