Roboguru

Bagaimana hasil perkalian skalar antara dua vektor kolom  dan ?

Pertanyaan

Bagaimana hasil perkalian skalar antara dua vektor kolom open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell row cell a subscript 3 end cell end table close parentheses dan open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell row cell b subscript 3 end cell end table close parentheses?

Pembahasan Soal:

Hasil perkalian skalar dua vektor dapat ditentukan dengan operasi dot product vektor sebagai berikut. 

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 open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell row cell a subscript 3 end cell end table close parentheses times open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell row cell b subscript 3 end cell end table close parentheses end cell row blank equals cell a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3 end cell end table 

Dengan demikian, hasil perkalian skalar antara dua vektor kolom yang diberikan adalah a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Afrisno

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 07 Juni 2021

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Pertanyaan yang serupa

Diketahui  dan . Jika  dan , maka panjang

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top. b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar squared end cell row cell left parenthesis 4 right parenthesis left parenthesis 9 right parenthesis plus left parenthesis negative 2 right parenthesis left parenthesis 3 right parenthesis plus left parenthesis x right parenthesis left parenthesis 3 right parenthesis end cell equals cell open parentheses square root of 4 squared plus left parenthesis negative 2 right parenthesis squared plus x to the power of 2 end exponent end root close parentheses squared end cell row cell 36 minus 6 plus 3 x end cell equals cell 16 plus 4 plus x squared end cell row cell 30 plus 3 x end cell equals cell 20 plus x squared end cell row cell x squared minus 3 x plus 20 minus 30 end cell equals 0 row cell x squared minus 3 x minus 10 end cell equals 0 row cell left parenthesis x minus 5 right parenthesis left parenthesis x plus 2 right parenthesis end cell equals 0 end table end style 
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank x 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 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank atau end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x 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 minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table end style

Karena  begin mathsize 14px style x less than 0 end style, maka yang memenuhi adalah begin mathsize 14px style x equals negative 2 end style, maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis end cell equals cell open parentheses table row 4 row cell negative 2 end cell row cell negative 2 end cell end table close parentheses minus open parentheses table row 9 row 3 row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 5 end cell row cell negative 5 end cell row cell negative 5 end cell end table close parentheses end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top minus b with rightwards arrow on top close vertical bar end cell equals cell square root of left parenthesis negative 5 right parenthesis squared plus left parenthesis negative 5 right parenthesis squared plus left parenthesis negative 5 right parenthesis squared end root end cell row blank equals cell square root of 25 plus 25 plus 25 end root end cell row blank equals cell square root of 75 end cell row blank equals cell 5 square root of 3 space satuan space panjang space end cell end table end style

Jadi, panjang dari begin mathsize 14px style left parenthesis a with rightwards arrow on top minus b with rightwards arrow on top right parenthesis end style adalah begin mathsize 14px style 5 square root of 3 end style satuan panjang

0

Roboguru

Diketahui vektor , Tentukan .

Pembahasan Soal:

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 2 comma 1 comma 2 close parentheses times open parentheses 4 comma negative 1 comma 3 close parentheses end cell row blank equals cell 2 open parentheses 4 close parentheses plus 1 open parentheses negative 1 close parentheses plus 2 open parentheses 3 close parentheses end cell row blank equals cell 8 minus 1 plus 6 end cell row blank equals 13 end table

Jadi, nilai table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with bar on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with bar on top end cell end table adalah 13

0

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.

0

Roboguru

Diketahui vektor , , dan . Tentukan hasil dari: c.

Pembahasan Soal:

Dengan menerapkan konsep perkalian vektor dot product, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell q with rightwards arrow on top times r with rightwards arrow on top end cell equals cell open parentheses negative 1 close parentheses open parentheses 2 close parentheses plus open parentheses 3 close parentheses open parentheses 1 close parentheses end cell row blank equals cell negative 2 plus 3 end cell row blank equals 1 end table end style 

Selanjutnya, dengan menerapkan perkalian skalar dengan vektor diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top open parentheses q with rightwards arrow on top times r with rightwards arrow on top close parentheses end cell equals cell p with rightwards arrow on top times open parentheses 1 close parentheses end cell row blank equals cell p with rightwards arrow on top end cell row blank equals cell negative 4 i with rightwards arrow on top minus 7 j with rightwards arrow on top end cell end table end style 

Dengan demikian, hasil dari begin mathsize 14px style p with rightwards arrow on top open parentheses q with rightwards arrow on top times r with rightwards arrow on top close parentheses end style adalah begin mathsize 14px style negative 4 i with rightwards arrow on top minus 7 j with rightwards arrow on top end style.

0

Roboguru

Jika vektor saling tegak lurus, maka nilai dari adalah ....

Pembahasan Soal:

Ingat bahwa jika vektor begin mathsize 14px style a with rightwards arrow on top space d a n space b with rightwards arrow on top end stylesaling tegak lurus, maka begin mathsize 14px style a with rightwards arrow on top times b with rightwards arrow on top equals 0 end style.
Oleh karena itu, jika vektor begin mathsize 14px style u with rightwards arrow on top space d a n space v with rightwards arrow on top end style saling tegak lurus, maka begin mathsize 14px style u with rightwards arrow on top times v with rightwards arrow on top end style = 0.

Jadi, jawaban yang tepat adalah C.

 

0

Roboguru

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