Pertanyaan

Dua vektor A = 6 i + 8 j + 10 k d an B = 5 i + 12 j + 13 k membentuk sudut α .Besar cos adalah ....

Dua vektor $A = 6 i + 8 j + 10 k d an B = 5 i + 12 j + 13 k$ membentuk sudut $α$ . Besar cos alpha adalah ....

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Diketahui : Ditanyakan : Besar cos ? Jawab : Pada perkalian titik antara dua vektor a dan b ( ), berlaku persamaan : Kemudian, nilaicos dapat dihitung : Dengan demikian, besarcos adalah . Jadi, jawaban yang tepat adalah E.

Diketahui :

Ditanyakan : Besar cos alpha ?

Jawab :

Pada perkalian titik antara dua vektor a dan b ( table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top space times space b with rightwards arrow on top end cell end table ), berlaku persamaan :

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 A with rightwards arrow on top close vertical bar space equals space open vertical bar 6 straight i with hat on top space plus space 8 straight j with hat on top space plus space 10 straight k with hat on top close vertical bar end cell row blank equals cell space square root of 6 squared space plus space 8 squared space plus space 10 squared end root end cell row blank equals cell space square root of 36 space plus space 64 space plus space 100 end root end cell row blank equals cell space square root of 200 space equals space 10 square root of 2 end cell row blank blank blank row blank rightwards arrow cell space open vertical bar B with rightwards arrow on top close vertical bar space equals space open vertical bar 5 stack straight i space with hat on top plus space 12 stack straight j space with hat on top 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 stack P space with rightwards arrow on top times space Q with rightwards arrow on top space equals space open parentheses 6 straight i with hat on top space plus space 8 straight j with hat on top space plus space 10 straight k with hat on top close parentheses space times space open parentheses 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 parentheses end cell row blank equals cell space open parentheses 6 stack straight i space with hat on top times space 5 straight i with hat on top close parentheses space plus space open parentheses 8 straight j with hat on top space times space 12 straight j with hat on top close parentheses space plus space open parentheses 10 straight k with hat on top space times space 13 straight k with hat on top close parentheses end cell row blank equals cell space 30 space plus space 96 space plus space 130 end cell row blank equals cell space 256 end cell row blank blank blank row blank rightwards arrow cell stack 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 space 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 256 over denominator open parentheses 10 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 64 over bold 65 end cell end table$

Dengan demikian, besar cos adalah .

Jadi, jawaban yang tepat adalah E.