Solving Inner Product Equations I'm trying to solve an exercise from Cheney's Analysis for Applied Mathematics. Let $X$ be a normed linear space with $a,b,c\in X$ taken as fixed vectors, and consider the equation
$x+\langle x, a\rangle c = b$
The goal being to find a general solution $x$. There are obvious case by case trivial solutions (i.e. when $b\perp a$, $x=b$). Based on the fact that this exercise is in the first section on Hilbert spaces, I'm inclined to think I should be able to be solved by applying axioms for an inner product space. I've tried taking the inner product of both sides with some combination of $a,b,$ or $c$, but thus far, to no success.
 A: Starting from  
$$x+\langle x,a\rangle c=b \tag 1$$
we form the inner product with $a$ to obtain  
$$\langle x,a\rangle+\langle x,a\rangle \langle c,a\rangle =\langle b,a\rangle \tag 2$$
Solving $(2)$ for $\langle x,a\rangle$ yields
$$\langle x,a\rangle=\frac{\langle b,a\rangle }{1+\langle c,a\rangle } \tag 3$$
for $1+\langle c,a\rangle \ne0$.  Substituting $(3)$ into $(1)$ and solving for $x$ reveals
$$\bbox[5px,border:2px solid #C0A000]{x=b-\frac{\langle b,a\rangle }{1+\langle c,a\rangle }c }$$

As @Chappers pointed out, for the case $1+\langle c,a\rangle =0$, we have that
$$x=b+sc \tag 4$$
for any complex valued $s$.  We can easily verify consistency of this solution by substituting $(4)$ into $(1)$ and using the fact that $1+\langle c,a\rangle =0 \implies \langle b,a\rangle =0$.
A: Supplementing the other answer, if $\langle c,a \rangle = -1$, the solution is more degenerate: first, we have, taking the inner product with $a$,
$$ \langle x,a \rangle(1+\langle c,a \rangle) = 0 = \langle b,a \rangle. $$
Now, taking a vector of the form $x=b+tc+y$, with $t$ real and $\langle y,c \rangle = 0$, we have
$$ b+tc+y + \langle b+tc+y,a \rangle c = b, $$
which using what we have so far cancels to
$$ y+\langle y,a \rangle c = 0 $$
But now taking the inner product with $y$, we find
$$ 0 = \langle y,y \rangle + \langle y,a \rangle \langle y,c \rangle = \langle y,y \rangle, $$
so $y=0$ and we conclude that the most general possible solution is the line
$$ \{b+tc : t \in \mathbb{R}\}. $$
(If the space is complex, it works for any complex $t$ if the linearity is in the first argument, but otherwise only for real $t$.)
