besar sudut antara vektor a=2i - j +3k dan b=i + 3j - 2k adalah

diketahui: a=(2,-1,3) b=(1,3,-2)
penyelesaian:
a.b= (2,-1,3)x(1,3,-2)
=(2x1)+(-1x1)+(3x(-2))
=2+(-1)+(-6)=-5
|a|=√2²+(-1)²+3²= √4+1+9=√14
|b|=√1²+3²+(-2)²= √1+9+4=√14
cos β= a.b/|a|.|b|
cos β= -5/√14.√14
cos β=-5/14
arc cos -5/14= 110° 55' 28.766"
= 110.92465717°=>111° (dibulatkan)

maaf jika salah

a = 2i - j + 3k => |a| = √(2^2 + (-1)^2 + 3^2) = √(4 + 1 + 9) = √14
b = i + 3j - 2k => |b| = √(1^2 + 3^2 + (-2)^2) = √(1 + 9 + 4) = √14
a . b = |a| |b| cos @
2(1) + (-1)(3) + 3(-2) = √14 . √14 cos @
2 - 3 - 6 = 14 cos @
cos @ = -7/14
Cos @ = -1/2
Cos @ = cos 120°
@ = 120°