# 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.

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 $$|\vec{v}|=\sqrt{3^2+(-1)^2+1^2}$$

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

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

2. 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: $$\\\frac{<3,-1,1>}{\sqrt{3^2+(-1)^2+1^2}}=\frac{\vec{v}}{|\vec{v}|}=\frac{<12,-4,4>}{\sqrt{12^2+(-4)^2+4^2}}.$$

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.