How the author get from one step to another?

Question

Target paper: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.480.10&rep=rep1&type=pdf (page #14)

$uU\cdot N_2 + vV\cdot N_2 = AP\cdot N_2$
This is the familiar equation of a line in the uv-plane for real variables $u, v$. The vector equation using real parameter $\lambda$ becomes
$(u, v) = AP \cdot N_2 \frac{(U\cdot N_2, V \cdot N_2)}{(U\cdot N_2)^2 + (V\cdot N_2)^2} + \lambda (V\cdot N_2, -U\cdot N_2)$

My question is:

How does the author get the second equation from the first one?

Answer 1

This is indeed the familiar equation for a line $$ Au+Bv=C, $$ assuming $A,B$ not both zero. The normal to the line is $(A,B)$ so a direction vector is $(B,-A)$. The point $\frac{C}{\lvert(A,B)\rvert^2}(A,B)$ lies on the line so you get parametric equation $$ (u,v)=\frac{C}{\lvert(A,B)\rvert^2}(A,B)+\lambda(B,-A). $$

Answer 2

Just let's write first the equation in a more convenient way:

$$au+bv = d$$

with $a=U\cdot N_2, b= V\cdot N_2, d = AP\cdot N_2 $.

Since this is a linear equation the solution is the sum of one particular solution $(u_p, v_p)$ and the general solution $(u_h,v_h)$ of the corresponding homogeneous equation $au_h+bv_h=0$.

The homogeneous equation has the obvious solution

$$au+bv=0:\:(u_h,v_h) = \lambda (b,-a) =\lambda (V\cdot N_2,-U\cdot N_2)$$

Now, the particular solution is constructed as follows

$$d = d\cdot \frac{a^2+b^2}{a^2+b^2} = \frac{d}{a^2+b^2}(a,b)\cdot (a,b)$$ where $(a,b)\cdot (a,b)$ is the scalar product of $(a,b)$ with itself.

Hence,

$$(u_p,v_p) = \frac{d}{a^2+b^2}(a,b) = AP \cdot N_2 \frac{(U\cdot N_2, V \cdot N_2)}{(U\cdot N_2)^2 + (V\cdot N_2)^2}$$

is a particular solution.