# Finding the unit vectors parallel to a tangent line

This is the solution to the problem. However, I dont understand how they got to the part that says "and the parallel vector is i+4j..". What does this mean and how did they derive that?

-

## 1 Answer

For any real number $m$, the vector $(1,m)$ determines a line of slope $m$ through the origin: simply note that the line through $(0,0)$ and $(1,m)$ has rise $m$ and run $1$.

In this case, your $m=4$, so the vector $(1,4) = 1i + 4j = i+4j$ is a vector of the appropriate slope, hence parallel to the tangent.

-
So only if the x component is 1 and you have (1,m), the slope is m? Whats so special about 1? – maq Feb 7 '11 at 1:55
@mohabitar: That it makes things easy! Or, that it is pretty much the easiest number to divide by. The vector $(x,y)$, with $x\neq 0$, determines a line of slope $\frac{y}{x}$. So you want to find any $x$ and $y$ with $\frac{y}{x}=m$. You can certainly find others, but isn't the simplest thing to take $x=1$? – Arturo Magidin Feb 7 '11 at 1:58
Ohh ok didnt realize that. And so is there any such/similar rule for 3D vectors? – maq Feb 7 '11 at 2:00
It depends on how you "know" the line. Lines through the origin in 3D need more than a single number to be determined (while in 2D the slope does it). If you know two points on the line, then the vector determined by their difference gives you a direction. – Arturo Magidin Feb 7 '11 at 2:21