I am trying to understand a few details in Jürgen Richter-Gebert's book "Perspectives on Projective Geometry" when he explains Plucker’s $\mu$ in Section 6.3. He gives a general formulation of the trick and some specific examples. My problem is that I sort of understand the general formulation, and fully understand each of the examples separately, but I don't quite see how the general rule maps onto each specific problem instances (instead each is verified by simple algebra). As such it isn't computational for me yet, and I want to really understand how to take a new problem, map it onto the general setting in order to solve the specific instance. I'm going to try to explain where the gaps in my understanding are while at the same time being faithful to the explanation in the book. I'll add questions illustrating my misunderstandings/gaps (in italicized parenthetical remarks).
The trick is described as follows. Suppose you have functions $f:\mathbb{R}^d\rightarrow\mathbb{R}$ and $g:\mathbb{R}^d\rightarrow\mathbb{R}$ whose zero sets describe two geometric objects. We're interested in intersections of $f$ and $g$ as well as objects that share the intersection of $f$ and $g$ with some third point. To make things concrete, let's assume $f$ and $g$'s zero-sets are two 2D lines, so $f(x,y)=a_1 x + b_1 y + c_1$ and $g(x, y) = a_2 x + b_2 y + c_2$. I'm going to use the author's notation and use $f(p)$ to denote the value of $f$ evaluated on the coordinates of the point $p=(x, y)$.
First he says, "If the linear combination $\lambda\cdot f(p)+\mu\cdot g(p)$ again describes an object of the same type then one can apply Plücker's $\mu$." (Q1: I am assuming $\lambda, \mu\in\mathbb{R}$. In that case, isn't the linear combination just an element of $\mathbb{R}$ and if so, what does he mean by this being the same type of object? Maybe he means the zero-set of the linear combination needs to be the same type of geometric object as $f$ and $g$? I think this is right.)
"All objects described by the map $p\mapsto\lambda\cdot f(p)+\mu\cdot g(p)$ will pass through the common zeros of $f$ and $g$."
Then, "If one in addition wants the object to pass through a specific point $q$ then the linear combination $p\mapsto g(q)\cdot f(p) - f(q) \cdot g(p)$ is the desired equation."
Ok, so all of that above seems to make sense and I understand his first example for obtaining the intersection of two lines. What I don't understand is how the second example maps onto Plücker's method.
The set-up is this. Let $a, b, c, d$ be four points, no three of which are collinear, in the plane. We want to find the point of intersection between the lines $a\vee b$ and $c\vee d$. Obviously, we could do the normal algebraic approach here, but the point is to show how to use Plücker's trick to solve it easily. Also, by way of notation we're going to represent our 2D projective points using homogeneous coordinates and use $[a, b, c]$ to denote the determinant of the 3x3 matrix with rows given by the coordinates of points a, b, and c.
He says, "The property that encodes that a point $p$ is on the line $c\vee d$ is simply $[c, d, p] = 0$. Thus we immediately obtain that the point $[c, d, b]\cdot a - [c, d, a]\cdot b$ must be the desired intersection." He goes on to verify all of this algebraically.
The verification is clear (just basic algebra, so I believe the result), what is not clear is how we "immediately obtain" the result from Plücker's trick. First, we don't have a third object $q$. Second, what are $f$ and $g$? I am at a loss for how to set $f$ and $g$ and apply Plücker's trick to obtain $[c, d, b]\cdot a - [c, d, a]\cdot b$.
If I set $f(p) = [c, d, p]$ and $g(p) = p$, then at least $f(p) g(q) - f(q) g(p)$ becomes $[c, d, p]\cdot q - [c, d, q]\cdot p$ so that looks like the result, but why are we evaluating it at $a$ and $b$? And more importantly why do we not care that $g(p)$ doesn't have a zero set anymore? I can easily see that the algebra works out, but I cannot see how Plücker's method had anything to do with this.
I agree that the book is being rather obscure here. The clue is (I'm quoting from the book here) that it is "using (a dual version of) Plücker's $\mu$".
So points become lines and vice versa, and $a$ is represented by $f(\mathscr{l})=0$, and $b$ is represented by $g(\mathscr{l})=0$, where $\mathscr{l}$ is a line. Dualizing/paraphrasing some remarks in the book: All points described by $$ \mathscr{l}\mapsto\lambda\cdot f(\mathscr{l})+\mu\cdot g(\mathscr{l}) $$ will be on the common zero of $f$ and $g$ (which in this case is the line $a\vee b$) ... If one wants in addition to have the point be on a specific line $\mathscr{m}$ then the linear combination $$ \mathscr{l}\mapsto g(\mathscr{m})\cdot f(\mathscr{l})+f(\mathscr{m})\cdot g(\mathscr{l}) \tag{1} $$ is the desired equation.
Getting back to the intersection of $a\vee b$ and $c\vee d$, let $\mathscr{m}=c\vee d$. Then (1) becomes $$ \mathscr{l}\mapsto [c,d,b]\cdot f(\mathscr{l})+[c,d,a]\cdot g(\mathscr{l}) $$ which, after translating from equations back to objects, gives us the intersection point $$ [c,d,b]\cdot a+[c,d,a]\cdot b. $$
Update: I skipped a few details in the above, which @John has expanded in the comments below.