Question

Resolver el ejercicio en la imagen. Gracias.

Answer 1

$\texttt{Asumir\'e que los datos son }\vec a =3i+j\texttt{ y }\vec b = 3i-2j\\ \\ (a) \text{Pr }_{\vec b}~\vec a=\left(\dfrac{\vec a\cdot \vec b}{\|\vec b\|^2}\right)\vec b=\left[\dfrac{(3,1)\cdot (3,-2)}{3^2+(-2)^2}\right](3,-2)\\ \\ \\ \text{Pr }_{\vec b}~\vec a=\dfrac{7}{13}(3,-2)\\ \\ \\ \text{Pr }_{\vec b}~\vec a=\left(\dfrac{21}{13},-\dfrac{14}{13}\right)\\ \\ \\ \left\|~\text{Pr }_{\vec b}~\vec a~\right\|=\dfrac{7}{\sqrt{13}}$



$(b) \cos \theta = \dfrac{\vec a\cdot \vec b}{\|\vec a\|\cdot \|\vec b\|}\\ \\ \\ \cos \theta =\dfrac{(3,1)\cdot(3,-2)}{\|(3,1)\|\cdot\|(3,-2)\|}\\ \\ \\ \cos \theta =\dfrac{7}{\sqrt{10}\cdot\sqrt{13}}\\ \\ \\ \cos \theta =\dfrac{7}{\sqrt{130}}\\ \\ \\ \theta=\arccos\left(\dfrac{7}{\sqrt{130}}\right)\\ \\ \\ \boxed{\theta \approx 52\°~7'~30.06''}$