## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

(a) $6$ (b) $18.4^{\circ}$ (c) Neither
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.