In questions 1 to 7, find the resultant of the given vectors. Give also both the magnitude and the direction of the resultants.

The first displacement is $12 \text{ km}$ south. Let's represent South as the negative y-direction.
$\vec{d_1} = (0, -12) \text{ km}$

The second displacement is $5 \text{ km}$ east. Let's represent East as the positive x-direction.
$\vec{d_2} = (5, 0) \text{ km}$

The resultant displacement $\vec{R}$ is the sum of $\vec{d_1}$ and $\vec{d_2}$.

$$\vec{R} = \vec{d_1} + \vec{d_2} = (0+5, -12+0) = (5, -12) \text{ km}$$

This means the resultant displacement is $5 \text{ km}$ East and $12 \text{ km}$ South.

The magnitude of the resultant vector is found using the Pythagorean theorem:

$$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$$ $$|\vec{R}| = \sqrt{(5)^2 + (-12)^2}$$ $$|\vec{R}| = \sqrt{25 + 144}$$ $$|\vec{R}| = \sqrt{169}$$ $$|\vec{R}| = 13 \text{ km}$$

The resultant vector is in the fourth quadrant (East and South). Let $\theta$ be the angle measured from the East direction towards the South.

$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{|R_y|}{|R_x|} = \frac{12}{5}$$ $$\theta = \arctan\left(\frac{12}{5}\right)$$ $$\theta \approx 67.38^\circ$$

The direction is $67.38^\circ$ South of East.

The magnitude of the resultant displacement is $\boxed{13 \text{ km}}$ and the direction is $\boxed{67.38^\circ \text{ South of East}}$.