Suppose we have a particle which starts from a point $A$ and moves with constant speed $u$ along the line $AB$.
One wants to show that the position vector $\mathbf{x}$ of the particle at time $t$ is $$\mathbf{x}=\mathbf{a}+ut\frac{\mathbf{b}-\mathbf{a}}{|\mathbf{b}-\mathbf{a}|},$$ where $\mathbf{a}$ and $\mathbf{b}$ are the position vectors of $A$ and $B$ relative to an arbitrary origin.
So, first we write down the vector equation of the straight line passing through two points $A$ and $B$, that is $$\mathbf{x}=\mathbf{a}+k(\mathbf{b}-\mathbf{a}),$$ where $k$ is a parameter. From the position equation of the particle $$\mathbf{x}(t)=\mathbf{v}_0t+\mathbf{x}_0,$$ at any time $t$, we have $\mathbf{x}_0=\mathbf{a}$ at $t=0$ .
But I am stuck on how to derive that $\mathbf{v_0}$ is equal to $u\frac{\mathbf{b}-\mathbf{a}}{|\mathbf{b}-\mathbf{a}|}$.
I would appreciate any help or suggestion. Thank you.
Best Answer
$\mathbf{b-a}$ is a vector in the direction from $\mathbf{a}$ to $\mathbf{b}$. Dividing by its length produces a unit vector in that direction.
Then multiplying by the (presumably positive) scalar $u$ produces a vector of magnitude $u$ in the direction from $\mathbf{a}$ to $\mathbf{b}$, which is precisely what you want as the velocity vector.