Operaciones de vectores, cálculo vectorial

$$\begin{align}& \end{align}$$

¡Hola Francisco!

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$$\begin{align}&p_1(2,3)\quad p_2(-1,5)\quad p_3 \left(\frac 12,-4  \right)\\&k_1=-2\quad k_2=\frac 13\\&\\&a)\quad v_1=p_1\to p_2 =(-1,5)-(2,3)=(-3,2)\\&\\&b)\quad v_2=p_1\to p_3= \left(\frac 12,-4  \right)-(2,3)=\left(-\frac 32,-7  \right)\\&\\&c) \quad v_3=p_2\to p_3= \left(\frac 12,-4  \right)-(-1,5)=\left(\frac 32,-9  \right)\\&\\&d)\quad v_1+v_2=(-3,2)+\left(-\frac 32,-7  \right)=\left(-\frac 92,-5  \right)\\&\\&e)\quad v_1-v_2=(-3,2)-\left(-\frac 32,-7  \right)=\left(-\frac 32,9  \right)\\&\\&f)\quad k_1·v_1=-2(-3,2) = (6,-4)\\&\\&g)\quad (k_1+k_2)·v_2=\left(-2+\frac 13\right)\left(-\frac 32,-7  \right)=\\&\qquad -\frac 53\left(-\frac 32,-7  \right)=\left(\frac 52,\frac{35}{3}  \right)\\&\\&h)\quad (v_1+v_2)+v_3=\\&\qquad v_1+v_2 \text{ ya se calculó antes}\\&\qquad =\left(-\frac 92,-5  \right)+\left(\frac 32,-9  \right)=\left(-3,-14  \right)\\&\\&i)\quad k_2(v_1-v_2)=\\&\qquad v_1-v_2\text{ ya se calculó antes}\\&\qquad \frac 13\left(-\frac 32,9  \right)=\left(-\frac 12,3  \right)\\&\\&j)\quad ||v_2||= \sqrt{\frac 94+49}=\sqrt{\frac{205}{4}}=\frac{\sqrt{205}}{2}\\&\\&k) 3 ||v_2||= 3·\frac{\sqrt{205}}{2}=\frac{3 \sqrt{205}}{2}\\&\\&l)\quad \text{Unitario de }v_3. \quad v_3=\left(\frac 32,-9  \right)\\&\\&u=\left(\frac{\frac 32}{\sqrt{\frac 94+81}},\frac{-9}{\sqrt{\frac 94+81}}  \right)=\left(\frac{\frac 32}{\frac{\sqrt{333}}{2}}  ,\frac{-9}{\frac{\sqrt{333}}{2}}  \right)=\\&\\&\left(\frac{3}{\sqrt{333}},\frac{-18}{\sqrt{333}}  \right)=\left(\frac{3}{ 3·\sqrt{37}},\frac{-18}{3·\sqrt{37}}  \right)=\\&\\&\left(\frac{1}{\sqrt{37}},\frac{-6}{\sqrt{37}}  \right)=\left(\frac{\sqrt{37}}{37},\frac{-6 \sqrt {37}}{37}  \right)\\&\\&m)\quad v_1=(-3,2)=-3(1,0)+2(0,1)\\&\qquad\; v_2=\left(-\frac 32,-7  \right)=-\frac 32(1,0)-7(0,1)\\&\qquad\; v_3=\left(\frac 32,-9  \right)=\frac 32(1,0)-9(0,1)\\&\end{align}$$

Y eso es todo.