the points given in the question is [5,-1,6] and [-2,6,3]
the position vectors associated with these points are given as
a=5i^-j^+6k^,b= -2i^+6j^+3k^
the dot product of these vectors are given as -[tex]a.b=ab*cos\alpha[/tex]
where α is the angle between them .
hence we have cosα=[tex]\frac{a.b}{ab}[/tex]
here [tex]a=\sqrt{5^2+[-1]^2+6^2}[/tex]
and [tex]b=\sqrt{[-2]^2+6^2+3^2}[/tex]
hence a=[tex]\sqrt{62}[/tex] and b=[tex]\sqrt{49}[/tex]
hence cosα=[tex]\frac{a.b}{ab}[/tex]
=5×[-2]+[-1]×6+6×3[tex]\frac{1}{\sqrt{62} *7}[/tex]
=[tex]\frac{2}{\sqrt{62}*7 }[/tex]
=0.0362857507
hence α=2.07947 degree
