Question

Найдите угол между векторами a и b

Answer 1

1) a(1; 2), b(1; -3)
cosa = (a*b)/(|a|*|b|)
a*b = 1*1 + 2*(-3) = 1 - 6 = -5
$|a| = \sqrt{ {1}^{2} + {2}^{2} } = \sqrt{1 + 4} = \sqrt{5}$
$|b| = \sqrt{ {1}^{2} + {( - 3)}^{2} } = \sqrt{1 + 9 } = \sqrt{10}$
$cos \alpha = \frac{ - 5}{ \sqrt{5} \times \sqrt{10} } = \frac{ - 5}{ \sqrt{5} \times \sqrt{5} \times \sqrt{2} } = \frac{ - 5}{5 \sqrt{2} } = - \frac{1}{ \sqrt{2} }$
=> угол а = 135°

2) a(-3; 4), b(5; 12)
cosa = (a*b)/(|a|*|b|)
a*b = -3*5 + 4*12 = -15 + 48 = 33
$|a| = \sqrt{ {( - 3)}^{2} + {4}^{2} } = \sqrt{9 + 16} = \sqrt{25} = 5$
$|b| = \sqrt{ {5}^{2} + {12}^{2} } = \sqrt{25 + 144} = \sqrt{169} = 13$
$cos \alpha = \frac{33}{5 \times 13} = \frac{33}{65}$
a = arccos33/65

3) a(0; 2), b(-3; 4)
cosa = (a*b)/(|a|*|b|)
a*b = 0*(-3) + 2*4 = 8
$|a| = \sqrt{ {0}^{2} + {2}^{2} } = \sqrt{4} = 2$
$|b| = \sqrt{ {( - 3)}^{2} + {4}^{2} } = \sqrt{9 + 16} = \sqrt{25} = 5$
$cos \alpha = \frac{8}{2 \times 5} = \frac{4}{5}$
угол a = arccos4/5

4) a(4; 0), b(0; 4)
cosa = (a*b)/(|a|*|b|)
a*b = 4*0 + 0*4 = 0
=> cosa = 0
угол а = 90°