Consider the vectors $\overrightarrow{{\mathrm{r}}_1}$and $\overrightarrow{{\mathrm{r}}_2}$represented by arrows in Figure 1.20. Are these two vectors equal? (b) If $\overrightarrow{a}=<400,200-100>m/s^2$,and $\overrightarrow{c}=\overrightarrow{a}$, what is the unit vector $\hat{c}$in the direction of $\hat{c}$?

Figure 1.20

• A vectoris a quantity that contains magnitude along with the direction.
• It's usually represented (drawn) by an arrow with the same direction as the amount and a length proportionate to the magnitude of the quantity.
• A vector does not have only a position, even though it has magnitude and direction.

For two vectors to be equal, they must have the same magnitude and direction.

From the given figure, it is analyzed that$\overrightarrow{r_1}$ and$\overrightarrow{r_2}$ are in opposite directions.

Hence, the two given vectors$\overrightarrow{r_1}$ and $\overrightarrow{r_2}$are not equal.

Unit vector or $\overrightarrow{c}$is given by,

$\hat{c}=\frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}.$

compute the magnitude of $\overrightarrow{c}$.

$$\begin{gathered} \left|\overrightarrow{c}\right|=\sqrt{{\left(400\right)}^2+{\left(200\right)}^2+{(-100)}^2} \\ =\sqrt{16\times {10}^4+4\times {10}^4+1\times {10}^4} \\ =\sqrt{21\times {10}^4} \\ =\sqrt{21}\times {10}^2 \end{gathered}$$

Substitute$\overrightarrow{c}$and its magnitude into the formula of unit vector.

$$\begin{gathered} \hat{\mathrm{c}}=\frac{400\hat{\mathrm{i}}+200\mathrm{j}+100\hat{\mathrm{k}}}{\sqrt{21}\times {10}^2} \\ =\frac{4}{\sqrt{21}}\hat{\mathrm{i}}+\frac{2}{\sqrt{21}}\hat{\mathrm{j}}-\frac{1}{\sqrt{21}}\hat{\mathrm{k}} \\ =0.873\hat{\mathrm{i}}+0.436\hat{\mathrm{j}}-0.218\hat{\mathrm{k}} \end{gathered}$$

Thus, the unit vector$\hat{c}$ in the direction of$\overrightarrow{c}$ is $(0.873,0.436,-0.218)$.