Question

diberikan vektor u( 2 3 -6) dan (v -3 2 0 )tentukan besar sudut antara vektor u dan v​

Answer 1

jwb

cos(theta) = u . v : |u| x |v| (dot product)

ket:

u . v = dot product vektor u dan v

|u| dan |v| = magnitudo

u . v = (2 x -3) + (3 x 2) + (-6 x 0)

= -6 + 6 + 0

= 0

|u| = [tex]\sqrt{}[/tex](2^2 + 3^2 + (-6)^2) (magnitudo)

= [tex]\sqrt{}[/tex](49)

= 7

|v| = [tex]\sqrt{}[/tex](-3)^2 + 2^2 + 0^2)

= [tex]\sqrt{}[/tex](13)

cos(theta) = 0 : 7 x [tex]\sqrt{}[/tex](13)

= 0

theta = arccos(0)

= 90

besar sudut antara vektor u dan v = 90 derajat