Question

The position vector for a proton is initially $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$and then later isrole="math" localid="1657003791208" $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$all in meters. (a) What is the proton’s displacement vector, and (b) to what plane is that vector parallel?

Short answer

(a) The proton’s displacement vector is$\left(-7\hat{i}+12\hat{j}\right)m$

(b) $\overrightarrow{d}$is parallel to the X-Y plane

Step-by-step solution

Given data

1) Initial position vector${\overrightarrow{\mathrm{r}}}_{\mathrm{i}}=5.0\hat{\mathrm{i}}-6.0\hat{\mathrm{j}}+2.0\hat{\mathrm{k}}$

2) Final position vector$\overrightarrow{{\mathrm{r}}_{\mathrm{f}}}=-2.0\hat{\mathrm{i}}-6.0\hat{\mathrm{j}}+2.0\hat{\mathrm{k}}$

Understanding the concept of vector addition and subtraction

Taking the difference between position vectors of proton, we can find the displacement vector. We have to use vector laws of addition for this.

Formula:

$\overrightarrow{d}={\overrightarrow{r}}_f-{\overrightarrow{r}}_i$

(a) Determining the displacement vector

To determine the displacement$\overrightarrow{d}$, we need to use the equation,

$$\begin{gathered} \overrightarrow{d}={\overrightarrow{r}}_f-{\overrightarrow{r}}_i \\ =\left(-2.0\hat{i}+6.0\hat{j}+2.0\hat{k}\right)m-\left(5.0\hat{i}-6.0\hat{j}+2.0\hat{k}\right)m \\ =\left(-7.0\hat{i}+12\hat{j}\right)m \end{gathered}$$

Therefore, the displacement vector is $\left(-7.0\hat{i}+12\hat{j}\right)m$

(b) Determining the direction of the plane

The displacement vector $\overrightarrow{d}$does not have any component along Z-axis. It means that the displacement vector lies in XY plane.Therefore, the displacement vector is parallel to the XY plane.