91Ó°ÊÓ

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

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

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

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

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

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

The unit vector along a vector $\stackrel{→}{r}$ is

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

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

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

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

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

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

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

From equation (III), the unit vector along $\stackrel{â‡\P Ä}{r}$is

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

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