91Ó°ÊÓ

Using the standard concept of magnitude and direction, the magnitude and the direction of the given vector can be calculated. In this problem the vector is pointing towards south.

Thus vector $\stackrel{→}{\mathrm{d}}$is represented by

$\stackrel{→}{d}=\left(3m\right)\left(-\hat{j}\right)$

Using equation (i), the magnitude of $5.0\stackrel{→}{d}$can be written as

$$\begin{gathered} 5.0\stackrel{→}{d}=5\left(3m\right)\left(-\hat{j}\right) \\ 5.0\stackrel{→}{d}\left(-15\right)\hat{j} \end{gathered}$$

It is known that positive scalars affect the direction but not the magnitude.

So the magnitude of $5.0\hat{d}$is$15\mathrm{m}$ .

Again, positive scalars affect the direction, the direction of the vector$5.0\stackrel{→}{d}$ is south.

Using equation (i), the magnitude of $-2.0\stackrel{→}{d}$is given by

$$\begin{gathered} -2\stackrel{→}{\mathrm{d}}=\left(3\mathrm{m}\right)\left(-\hat{\mathrm{j}}\right) \\ -2\stackrel{→}{\mathrm{d}}=-2\left(3\mathrm{m}\right)\left(-\hat{\mathrm{j}}\right) \\ -2\stackrel{→}{\mathrm{d}}=6\mathrm{m} \end{gathered}$$

Here the scalar is negative. Thus the negative sign reverses the direction. Therefore, the new direction of the vector $-2.0\stackrel{→}{d}$ is north.