How can I intuitively interpret this vector operation? In reading through some very old source code that I inherited and came across a three-dimensional Euclidean vector operation that I can't seem to gain an intuition for. Transcribing the program code into mathematics, the operation (which has three inputs, vectors $a$, $b$, and $c$) is as follows:
$$
a - \frac{a \cdot b}{b \cdot c}c
$$
The comment alongside the source code advertises that the function "projects a vector into a plane along a third vector" and that the result is "vector a projected into the plane defined by b along c". 
My geometric intuition is a bit rusty, but I've been unable to put together any combination of vector and plane projections that would yield the expression above. The comments are a bit ambiguous and can be interpreted in multiple ways, so I suppose that I'm not looking at it in the right way. Could someone elaborate on how this sort of operation might be described qualitatively?
 A: notice that the expression for your vector does not depend on the magnitudes of $\vec b$ or $\vec c$ . It can be written in terms of unit vectors.
$$ \vec v = \vec a - \frac{\vec a \cdot \hat b}{\hat b \cdot \hat c}\hat c $$
imagine that $\hat b$ defines the normal vector to the ground 
and $\hat c$ defines the direction of the sun's rays , then $\vec v$ will be exactly the vector describing the shadow cast on the ground by the vector $\vec a$ 
A: If you take the scalar product with $b$ you get zero, so the vector constructed here is perpendicular to $b$, so lies in a plane perpendicular to $b$ (defined by $b$)
You obtain the vector by taking $a$ and adjusting it by a multiple of $c$, which might be what is meant by a projection along the vector $c$.
A: The vector is co-planar with $\underline {a}$ and $\underline {c}$ and is perpendicular to $\underline {b}$
A: You can solve for the formula given the description.
Projecting into the plane perpendicular to $b$ means you have
$$v \cdot b = 0$$
That you're projecting $a$ in the direction given by $c$ means you have
$$ v = a + t c $$
for some scalar $t$.
To make them both true (so that you're projecting to the plane defined by $b$), you solve the system of equations. Plugging the latter into the former gives
$$ a \cdot b + t (c \cdot b) = 0 $$
and thus
$$t = -\frac{a \cdot b}{c \cdot b}$$
