Z treści zadania wiemy, że:

$\overrightarrow{AB}=\left[1,3\right]$  

$\overrightarrow{CA}=\left[-\sqrt{2},2\sqrt{2}\right]$  

Zatem:

$\overrightarrow{AC}=\left[\sqrt{2},-2\sqrt{2}\right]$  

 

Wyznaczmy kąt między wektorami AB i AC. 

$\cos \alpha =\frac{u_1{v}_1+u_2{v}_2}{\sqrt{u_1^2+u_2^2}\cdot \sqrt{v_1^2+v_2^2}}$  

$\cos \alpha =\frac{1\cdot \sqrt{2}+3\cdot \left(-2\sqrt{2}\right)}{\sqrt{1^2+3^2}\cdot \sqrt{{\left(-\sqrt{2}\right)}^2+{\left(2\sqrt{2}\right)}^2}}$  

$\cos \alpha =\frac{\sqrt{2}-6\sqrt{2}}{\sqrt{1+9}\cdot \sqrt{2+8}}$  

$\cos \alpha =\frac{-5\sqrt{2}}{\sqrt{10}\cdot \sqrt{10}}$  

$\cos \alpha =\frac{-5\sqrt{2}}{10}$  

$\cos \alpha =-\frac{\sqrt{2}}{2}$  

$\cos \alpha =-\cos {45}^{\circ }$  

$\cos \alpha =\cos \left({180}^{\circ }-{45}^{\circ }\right)$  

$\cos \alpha =\cos {135}^{\circ }$  

$\alpha ={135}^{\circ }$  

 

Odp. D