91Ó°ÊÓ

The given two vectors are,

A vector quantity has magnitude as well as direction. To add two vector quantities, we should use the vector law of addition. If two vectors are exactly in the same direction, we can add them using normal addition rules. But if the directions of two vectors are different, then we need to find the component of the vectors along the unit vector direction and then add them.

In the problem, vector quantities are given in terms of unit vectors. Therefore, we can add them using the vector law of addition. Using the resultant vector in terms of the unit vector, we can find the direction of the resultant vector.

Formula:

Considering the sum of $\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}$is$\stackrel{→}{\mathrm{r}}$, the magnitude of the vector is given by,

localid="1657014645424" $\mathrm{r}=\sqrt{{{\mathrm{r}}_{\mathrm{x}}}^2+{{\mathrm{r}}_{\mathrm{y}}}^2}.........\left(\mathrm{i}\right)$

The direction is given by,

localid="1657014745273" $\mathrm{θ}={\tan }^{-1}\left(\frac{{\mathrm{r}}_{\mathrm{y}}}{{\mathrm{r}}_{\mathrm{x}}}\right)........\left(\mathrm{ii}\right)$

The x, y components of $\stackrel{→}{r}$are

role="math" localid="1657015061990" $$\begin{gathered} {\mathrm{r}}_{\mathrm{x}}={\mathrm{a}}_{\mathrm{x}}+{\mathrm{b}}_{\mathrm{x}} \\ =\left(4\mathrm{m}\right)\hat{\mathrm{i}}+\left(-13\mathrm{m}\right)\hat{\mathrm{i}} \\ =\left(-9\mathrm{m}\right)\hat{\mathrm{i}} \end{gathered}$$

And

$$\begin{gathered} {\mathrm{r}}_{\mathrm{y}}={\mathrm{a}}_{\mathrm{y}}+{\mathrm{b}}_{\mathrm{y}} \\ =\left(3\mathrm{m}\right)\hat{\mathrm{j}}+\left(7\mathrm{m}\right)\hat{\mathrm{j}} \\ =\left(10\mathrm{m}\right)\hat{\mathrm{j}} \end{gathered}$$

Thus,

$$\begin{gathered} \stackrel{→}{\mathrm{r}}={\mathrm{r}}_{\mathrm{x}}+{\mathrm{r}}_{\mathrm{y}} \\ =\left(-9\mathrm{m}\right)\hat{\mathrm{i}}+\left(10\mathrm{m}\right)\hat{\mathrm{j}} \end{gathered}$$

Using equation (i), the magnitude will be,

$\mathrm{r}=\sqrt{{{\mathrm{r}}_{\mathrm{x}}}^2+{{\mathrm{r}}_{\mathrm{y}}}^2}$

Substitute the values in the above equation,

$$\begin{gathered} \left|\stackrel{→}{\mathrm{r}}\right|=\sqrt{\left(-9\mathrm{m}\right)+{\left(10\mathrm{m}\right)}^2} \\ =13\mathrm{m} \end{gathered}$$

Thus, the magnitude is $13\mathrm{m}$

Using equation (ii), the magnitude is given by,

$\mathrm{θ}={\tan }^{-1}\left(\frac{10 \mathrm{m}}{-9\mathrm{m}}\right)$

Thus, the magnitude is $-{48}^{°}$or${132}^{°}$ .

Since the x component of the resultant is negative and y component is positive, it is found the magnitude of the given vector to be${132}^{°}$ .