91Ó°ÊÓ

The problem involves vector algebra. In vector algebra, various algebraic operations can be performed on vectors. To understand this concept, a vector algebraic diagram can be drawn in which the motion is shown. In this diagram, all the vectors in positive x-direction are added and subtracted the vectors which are in negative x-direction. Similarly, all the vectors in positive y-direction are added and subtracted the vectors which are in negative y-direction. Thus, we would get two components of the resultant vectors, one along x and other along y (it may be positive or negative). Using the Pythagoras theorem, the resultant vectors can be found. Using trigonometry, the angle made by the resultant vector with the x-axis cab be calculated.

Given,

$$\begin{gathered} A=\left(3.1\mathrm{km}\right)\stackrel{â‡\P Ä}{\mathrm{j}} \\ \mathrm{B}=-\left(2.4\mathrm{km}\right)\stackrel{â‡\P Ä}{\mathrm{i}} \\ \mathrm{C}=-\left(5.2\mathrm{km}\right)\stackrel{â‡\P Ä}{\mathrm{j}} \end{gathered}$$

The resultant vector can be written as

$R=\sqrt{A^2+B^2}$

Vector diagram can be drawn as follows.

In the above figure, as the person moves first 3.1 km in the north direction, it is shown in the positive y- direction. Then, the person moves in the west 2.4 km, so it is shown in the negative x- direction, and finally, he is moving in the south 5.2 km, so it is shown in the negative y- direction as shown in the figure.

As A and C are parallel but opposite in direction, resultant of those two, D can be found as follows:

$$\begin{gathered} D=A+C \\ D=\left(3.0\mathrm{km}\right)\stackrel{â‡\P Ä}{\mathrm{j}}-\left(5.2\mathrm{km}\right)\stackrel{â‡\P Ä}{\mathrm{j}} \\ \mathrm{d}=-\left(2.1\mathrm{km}\right)\stackrel{â‡\P Ä}{\mathrm{j}} \end{gathered}$$

The negative sign implies that the resultant of A and C is along the negative y-direction along the south.

Vector B is along the negative x- direction.

Now, the resultant of D and B is given by,

$R=\sqrt{D^2+B^2}$

$$\begin{gathered} R=\sqrt{{\left(-2.1\right)}^2+{\left(-2.4\right)}^2} \\ R=3.2\mathrm{km} \end{gathered}$$

Direction of the resultant is given by,

$$\begin{gathered} \tan \mathrm{θ}=\frac{D}{B} \\ \tan \mathrm{θ}=\frac{-2.1}{-2.4} \end{gathered}$$

So,

$\mathrm{θ}=41°\mathrm{or}221°$

Value of$\mathrm{θ}$can be, $41°\mathrm{or}{221}^{°}$but as we can see in the vector diagram, the resultant is in the third quadrant, $41°$so cannot be correct. Hence, the angle is$221°$ with the positive x axis or with the east direction.