Answer
The angle between the direction of the two vectors is $~~90^{\circ}$
Work Step by Step
We can express the particle's position in unit-vector notation:
$r = (3.0~m)~\hat{i}+ (4.0~m)~\hat{j}$
We can express the force in unit-vector notation:
$F = (-8.0~N)~\hat{i}+(6.0~N)~\hat{j}$
We can find the angle between the two vectors:
$cos~\theta = \frac{r\cdot F}{\vert r \vert \cdot \vert F \vert}$
$cos~\theta = \frac{(3.0)(-8.0)+(4.0)(6.0)}{\vert r \vert \cdot \vert F \vert}$
$cos~\theta = \frac{0}{\vert r \vert \cdot \vert F \vert}$
$cos~\theta = 0$
$\theta = 90^{\circ}$
The angle between the direction of the two vectors is $~~90^{\circ}$
