Vektorsudut dua vektor
[tex]\sf \cos (\angle {AB, AC}) = \dfrac{AB.AC}{|AB|.|AC|}[/tex]
AB = b - a = (4- 2) , (-4+3) , (3- 4)AB = {2, - 1, - 1}[tex]\sf |AB| = \sqrt{2^2 + (-1)^2 +(-1)^2} = \sqrt 6[/tex]
AC = c - a = (3-2) , (-5 + 3) , (5- 4)AC = (1, - 2, 1)[tex]\sf |AC| = \sqrt{1^2 + (-2)^2 +(1)^2} = \sqrt 6[/tex]
AB. AC = 2 (1) - 1(-2) - 1(1)AB. AC =2+2 - 1 = 3
[tex]\sf \cos (\angle {AB, AC}) = \frac{3}{\sqrt6 . \sqrt 6} = \frac{3}{6} =\frac{1}{2}[/tex]
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Vektor
sudut dua vektor
[tex]\sf \cos (\angle {AB, AC}) = \dfrac{AB.AC}{|AB|.|AC|}[/tex]
AB = b - a = (4- 2) , (-4+3) , (3- 4)
AB = {2, - 1, - 1}
[tex]\sf |AB| = \sqrt{2^2 + (-1)^2 +(-1)^2} = \sqrt 6[/tex]
AC = c - a = (3-2) , (-5 + 3) , (5- 4)
AC = (1, - 2, 1)
[tex]\sf |AC| = \sqrt{1^2 + (-2)^2 +(1)^2} = \sqrt 6[/tex]
AB. AC = 2 (1) - 1(-2) - 1(1)
AB. AC =2+2 - 1 = 3
[tex]\sf \cos (\angle {AB, AC}) = \frac{3}{\sqrt6 . \sqrt 6} = \frac{3}{6} =\frac{1}{2}[/tex]