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?

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,$z=0.0\mathrm{m}$

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)

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$\overrightarrow{r}=\left(-5.0\mathrm{m}\right)\hat{\mathrm{i}}+\left(8.0\mathrm{m}\right)\hat{\mathrm{j}}$.

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.

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

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.

The position vector is drown in the figure below.

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 ${}^{∆\overrightarrow{r}=\left(8.0\mathrm{m}\right)\hat{i}-\left(8.0\mathrm{m}\right)\hat{j}}$.

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.

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.