Answer
(a) $6$
(b) $18.4^{\circ}$
(c) Neither
Work Step by Step
Let us consider two vectors $v=pi+qj$ and $w=xi+yj$
Then we have:
$v \cdot w=(pi+qj) \cdot (xi+yj) =px+qy$
The magnitude of any vector (let us say $v$) can be determined using the formula:
$||v||=\sqrt{p^2+q^2+r^2} $
a)
We calculate the dot product of the two given vectors:
$v \cdot w=(2i +2j) \cdot (i +2j) = (2)(1) +(2)(2) \\= 6$
b)
Let us consider two vectors $v=pi+qj+rk$ and $w=xi+yj+zk$. If $\theta$ is the angle between the two vectors, then we have: $ \cos \theta =\dfrac{v \cdot w}{|v||w|}$
We plug in dot product from (a) and our vector lengths (calculated with the magnitude formula):
$\theta=\cos^{-1} (\dfrac{6}{\sqrt 8 \sqrt {5}}) \approx 18.4^{\circ}$
c)
When the dot product between the two vectors is $0$, then they are perpendicular or orthogonal. For the vectors to be parallel, the angle has to be $0^{\circ}$ or $180^{\circ}$. In this case, both vectors are neither orthogonal nor parallel.
