En nuestro caso tenemos los dos vectores

$\text{}\vec{v}=\left(\begin{array}{c}3\\ 9\end{array}\right)\backslash vec\; v\; =\; \backslash begin\{pmatrix\}3\backslash \backslash 9\backslash end\{pmatrix\}$ y $\vec{w}=\left(\begin{array}{c}2\\ 1\end{array}\right)\backslash vec\; w\; =\; \backslash begin\{pmatrix\}2\backslash \backslash 1\backslash end\{pmatrix\}$

Utiliza estos y obtén:

$\cos (\phi )=\frac{\left(\begin{array}{c}3\\ 9\end{array}\right)\circ \left(\begin{array}{c}2\\ 1\end{array}\right)}{\mid \left(\begin{array}{c}3\\ 9\end{array}\right)\mid \cdot \mid \left(\begin{array}{c}2\\ 1\end{array}\right)\mid }\backslash cos(\backslash varphi)\; =\; \backslash frac\{\backslash begin\{pmatrix\}3\backslash \backslash 9\backslash end\{pmatrix\}\; \backslash circ\; \backslash begin\{pmatrix\}2\backslash \backslash 1\backslash end\{pmatrix\}\}\{\backslash left|\backslash begin\{pmatrix\}3\backslash \backslash 9\backslash end\{pmatrix\}\backslash right|\; \backslash cdot\; \backslash left|\backslash begin\{pmatrix\}2\backslash \backslash 1\backslash end\{pmatrix\}\backslash right|\}$

Resuelve la fórmula según $\phi \backslash varphi$:

$\phi ={\cos }^{-1}\left(\frac{\left(\begin{array}{c}3\\ 9\end{array}\right)\circ \left(\begin{array}{c}2\\ 1\end{array}\right)}{\mid \left(\begin{array}{c}3\\ 9\end{array}\right)\mid \cdot \mid \left(\begin{array}{c}2\\ 1\end{array}\right)\mid }\right)\backslash \; \backslash varphi\; =\; \backslash cos^\{-1\}\backslash left(\; \backslash frac\{\backslash begin\{pmatrix\}3\backslash \backslash 9\backslash end\{pmatrix\}\; \backslash circ\; \backslash begin\{pmatrix\}2\backslash \backslash 1\backslash end\{pmatrix\}\}\{\backslash left|\backslash begin\{pmatrix\}3\backslash \backslash 9\backslash end\{pmatrix\}\backslash right|\; \backslash cdot\; \backslash left|\backslash begin\{pmatrix\}2\backslash \backslash 1\backslash end\{pmatrix\}\backslash right|\}\backslash right)$

Por último, determina el producto escalar para el numerador y las dos longitudes para el denominador.

$\vec{v}\circ \vec{w}=\left(\begin{array}{c}3\\ 9\end{array}\right)\circ \left(\begin{array}{c}2\\ 1\end{array}\right)=3\cdot 2+9\cdot 1=15\backslash vec\; v\; \backslash circ\; \backslash vec\; w\; =\; \backslash begin\{pmatrix\}3\backslash \backslash 9\backslash end\{pmatrix\}\; \backslash circ\; \backslash begin\{pmatrix\}2\backslash \backslash 1\backslash end\{pmatrix\}\; =\; 3\; \backslash cdot\; 2\; +\; 9\; \backslash cdot\; 1\; =\; 15$

$\text{}\mathrm{\mid }\vec{v}\mathrm{\mid }\cdot \mathrm{\mid }\vec{w}\mathrm{\mid }=(\sqrt{3^2+9^2})\cdot (\sqrt{2^2+1^2})|\backslash vec\; v|\; \backslash cdot\; |\backslash vec\; w|\; =\; (\backslash sqrt\{3^2\; +\; 9^2\})\; \backslash cdot\; (\backslash sqrt\{2^2\; +\; 1^2\})$

$=\sqrt{90}\cdot \sqrt{5}=\; \backslash sqrt\{90\}\; \backslash cdot\; \backslash sqrt\{5\}$

$=15\sqrt{2}=\; 15\backslash sqrt\{2\}$

De la división de los dos resultados obtienes ahora el factor que introduces en el arco coseno.

${\displaystyle \phi ={\cos }^{-1}\left(\frac{15}{15\sqrt{2}}\right)}\backslash displaystyle\; \backslash \; \backslash varphi\; =\; \backslash cos^\{-1\}\; \backslash left(\; \backslash frac\{15\}\{15\backslash sqrt\{2\}\}\; \backslash right)$

Por lo tanto, el ángulo de intersección $\phi ={45}^{\circ }\backslash \; \backslash varphi\; =\; 45^\backslash circ$