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

(a) $1-\sqrt 3$ (b) $105^{\circ}$ (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 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.