Answer:
[tex]\theta= \frac{\pi}{2} +\pi \cdot i[/tex], for all [tex]i = \mathbb{Z} \cup\{0\}[/tex]
Step-by-step explanation:
The velocity vector is found by deriving the position vector depending on the time:
[tex]\dot r(t)= v (t) = 3 \cdot i +\sqrt{2} \cdot j + 2\cdot t \cdot k[/tex]
In turn, acceleration vector is found by deriving the velocity vector depending on time:
[tex]\ddot r(t) = \dot v(t) = a(t) = 2 \cdot k[/tex]
Velocity and acceleration vectors at [tex]t = 0[/tex] are:
[tex]v(0) = 3\cdot i + \sqrt{2} \cdot j\\a(0) = 2 \cdot k\\[/tex]
Norms of both vectors are, respectively:
[tex]||v(0)||\approx 3.317\\||a(0)|| \approx 2[/tex]
The angle between both vectors is determined by using the following characteristic of a Dot Product:
[tex]\theta = \cos^{-1}(\frac{v(0) \bullet a(0)}{||v(0)||\cdot ||a(0)||})[/tex]
Given that cosine has a periodicity of [tex]\pi[/tex]. There is a family of solutions with the form:
[tex]\theta= \frac{\pi}{2} +\pi \cdot i[/tex], for all [tex]i = \mathbb{Z} \cup\{0\}[/tex]
