From Massey's Algebraic topology:
Any continuous map of $E^n$ into itself has at least one fixed point.
The proof starts by trying showing that if there exists a continuous function $f(x)$ from $E^2$ into itself that has no fixed points then $S^1$ is a retract of $E^2$
He then constructs $r(x)$ by drawing a point $x=(x_0,y_0)$ and creating a line from $f(x)$ through $x$, and into $r(x)$ on $S^1$.
He then says that, using vector notation, we can easily write a formula for $r(x)$ in terms of $f(x)$ and uses this formula in the proof, but I have no idea how to write this using vector notation. I see that $v^{\rightarrow} = f(x) + t(f(x) - x)$ would be a line that will eventually intersect $S^1$, but I don't see how I could write this function so that the final point would be in $S^1$.
Anyone have any ideas?
I'm going to modify your formula for $v$ slightly, writing $$ v(t) = f(x) + t (x - f(x)) $$ so that $v(0) = f(x)$ and $v(1) = x$, and as $t$ ranges over nonnegative values, it describes a ray from $f(x)$, through $x$, one that hits the unit sphere at some point, and continues onward.
Writing $u$ for $f(x)$ to simplify things, and $v$ for $v(t)$, if you want $\|v\| = 1$, i.e., you want $v(t)$ to be on the unit sphere, you need $ v\cdot v = 1$, i.e., $$ (u + t(x-u)) \cdot (u + t (x-u)) = 1 $$ which turns into $$ u \cdot u + 2t (x-u) \cdot u + t^2 (x-u) \cdot (x-u) = 1 $$ which is a quadratic in $t$ that has only one positive solution, which is $$ t = \frac{ - (x-u) \cdot u + \sqrt{ ((x-u) \cdot u)^2 - 4 \|x-u\|^2 (\|u\|^2 - 1)}} {2 \|x-u\|^2} $$
Yeah, it's a mess. But it's the answer you need.