How can I find the equation for the position vector of a particle at time t? Question: At time $t = 0$, a particle is located at the point (I, 2, 3). It travels in a straight line to the point (4, 1,4), has speed 2 at (I, 2, 3) and constant acceleration $3i - j + k$. Find an equation for the position vector $r(t)$ of the particle at time $t$.
The solution says to find the velocity vector by integrating the acceleration vector. When I do this, I get
$\vec{v}=3t\vec{i}-t\vec{j}+t\vec{k}+C$
The solution also says the the vector of the particle path is $<3,-1,1>$ since it's just $<4-1,1-2,4-3>$. I understand this part as well.
I'm confused about this part:
The solution says that since the speed is 2, this means $|\vec{v}|=2$ and
$\frac{\vec{v}}{|\vec{v}|}=\frac{<3,-1,1>}{\sqrt{3^2+(-1)^2+1^2}}$
The velocity is a product of direction and speed so
$\vec{v}=|\vec{v}|*\frac{\vec{v}}{|\vec{v}|}=2*\frac{<3,-1,1>}{\sqrt{3^2+(-1)^2+1^2}}$
I don't understand why I'm using
$|\vec{v}|=2$ and $|\vec{v}|=\sqrt{3^2+(-1)^2+1^2}$
 A: 
The solution also says the the vector of the particle path is $<3,-1,1>$
The solution says that since the speed is 2, this means $|\vec{v}|=2$ and $\frac{\vec{v}}{|\vec{v}|}=\frac{<3,-1,1>}{\sqrt{3^2+(-1)^2+1^2}}.$
The velocity is a product of direction and speed so $\vec{v}=|\vec{v}|*\frac{\vec{v}}{|\vec{v}|}=2*\frac{<3,-1,1>}{\sqrt{3^2+(-1)^2+1^2}}$

The above is totally correct. Your confusion

I don't understand why I'm using $|\vec{v}|=\sqrt{3^2+(-1)^2+1^2}$

stems from

*

*thinking that \begin{gather}\frac{\vec{v}}{|\vec{v}|}=\frac{<3,-1,1>}{\sqrt{3^2+(-1)^2+1^2}}\tag{*}\\\iff \vec{v}=<3,-1,1>,\end{gather}


*and conflating path vector and velocity vector.
The given path/direction vector $<3,-1,1>$ (let's call it $\vec{h})$ is unique up to a positive scalar multiple: after all, $<9,-3,3>$  and $<12,-4,4>$ too can be considered the particle's path vector. Each of these path vectors has the same unit vector equalling that of the velocity vector $\vec{v};$ equation $(*)$ is stating this relationship between $\vec{v}$ and $\vec{h}.$
To be clear: $$\\\frac{<3,-1,1>}{\sqrt{3^2+(-1)^2+1^2}}=\frac{\vec{v}}{|\vec{v}|}=\frac{<12,-4,4>}{\sqrt{12^2+(-4)^2+4^2}}.$$
A: You seem to be using the notation $|\overrightarrow{v}|$ in two different ways, which is confusing.
The direction of motion is $\left(\begin{matrix}3\\-1\\1\end{matrix}\right)$, and the unit vector in this direction is $$\frac{1}{\sqrt{11}}\left(\begin{matrix}3\\-1\\1\end{matrix}\right).$$So at the point $(1,2,3)$, when the magnitude of the velocity (i.e. the speed) is $2$, the velocity of the particle is $$\frac{2}{\sqrt{11}}\left(\begin{matrix}3\\-1\\1\end{matrix}\right).$$
I hope this helps.
