Question

A watermelon seed has the following coordinates:${}^{\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$,${}^{\mathbf{y}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$, and${}^{\mathbf{z}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{m}}$. Find its position vector (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the${}^{\mathbf{x}}$axis. (d) Sketch the vector on a right-handed coordinate system. If the seed is moved to the${}^{\mathbf{x}\mathbf{y}\mathbf{z}}$coordinates${}^{\left(3.00m,0m,0m\right)}$, what is its displacement (e) in unit-vector notation and as (f) a magnitude and (g) an angle relative to the positive${}^{\mathbf{x}}$ direction?

Short answer

(a) The position vector of seed in unit vector notation is $\overrightarrow{r}=\left(-5\mathrm{m}\right)\hat{\mathrm{i}}+\left(8\mathrm{m}\right)\hat{\mathrm{j}}.$

(b) The magnitude of the position vector is 9.4 m.

(c) The direction of position vector relative to positive x-direction is$-58°$

(d) The position vector on a right handed coordinate system is drawn in figure.

(e) The displacementin unit vector notation is$∆\overrightarrow{r}=\left(8.0\mathrm{m}\right)\hat{\mathrm{i}}-\left(8.0\mathrm{m}\right)\widehat{\mathrm{j}.}$

(f) The magnitude of displacementis$11\mathrm{m}$.

(g) The angle of displacement vector relative to positive x-direction is$-45°$.

Step-by-step solution

Given data

The x-coordinates of the seed,$x=-5.0\mathrm{m}$

The y-coordinates of the seed,$y=8.0\mathrm{m}$

The z-coordinates of the seed,$z=0.0\mathrm{m}$

Understanding the vector magnitude and displacement vector

A vector is a geometrical object with both magnitude and direction. The magnitude is the length of the vector. The displacement vector gives the change in the position of an object.

The expression for the magnitude of a vector ${}^r$is given as:

$r=\sqrt{x^2+y^2+z^2}$ … (i)

Here, $x,y$ and z are the components of vectors in X, Y and Z-directions.

The direction of the vector is given as:

$\tan \theta =\frac{y}{x}$… (ii)

(a) Determination of the position vector

The position vector can be written as:

$\overrightarrow{r}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}$

Substitute the values in the above expression.

$$\begin{gathered} \overrightarrow{r}=\left(-5.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(8.0\mathrm{m}\right)\hat{\mathrm{j}}+\left(0\mathrm{m}\right)\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{r}}=\left(-5.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(8.0\mathrm{m}\right)\hat{\mathrm{j}} \end{gathered}$$

Thus, the position vector of seed is$\overrightarrow{r}=\left(-5.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(8.0\mathrm{m}\right)\hat{\mathrm{j}}$.

(b) Determination of the magnitude of position vector

Using equation (i), the magnitude of unit vector is calculated as:

$$\begin{gathered} \left|\overrightarrow{r}\right|=\sqrt{{\left(-5.0\mathrm{m}\right)}^2+{\left(8.0\mathrm{m}\right)}^2} \\ =9.4\mathrm{m} \end{gathered}$$

Thus, the magnitude of unit vector is 9.4 m.

(c) Determination of the angle of position vector

Using equation (ii), the angle of position vector iscalculated as:

$$\begin{gathered} \tan \theta =\frac{y}{x} \\ \theta ={\tan }^{-1}\left(\frac{8.0\mathrm{m}}{-5.0\mathrm{m}}\right) \\ =-58°\mathrm{or}122° \end{gathered}$$

If ${}^{\theta }$is taken as ${}^{-58°}$, it is measured in the clockwise direction with respect to the negative x axis, and if it is taken as ${}^{122°}$measured in the counterclockwise direction with respect to the positive x axis. This can be concluded from the fact that the vector is in second quadrant.

(d) To draw the vector diagram

The position vector is drown in the figure below.

(e) Determination of the displacement in unit vector notation

Since seed is moved to new position ${}^{\left(3.00\mathrm{m},0,0\right)}$

$$\begin{gathered} {\overrightarrow{r}}_n=\left(3.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(0\mathrm{m}\right)\hat{\mathrm{j}}+\left(0\mathrm{m}\right)\hat{\mathrm{k}} \\ {\overrightarrow{r}}_n=\left(3.0\mathrm{m}\right)\hat{\mathrm{i}} \end{gathered}$$

The displacement of the seed in unit vector notation is given by,

role="math" localid="1654597211060" $$\begin{gathered} ∆\overrightarrow{r}={\overrightarrow{r}}_n-\overrightarrow{r} \\ =\left(3.0\mathrm{m}\right)\hat{\mathrm{i}}-\left\{\left(-5.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(8.0\mathrm{m}\right)\hat{\mathrm{j}}\right\} \\ =\left(8.0\mathrm{m}\right)\hat{\mathrm{i}}-\left(8.0\mathrm{m}\right)\hat{\mathrm{j}} \end{gathered}$$

Thus, the displacement of seed in unit vector notation is ${}^{∆\overrightarrow{r}=\left(8.0\mathrm{m}\right)\hat{i}-\left(8.0\mathrm{m}\right)\hat{j}}$.

(f) Determination of the magnitude of displacement

Using equation (i), the magnitude of displacement is calculated as:

$$\begin{gathered} \left|∆\overrightarrow{r}\right|=\sqrt{{\left(8.0\mathrm{m}\right)}^2+{\left(-8.0\mathrm{m}\right)}^2} \\ =11.3\mathrm{m} \end{gathered}$$

Thus, the magnitude of displacement is 11.3 m.

(g) Determination of the angle of displacement

Using equation (ii), the angle of new displacement is,

$$\begin{gathered} \tan \theta =\frac{y}{x} \\ \theta ={\tan }^{-1}\left(\frac{-8.0\mathrm{m}}{8.0\mathrm{m}}\right) \\ =-45° \end{gathered}$$

Thus, the angle of displacement is${}^{-45°.}$It means that it is measured in the clockwise direction with respect to positive x axis as the vector is in the fourth quadrant.