Question

A proton is located at $<3\times {10}^{-10},-3\times {10}^{-10},8\times {10}^{-10}>\mathbf{m}$.

(a)What is $\overrightarrow{\mathbf{r}}$, the vector from the origin to the location of the proton?

(b)What is $\left|\overrightarrow{r}\right|$?

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

Short answer

(a) The vector from the origin to the location of the proton is $\overrightarrow{\mathbf{r}}\mathbf{=}\left(3\times {10}^{-10}\hat{i}-3\times {10}^{-10}\hat{j}+8\times {10}^{-10}\hat{k}\right)\mathbf{}\mathbf{m}$.

(b) The magnitude of$\overrightarrow{\mathbf{r}}$is$\left|\overrightarrow{r}\right|\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{m}$.

(c) The unit vector along $\overrightarrow{\mathbf{r}}$is $\hat{\mathbf{r}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{33}\hat{\mathbf{i}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{33}\hat{\mathbf{j}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{883}\hat{\mathbf{k}}$ .

Step-by-step solution

Given data

Location of the proton is$r=<3\times {10}^{-10},-3\times {10}^{-10},8\times {10}^{-10}>\mathrm{m}$

Position vector, magnitude of a vector and unit vector

The position vector of a location with respect to origin is

$\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{x}\hat{\mathbf{i}}\mathbf{+}\mathbf{y}\hat{\mathbf{j}}\mathbf{+}\mathbf{z}\hat{\mathbf{k}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}......\left(\mathrm{l}\right)$

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

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

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

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

Determining the position vector of the proton

From equation (I), the position vector of the proton is

$\overrightarrow{\mathrm{r}}=(3\times {10}^{-10}\hat{\mathrm{i}}-3\times {10}^{-10}\hat{\mathrm{j}}+8\times {10}^{-10}\hat{\mathrm{k}})\mathrm{m}$

Determining the magnitude of the position vector of the proton

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

$$\begin{gathered} \left|\overrightarrow{r}\right|=\sqrt{{\left(3\times {10}^{-10}\right)}^2+{\left(-3\times {10}^{-10}\right)}^2+{\left(8\times {10}^{-10}\right)}^2}\mathrm{m} \\ =\sqrt{3^2+3^2+8^2}\times {10}^{-10}\mathrm{m} \\ =9.06\times {10}^{-10}\mathrm{m} \end{gathered}$$

Thus, the required magnitude is$9.06\times {10}^{-10}\mathrm{m}$

Determining the unit vector along the position vector of the proton

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

$$\begin{gathered} \overrightarrow{\mathrm{r}}=\frac{(3\times {10}^{-10}\hat{\mathrm{i}}-3\times {10}^{-10}\hat{\mathrm{j}}+8\times {10}^{-10}\hat{\mathrm{k}})\mathrm{m}}{9.06\times {10}^{-10}\mathrm{m}} \\ =0.33\hat{i}-0.33\hat{j}+0.883\hat{k} \end{gathered}$$

Thus, the required unit vector is$0.33\hat{i}-0.33\hat{j}+0.883\hat{k}$