Answer
(a) $1-\sqrt 3$
(b) $105^{\circ}$
(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 for our vectors:
$v \cdot w=( i+\sqrt 3 j) \cdot (i -j) = (1) (\sqrt 3) +(\sqrt 3)(-1) \\=1-\sqrt 3$
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 into the above equation the dot product from (a) and the vector lengths that we calculate with the magnitude formula:
This yields: $\theta=\cos^{-1} (\dfrac{1-\sqrt 3}{ (2) (\sqrt 2)}) \approx 105^{\circ}$
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.
