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Dua buah vektor yang saling tegak lurus besarnya 5 N dan 15 N. Hasil perkalian titik kedua vektor tersebut adalah….

Pertanyaan

Dua buah vektor yang saling tegak lurus besarnya 5 N dan 15 N. Hasil perkalian titik kedua vektor tersebut adalah….undefined 

  1. 45 Nundefined 

  2. 20 Nundefined 

  3. 10 Nundefined 

  4. 3 Nundefined 

  5. 0 Nundefined 

Pembahasan Video:

Pembahasan Soal:

Perkalian vektor dibagi menjadi 2 jenis, yaitu Perkalian Titik (Dot Product) dan Perkalian Silang (Cross Product). Perkalian Titik dapat dihitung menggunakan rumus:

A times B equals open vertical bar A close vertical bar open vertical bar B close vertical bar cos open parentheses theta close parentheses

Sedangkan Perkalian Silang dapat dihitung dengan rumus:

A cross times B equals open vertical bar A close vertical bar open vertical bar B close vertical bar sin open parentheses theta close parentheses

 

Diketahui: F subscript 1 equals 5 space straight NF subscript 2 equals 15 space straight N, dan theta equals 90 to the power of 0
Ditanyakan: F subscript 1 times F subscript 2 equals ?

begin mathsize 14px style F subscript 1 times F subscript 2 equals open vertical bar F subscript 1 close vertical bar open vertical bar F subscript 2 close vertical bar cos theta space space space space space space space space space space space equals left parenthesis 5 right parenthesis left parenthesis 15 right parenthesis cos space 90 degree space space space space space space space space space space space equals 0 end style 

Jadi, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

Dua buah vektor masing-masing 10 satuan dan 5 satuan. Vektor tersebut satu sama lainnya saling tegak lurus, maka hasil perkalian titiknya adalah...

Pembahasan Soal:

Diketahui :

A with rightwards arrow on top equals 10 space satuan B with rightwards arrow on top equals 5 space satuan 

Ditanya : 

Hasil perkalian titik ?

Penyelesaian:

Perkalian titik atau dot product dua buah vektor merupakan besaran skalar yang sama besarnya dengan hasil kali kedua vektor itu terhadap cosinus sudut apitnya. Kedua vektor tersebut saling tegak lurus, maka sudut apitnya 90º. 

A times B equals open vertical bar A close vertical bar open vertical bar B close vertical bar space cos space alpha A times B equals 10 times 5 times cos space 90 degree A times B equals 50 times 0 A times B equals 0

Dengan demikian, hasil perkalian titiknya adalah nol.

Jadi, jawaban yang benar adalah E.

Roboguru

Jika diberikan vektor-vektor  dan ,  ...

Pembahasan Soal:

Diketahui :

a with rightwards arrow on top equals 4 i minus 8 j plus k b with rightwards arrow on top equals 2 i plus j minus 2 k 

Ditanya :

a.b ?

Penyelesaian :

Perkalian titik vektor untuk vektor satuan a with rightwards arrow on top space dan space b with rightwards arrow on top adalah

a with rightwards arrow on top equals open parentheses table row cell table row 4 row cell negative 8 end cell row 1 end table end cell end table close parentheses space dan space b with rightwards arrow on top equals open parentheses table row 2 row 1 row cell negative 2 end cell end table close parentheses 

Maka a.b

a. b equals open parentheses table row 4 row cell negative 8 end cell row 1 end table close parentheses open parentheses table row 2 row 1 row cell negative 2 end cell end table close parentheses a. b equals 8 minus 8 minus 2 a. b equals negative 2 

Dengan demikian, a.b = -2.

Jadi, jawaban yang benar adalah B.

Roboguru

Diketahui : maka adalah ....

Pembahasan Soal:

Diketahui :

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top space end cell equals cell space 2 straight i with hat on top space plus space 3 straight j with hat on top space minus space 2 straight k with hat on top end cell row cell b with rightwards arrow on top space end cell equals cell space 3 straight i with hat on top space minus space 2 straight j with hat on top space minus space 4 straight k with hat on top end cell end table 

Ditanyakan : a with rightwards arrow on top space • space b with rightwards arrow on top?

Jawab :

Perkalian titik antara dua vektor a dan b (a with rightwards arrow on top space • space b with rightwards arrow on top), menghasilkan besaran skalar sehingga disebut perkalian skalar. Maka nilai a with rightwards arrow on top space • space b with rightwards arrow on top pada soal dapat dihitung :

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top space times space b with rightwards arrow on top space end cell equals cell space open parentheses 2 straight i with hat on top space plus space 3 straight j with hat on top space minus space 2 straight k with hat on top close parentheses space times space open parentheses 3 straight i with hat on top space minus space 2 straight j with hat on top space minus space 4 straight k with hat on top close parentheses end cell row blank equals cell space open parentheses 2 straight i with hat on top space times space 3 straight i with hat on top close parentheses space plus space open parentheses 3 straight j with hat on top times space open parentheses negative space 2 straight j with hat on top close parentheses close parentheses space plus space open parentheses open parentheses negative space 2 straight k with hat on top close parentheses space times space open parentheses negative 4 straight k with hat on top close parentheses close parentheses end cell row blank equals cell space 6 space minus space 6 space plus space 8 end cell row blank equals cell bold space bold 8 bold space bold satuan bold space end cell end table 

Dengan demikian, nilai a with rightwards arrow on top space • space b with rightwards arrow on top adalah 8 satuan.

