What is the position vector?

Question a)

Applying the definition of position vector:

Since our body is located at point (- 1,2) then x=-1, y=2 and z=0. Substituting x, y and z in the definition of position vector for this point we get:

$$\begin{gathered} \overrightarrow{r_1}=-1·\overrightarrow{i}+ 2·\overrightarrow{j} + 0·\overrightarrow{k} m\Rightarrow \\ \boxed{\overrightarrow{r_1}= -\overrightarrow{i}+2·\overrightarrow{j}m} \end{gathered}$$

Once we have the position vector, to calculate its magnitude we just apply the equation:

and substituting the corresponding values:

$$\begin{gathered} \left|\overrightarrow{r_1}\right|=\sqrt{(-1{)}^2+(2{)}^2+(0{)}^2} m\Rightarrow \\ \boxed{\left|\overrightarrow{r_1}\right|=\sqrt{5} m} \end{gathered}$$

Question b)

Applying the same equations that in the previous question, we get that for point P₂ (-3,4), the position vector (r₂) is:

$$\begin{gathered} \overrightarrow{r_2}=-3·\overrightarrow{i}+4·\overrightarrow{j}+0·\overrightarrow{k} m\Rightarrow \\ \boxed{\overrightarrow{r_2}=-3·\overrightarrow{i}+4·\overrightarrow{j} m} \end{gathered}$$

and its modulus:

$$\begin{gathered} \left|\overrightarrow{r_2}\right|=\sqrt{(-3{)}^2+(4{)}^2+(0{)}^2} m\Rightarrow \\ \left|\overrightarrow{r_2}\right|=\sqrt{25} m\Rightarrow \\ \boxed{\left|\overrightarrow{r_2}\right|=5 m} \end{gathered}$$