Question

(a)On a piece of graph paper, draw the vector $\overrightarrow{\mathbf{g}}\mathbf{=}<4,7,0>\mathbf{m}$. Put the tail of the vector at the origin.

(b)Calculate the magnitude of$\overrightarrow{\mathbf{g}}$.

(c)Calculate$\hat{\mathbf{g}}$, the unit vector pointing in the direction of$\overrightarrow{\mathbf{g}}$.

(d)On the graph, draw$\overrightarrow{\mathbf{g}}$. Put the tail of the vector at $<1,0,0>$so you can compare$\hat{\mathbf{g}}$and$\overrightarrow{\mathbf{g}}$.

(e)Calculate the product of the magnitude$\left|\overrightarrow{\mathbf{g}}\right|$times the unit vector$\hat{\mathbf{g}}$,$\left(\left|\overrightarrow{g}\right|\right)\mathbf{\left(}\hat{\mathbf{g}}\mathbf{\right)}$.

Short answer

(a),(d) The graphical representation of the vector$\overrightarrow{\mathbf{g}}$and its unit vector$\hat{\mathbf{g}}$is

(b) The magnitude of $\overrightarrow{\mathbf{g}}$is $\mathbf{8}\mathbf{.}\mathbf{06}\mathbf{}\mathbf{m}$.

(c) The unit vector along $\overrightarrow{\mathbf{g}}$is $\mathbf{0}\mathbf{.}\mathbf{496}\hat{\mathbf{i}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{868}\hat{\mathbf{j}}$.

(e) The product of the magnitude$\left|\overrightarrow{\mathbf{g}}\right|$times the unit vector$\hat{\mathbf{g}}$is $\left(4\hat{i}+7\hat{j}\right)\mathbf{}\mathbf{m}$.

Step-by-step solution

Given data

The vector is provided.

Magnitude of a vector and unit vector

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

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

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

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

Drawing the vector  g→

The vector $\overrightarrow{g}$ in coordinate form is

$\overrightarrow{g}=\left(4\hat{i}+7\hat{j}\right)\mathrm{m}$

The graphical representation of the vector is as follows

Determining the magnitude of the vector g→

From equation (I) the magnitude of vector $\overrightarrow{g}$is

$$\begin{gathered} \left|\overrightarrow{g}\right|=\sqrt{4^2+7^2}\mathrm{m} \\ =8.06\mathrm{m} \end{gathered}$$

Thus, the required magnitude is $8.06\mathrm{m}$.

Determining the unit vector along g→

From equation (II) the unit vector along $\overrightarrow{g}$is

$$\begin{gathered} \hat{g}=\frac{\left(4\hat{\mathrm{i}}+7\hat{\mathrm{j}}\right)\mathrm{m}}{8.06\mathrm{m}} \\ =0.496\hat{i}+0.868\hat{j} \end{gathered}$$

Thus, the required unit vector is$0.496\hat{i}+0.868\hat{j}$

Drawing the unit vector g→

The graphical representation of $\hat{g}$ with the tail at is shown in the image in Step 3.

Determining the product of the magnitude of g→ and the unit vector along g→

From equation (II), the product of the magnitude and the unit vector will just return the original vector, that is

$$\begin{gathered} \left|\overrightarrow{g}\right|\hat{g}=\overrightarrow{g} \\ =\left(4\hat{i}+7\hat{j}\right)m \end{gathered}$$

Thus, the required product is$\left(4\hat{i}+7\hat{j}\right)m$