Answer:
Only at [tex]t = 0[/tex].
Step-by-step explanation:
Differentiate [tex]\vec{r}(t)[/tex] (the position vector at time [tex]t[/tex]) with respect to [tex]t\![/tex] to find expressions for the velocity vector [tex]\vec{v}(t)[/tex] and the acceleration vector [tex]\vec{a}(t)[/tex].
[tex]\begin{aligned} & \vec{v}(t) \\ &= \frac{d}{dt}\left[\vec{r}(t)\right] \\ &= \frac{d}{dt}\left[\left(2\, t^{3}\right)\, \vec{i} + (5\, t) \, \vec{j} - \left(t^{2}\right)\, \vec{k}\right] \\ &= \left(\frac{d}{dt}\left[2\, t^3\right]\right)\, \vec{i} + \left(\frac{d}{dt}\left[5\, t\right]\right)\, \vec{j} - \left(\frac{d}{dt}\left[t^2\right]\right)\, \vec{k} \\ &= \left(6\, t^2\right)\, \vec{i} + 5\, \vec{j} - 2\, t\, \vec{k} \end{aligned}[/tex].
[tex]\begin{aligned} & \vec{a}(t) \\ &= \frac{d}{dt}\left[\vec{v}(t)\right] \\ &= \frac{d}{dt}\left[\left(6\, t^{2}\right)\, \vec{i} + 5 \, \vec{j} - \left(2\, t\right)\, \vec{k}\right] \\ &= \left(\frac{d}{dt}\left[6\, t^2\right]\right)\, \vec{i} + \left(\frac{d}{dt}\left[5\right]\right)\, \vec{j} - \left(\frac{d}{dt}\left[2\, t\right]\right)\, \vec{k} \\ &= \left(12\, t\right)\, \vec{i} + 0\, \vec{j} - 2\, \vec{k} \end{aligned}[/tex].
Two vectors are orthogonal to one another when their dot product is zero.
Calculate the dot product of [tex]\vec{v}(t)[/tex] and [tex]\vec{a}(t)[/tex]:
[tex]\begin{aligned}& \left(\vec{v}(t)\right) \cdot \left(\vec{a}(t)\right) \\ &= \left(6 \, t^2\right) \cdot (12\, t) + 5 \times 0 + (-2\, t) \cdot (-2) \\ &= 72\, t^3 + 4\, t\end{aligned}[/tex].
Set the dot product of [tex]\vec{v}(t)[/tex] and [tex]\vec{a}(t)[/tex] to zero and solve for [tex]t[/tex]:
[tex]18\, t^3 + t = 0[/tex].
[tex]t\, \left(18\,t ^2 + 1\right) = 0[/tex]
The only real root of this equation is [tex]t = 0[/tex].
Therefore, [tex]\vec{v}(t)[/tex] and [tex]\vec{a}(t)[/tex] are orthogonal to one another only at [tex]t = 0[/tex].
