Question

A positron undergoes a displacement${}^{\mathbf{∆}\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}}$, ending with the position vector localid="1654329593013" ${}^{\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}}$, in meters. What was the positron’s initial position vector?

Short answer

The initial position vector of positron is$\overrightarrow{r_0}=\left(-2.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(6.0\mathrm{m}\right)\hat{\mathrm{j}}+\left(-10.0\mathrm{m}\right)\hat{\mathrm{k}}.$

Step-by-step solution

Given data

The displacement of the positron, $∆\overrightarrow{r}=\left(2.0\mathrm{m}\right)\hat{\mathrm{i}}-\left(3.0\mathrm{m}\right)\hat{\mathrm{j}}+\left(6.0\mathrm{m}\right)\hat{\mathrm{k}}$

The final position vector, $\hat{r}=\left(3.0\mathrm{m}\right)\hat{\mathrm{j}}-\left(4.0\mathrm{m}\right)\hat{\mathrm{k}}$

Understanding the position vector and displacement vector

In a coordinate system, the position vector gives the location of an object relative to the origin. The displacement vector gives the change in the position of an object.

The expression for the displacement vector is given as:

$∆\overrightarrow{r}=\overrightarrow{r}-∆\overrightarrow{r_0}$ … (i)

Here, ${}^{\overrightarrow{r_0}}$and ${}^{\overrightarrow{r}}$are the position vectors for initial and final positions respectively.

Determination of the initial position vector of positron

Using equation (i), the initial position vector can be calculated as:

$$\begin{gathered} {\overrightarrow{r}}_0=\overrightarrow{r}-∆\overrightarrow{r} \\ =\left\{\left(3.0\mathrm{m}\right)\hat{\mathrm{j}}-\left(4.0\mathrm{m}\right)\hat{\mathrm{k}}\right\}-\left\{\left(2.0\mathrm{m}\right)\hat{\mathrm{i}}-\left(3.0\mathrm{m}\right)\hat{\mathrm{j}}+\left(6.0\right)\hat{\mathrm{k}}\right\} \\ =\left(2.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(6.0\right)\hat{\mathrm{j}}+\left(-10.0\mathrm{m}\right)\hat{\mathrm{k}} \end{gathered}$$

Thus, the initial position vector of positron is $\overrightarrow{r_0}=\left(-2.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(6.0\mathrm{m}\right)\hat{\mathrm{j}}+\left(-10.0\mathrm{m}\right)\hat{\mathrm{k}}.$