Jadi, jawaban yang tepat adalah D.

Roboguru

Dua vektor saling tegak lurus, yaitu dan . Tentukan nilai b!

Pembahasan Soal:

Diketahui :

A with rightwards arrow on top space equals space 4 straight i with hat on top space minus space 6 straight j with hat on top space minus space 10 straight k with hat on top B with rightwards arrow on top space equals space 4 straight i with hat on top space minus space 4 straight j with hat on top space plus space b straight k with hat on top 

Ditanyakan : Nilai b?

Jawab :

Dua vektor yang saling tegak lurus, perkalian titik antara dua vektor tersebut bernilai nol. Maka didapatkan perhitungan :

table attributes columnalign right center left columnspacing 0px end attributes row cell A with rightwards arrow on top space times space B with rightwards arrow on top space end cell equals cell space open parentheses 4 straight i with hat on top minus space 6 straight j with hat on top space minus space 10 straight k with hat on top close parentheses space times space open parentheses 4 straight i with hat on top space minus space 4 straight j with hat on top space plus space b straight k with hat on top close parentheses end cell row cell 0 space end cell equals cell space open square brackets 4 straight i with hat on top space times space 4 straight i with hat on top close square brackets space plus space open square brackets open parentheses negative space 6 straight j with hat on top close parentheses space times space open parentheses negative space 4 straight j with hat on top close parentheses close square brackets space plus space open square brackets open parentheses negative space 10 straight k with hat on top space times space b straight k with hat on top close parentheses close square brackets end cell row cell 0 space end cell equals cell space 16 space plus space 24 space minus space 10 b end cell row cell 10 b space end cell equals cell space 40 end cell row cell b space end cell equals cell space 40 over 10 end cell row cell b space end cell equals cell space bold 4 end cell end table  

Dengan demikian, didapatkan nilai b adalah 4.

Roboguru

Dua vektor dan membentuk sudut . Hitung besar cos !

Pembahasan Soal:

Diketahui :

table attributes columnalign right center left columnspacing 0px end attributes row cell P with rightwards arrow on top space end cell equals cell space 3 straight i with hat on top space plus space 4 straight j with hat on top space plus space 5 straight k with hat on top end cell row cell Q with rightwards arrow on top space end cell equals cell space 5 straight i with hat on top space plus space 12 straight j with hat on top space plus space 13 straight k with hat on top end cell end table 

Ditanyakan : Besar cos alpha?

Jawab :

Pada perkalian titik antara dua vektor a dan b (ab), berlaku persamaan :

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top space times space b with rightwards arrow on top space end cell equals cell space b with rightwards arrow on top space times space a with rightwards arrow on top end cell row cell a with rightwards arrow on top space times space b with rightwards arrow on top space end cell equals cell space 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 alpha end cell end table 

Kemudian, nilai cos bold italic alpha dapat dihitung :

table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards arrow cell space open vertical bar P with rightwards arrow on top close vertical bar space equals space open vertical bar 3 straight i with hat on top space plus space 4 straight j with hat on top space plus space 5 straight k with hat on top close vertical bar end cell row blank equals cell space square root of 3 squared space plus space 4 squared space plus space 5 squared end root end cell row blank equals cell space square root of 9 space plus space 16 space plus space 25 end root end cell row blank equals cell space square root of 50 space equals space 5 square root of 2 end cell row blank blank blank row blank rightwards arrow cell space open vertical bar Q with rightwards arrow on top close vertical bar space equals space open vertical bar 5 straight i with hat on top space plus space 12 straight j with hat on top space plus space 13 straight k with hat on top close vertical bar end cell row blank equals cell space square root of 5 squared space plus space 12 squared space plus space 13 squared end root end cell row blank equals cell space square root of 25 space plus space 144 space plus space 169 end root end cell row blank equals cell space square root of 338 space equals space 13 square root of 2 end cell row blank blank blank row blank rightwards arrow cell space P space times space Q space equals space open parentheses 3 straight i space plus space 4 straight j space plus space 5 straight k close parentheses space times space open parentheses 5 straight i space plus space 12 straight j space plus space 13 straight k close parentheses end cell row blank equals cell space open parentheses 3 straight i space times space 5 straight i close parentheses space plus space open parentheses 4 straight j space times space 12 straight j close parentheses space plus space open parentheses 5 straight k space times space 13 straight k close parentheses end cell row blank equals cell space 15 space plus space 48 space plus space 65 end cell row blank equals cell space 128 end cell row blank blank blank row blank rightwards arrow cell space P with rightwards arrow on top space times space Q with rightwards arrow on top space equals space open vertical bar P with rightwards arrow on top close vertical bar open vertical bar Q with rightwards arrow on top close vertical bar space cos space alpha end cell row cell cos space alpha space end cell equals cell space fraction numerator P with rightwards arrow on top space times Q with rightwards arrow on top over denominator open vertical bar P with rightwards arrow on top close vertical bar open vertical bar Q with rightwards arrow on top close vertical bar end fraction end cell row cell cos space alpha space end cell equals cell space fraction numerator 128 over denominator open parentheses 5 square root of 2 close parentheses space open parentheses 13 square root of 2 close parentheses end fraction end cell row cell cos space alpha space end cell equals cell space bold 0 bold comma bold 98 end cell row cell cos space alpha space end cell equals cell space bold 1 end cell end table  

Dengan demikian, besar cos bold italic alpha adalah 0,98 atau apabila dibulatkan sama dengan 1.

Roboguru

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