besar sudut antara vektor a ke vektor b

a = (-3, 2, 4)
b = (2, -3, 3)

ditanyakan α ?

[tex]cos \: \alpha = \frac{ab}{ |a| |b| } [/tex]
[tex]cos \: \alpha \: = \frac{( - 3. \: 2. \: 4) \times (2 . - 3. \: 3)}{ (\sqrt{ { (- 3)}^{2} + {(2)}^{2} + ( {4}^{2} )} \times \sqrt{( {2)}^{2} + { (- 3)}^{2} + {(3)}^{2} } } \\ cos \: \alpha = \frac{( - 6 - 6 + 12)}{ \sqrt{9 + 4 + 16} \times \sqrt{4 + 9 + 9} } \\ cos \: \alpha = \frac{0}{ \sqrt{29} \times \sqrt{22} } \\ cos \: \alpha = 0 [/tex]

α = 90°

D