Помогите..................

(AB) = (3-4; -1-2; 6-4) = (-1; -3; 2); AC = (2-4; 1-2; 0-4) = (-2; -1; -4)
Угол между AB и AC
$cos(AB;AC)= \frac{X1*X2+Y1*Y2+Z1*Z2}{ \sqrt{X1^2+Y1^2+Z1^2}*\sqrt{X2^2+Y2^2+Z2^2} } =$
$= \frac{(-1)(-2)+(-3)(-1)+2(-4)}{ \sqrt{(-1)^2+(-3)^2+2^2}* \sqrt{(-2)^2+(-1)^2+(-4)^2} } = \frac{2+3-8}{ \sqrt{1+9+4}* \sqrt{4+1+16} } =$
$=\frac{-3}{ \sqrt{14}* \sqrt{21} } = \frac{-3}{ \sqrt{2}* \sqrt{7}* \sqrt{3}* \sqrt{7}}=- \frac{3}{7 \sqrt{6} } = - \frac{3 \sqrt{6} }{7*6}=- \frac{ \sqrt{6} }{14}$
Угол (AB; AC) = pi - arccos(√6/14)