Question

Найдите угол между прямыми l1 : 5x - 12y - 16 = 0 и l2 : 3x + 4y - 12 = 0

Answer 1

$l_1:\; 5x-12y-16=0\; \; \to \; \; \overline {n_1}=(5,-12)\\\\l_2:\; 3x+4y-12=0\; \; \to \; \; \overline {n_2}=(3,4)\\\\cos\varphi = \frac{\overline {n_1}\cdot \overline {n_2}}{|\overline {n_1}|\cdot |\overline n_2|} = \frac{5\cdot 3-12\cdot 4}{\sqrt{5^2+12^2}\cdot \sqrt{3^2+4^2}} = \frac{-33}{13\cdot 5} \approx -0,51\\\\\varphi =\pi -arccos0,51$

Answer 2

Tqα =(k₂ -k₁)/(1+k₁*k₂) .
5x - 12y -16 =0 ⇔y =(5/12)*x -16/5 ⇒ k₁ =5/12 .
3x +4y -12 =0⇔y = (-3/4)*x +3 ⇒ k₂ =(-3/4).
tqα = ( (-3/4) -(5/12) ) /(1 +(5/12)(-3/4))= -56/33 .
α = - arctq(56/33) .  ||  90°<α< 180° ||


tqα  =  -56/33 ⇒cosα =-1/√(1+tq²α) =-1/√(1+(-56/33)² =-33/√(1089 +3136) =
-33/√4225 = -33/65 .