A planet is located at $\mathbf{<}\mathbf{-}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathbf{,}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{>}$. A star is located at $\mathbf{<}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{,}\mathbf{-}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{,}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{>}$.

(a)What is $\overrightarrow{\mathbf{r}}$, the vector from the star to the planet?

(b)What is the magnitude of $\overrightarrow{\mathbf{r}}$?

(c)What is $\hat{\mathbf{r}}$, the unit vector (vector with magnitude 1) in the direction of $\overrightarrow{\mathbf{r}}$?

Location of the planet is $<-1\times {10}^{10},8\times {10}^{10},-3\times {10}^{10}>$.

Location of the star is $<6\times {10}^{10},-5\times {10}^{10},1\times {10}^{10}>$.

The position vector of a location $<x_1,y_1,z_1>$with respect to another location $<x_2,y_2,z_2>$is

$\overrightarrow{\mathbf{r}}\mathbf{=}\left(x_2-x_1\right)\hat{\mathbf{i}}\mathbf{+}\left(y_2-y_1\right)\hat{\mathbf{j}}\mathbf{+}\left(z_2-z_1\right)\hat{\mathbf{k}}\mathbf{}.....\left(\mathrm{l}\right)$

The magnitude of a vector $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$is

$\left|\overrightarrow{r}\right|\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\mathbf{}.....\left(\mathrm{II}\right)$

The unit vector along a vector $\overrightarrow{r}$ is

$\hat{\mathbf{r}}\mathbf{=}\frac{\overrightarrow{\mathbf{r}}}{\left|\overrightarrow{r}\right|}.....\left(\mathrm{III}\right)$

From equation (I), the position vector of the planet with respect to the star is

$$\begin{gathered} \overrightarrow{r}=\left(-1\times {10}^{10}-6\times {10}^{10}{10}^{10}\right)\hat{i}+\left(8\times {10}^{10}-\left(-5\times {10}^{10}\right)\right)\hat{j}+\left(-3\times {10}^{10}-1\times {10}^{10}\right)\hat{k} \\ =\left(-7\hat{i}+13\hat{j}-4\hat{k}\right)\times {10}^{10} \end{gathered}$$

Thus, the required position vector is $\left(-7\hat{i}+13\hat{j}-4\hat{k}\right)\times {10}^{10}$

From equation (II), the magnitude of $\overrightarrow{r}$is

$$\begin{gathered} \left|\overrightarrow{r}\right|=\sqrt{{\left(-7\right)}^2+{\left(13\right)}^2+{\left(-4\right)}^2}\times {10}^{10} \\ =\sqrt{49+169+16}\times {10}^{10} \\ =15.3\times {10}^{10} \end{gathered}$$

Thus, the required magnitude is $15.3\times {10}^{10}$

From equation (III), the unit vector along $\stackrel{\rightharpoonup }{r}$is

$$\begin{gathered} \hat{r}=\frac{\left(-7\hat{i}+13\hat{j}-4\hat{k}\right)\times {10}^{10}}{15.3\times {10}^{10}} \\ =-0.458\hat{i}+0.85\hat{j}-0.261\hat{k} \end{gathered}$$

Thus, the required unit vector is$-0.458\hat{i}+0.85\hat{j}-0.261\hat{k}$