Найти угол между векторами a= (2; -4), b= (-6; 2).

$cos \alpha = \frac{ab}{ |a| |b| } \\ ab = 2 \times ( - 6) - 4 \times 2 = - 12 - 8 = - 20 \\ |a| = \sqrt{ {2}^{2} + {( - 4)}^{2} } = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5} \\ |b| = \sqrt{ {( - 6)}^{2} + {2}^{2} } = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10} \\ cos \alpha = \frac{ - 20}{2 \sqrt{5} \times 2 \sqrt{10} } = \frac{ - 20}{4 \times 5 \times \sqrt{2} } = - \frac{1}{ \sqrt{2} }$
=> a = 135°.