The position vector, $\mathbf{r}$, of a particle at time $t$ is given by $$\mathbf{r} = t\mathbf{i} + (2-t)\mathbf{j} + (t+1)\mathbf{k}$$ Find the smallest distance of the particle from the origin and the time at which this occurs.
$t = \dfrac{1}{3}$ and $d = \dfrac{14}{3}$
A particle has velocity at time $t$ given by $(2t+1)\mathbf{i} + (3-t)\mathbf{j}$. Initially, it has position vector $\mathbf{i} - \mathbf{j}$. Find, in terms of $t$, the position vector of the particle.
$(t^2 + t + 1)\mathbf{i} + (3t - \frac{1}{2}t^2 - 1)\mathbf{j}$