Question

(a) Consider the vectors ${\overrightarrow{r}}_1$and ${\overrightarrow{r}}_2$represented by arrows in Figure 1.20. Are these two vectors equal? (b) If ,and $\overrightarrow{c}=\overrightarrow{a}$, what is the unit vector $\hat{\mathrm{c}}$in the direction of $\overrightarrow{\mathrm{c}}$?

Short answer

(a) The vectors${\overrightarrow{r}}_1$ and${\overrightarrow{r}}_2$ are equal in magnitude but opposite in direction.

(b) The unit vector$\hat{c}$ in the direction of is$\left(0.873,0.436,-0.218\right)$.

Step-by-step solution

Definition of a vector

• A vector is 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.

(a) Checking whether the vectors are equal or not

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.

(b) Determination of the magnitude

Unit vector or is given by,

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

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

$\left|\overrightarrow{c}\right|=\sqrt{{\left(400\right)}^2+{\left(200\right)}^2+{\left(-100\right)}^2}$

$=\sqrt{16\times {10}^4+4\times {10}^4+1\times {10}^4}$

$=\sqrt{21\times {10}^4}$

$=\sqrt{21}\times {10}^2$

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

$\hat{c}=\frac{400\hat{i}+200j-100\hat{k}}{\sqrt{21}\times {10}^2}$

$=\frac{4}{\sqrt{21}}\hat{i}+\frac{2}{\sqrt{21}}\hat{j}-\frac{1}{\sqrt{21}}\hat{k}$

role="math" localid="1653561209299" $=0.873\hat{i}+0.436\hat{j}-0.218\hat{k}$

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