Point of intersection in a poorly written question? Yep, I got burned again by doing some past papers and a really badly formulated question which will most likely appear on my exam paper has emerged once more. As I am studying Frenet Frames and now and the different curves and splines, I got to this question which is incredibly unclear on what it wants from me to do and more importantly it is bad enough to appear on the upcoming exam.
The question goes like this:

My first question is  some sort of a predefined equation which one has to know before tackling the probelm and how can it be related to computer graphics?
Secondly, how can I find the intersection point without being given any firm parameters? I cannot seem to find any good example to practice with in order to find a solution for it and the hint for parametric form of line has rather led me astray as I am reading mostly (I believe so) unrelated material to the overall problem I have on my hands here.
Any ideas on getting a direction on how and where to start?
 A: The idea is to write down the general formula for the intersection point of a plane with a line. In computer graphics this problem may emerge while trying to construct a polyhedral surface made out of little triangles, or other types of polygons. Could be also the solution to an optimization problem of some kind. It is sounds like a very reasonable basic geometric problem that one may need to solve.
The idea is that you should have a bunch of input data that defines a plane and a line in three space and you have to construct an equation that you will put in your computer function, which will spit out either the three coordinates of the intersection point, if it exists, or return a message that there is no intersection point, if there isn't one. 
Since there could be some ambiguity about how a plane in three space is defined, the question specifies what kind of data that determines a plane you need to consider. More precisely, given the equation $(p_s-x)\cdot \hat{n}=0$, which is written in vector form, you are told that your plane passes through a given point $c = (c_1, c_2, c_3)$ and is orthogonal to a given direction, determined by the vector $\hat{n} = (n_1, n_2, n_3)$, also called a normal vector. 
The equation $(p_s-x)\cdot \hat{n}=0$ says that a point $p_s = (x, y, z)$ in three space is a point on your specific plane, let's call it $P$ defined by the point $c$ and the orthogonal vector $\hat{n}$, if and only if it satisfies the equation $$(p_s-x)\cdot \hat{n}=0 \,\, \text{ or in coordinates}$$
$$n_1(x-c_1) + n_2(x-c_2) n_3(x-c_3) = 0.$$
Now, a line $L$ in three space is most frequently defined by a point $a=(a_1, a_2, a_3)$ it passes through and a direction, determined by a vector $\hat{v} = (v_1, v_2, v_3)$, that the line follows. This equation is usually written in vector form as 
$$p_s = a + t \hat{v},$$ or in coordinate form
\begin{align}
x &= a_1 + t\,v_1 \\ 
y &= a_2 + t\,v_2 \\ 
z &= a_3 + t\,v_3
\end{align}
where $p_s=(x,y,z)$ is a point from three space which lies on the line if and only if it satisfies the equation  $p_s = a + t \hat{v}$. In particular, this equation determines a one-to-one correspondence between the parameter $t$ and the points $p_s=a+t\hat{v}$ lying on the line $L$. If I have a parameter $t$ I have a point on $L$ and vice versa. 
And so: input data -- $P=[c,\hat{n}]$ and $L=[a,\hat{v}]$. Output data: $q_0 = (x_0, y_0, z_0)$ such that $q_0 \in L$ and $q_0 \in P$ which means that 
\begin{align} 
&(q_0-c)\cdot\hat{n}=0\\
&q_0 = a + t_0\hat{v}
\end{align}
Well, to solve this system of equations, simply take the second one and plug it into the first:
\begin{align} 
0&=(a + t\hat{v}-c)\cdot\hat{n}\\
&=\big((a-c) + t \hat{v}\big) \cdot \hat{n}\\
&=(a-c)\cdot\hat{n} + t(\hat{v}\cdot\hat{n})
\end{align}
Thus we obtain an equation for the parameter $t$. We solve
$$(a-c)\cdot\hat{n} + t(\hat{v}\cdot\hat{n})=0$$
and obtain $$t_0 = - \frac{(a-c) \cdot \hat{n}}{(\hat{v}\cdot\hat{n})}$$
Finally the intersection point you are looking for can be written as equation 
$$q_0 = a  - \left(\frac{(a-c)\cdot\hat{n}}{(\hat{v}\cdot\hat{n})}\right)\hat{v}$$ 
where everything is known because it's written in terms of the provided input data. Or in coordinate equations
\begin{align}
x_0 &= a_1 - \left(\frac{(a_1-c_1)n_1 + (a_2-c_2)n_2 + (a_3-c_3)n_3}{v_1n_1+v_2n_2+v_3n_3}\right)\, v_1\\
y_0 &= a_2 - \left(\frac{(a_1-c_1)n_1 + (a_2-c_2)n_2 + (a_3-c_3)n_3}{v_1n_1+v_2n_2+v_3n_3}\right)\, v_2\\
z_0 &= a_3 - \left(\frac{(a_1-c_1)n_1 + (a_2-c_2)n_2 + (a_3-c_3)n_3}{v_1n_1+v_2n_2+v_3n_3}\right)\, v_3.
\end{align}
So your computer function looks like this $[q_0] = \text{IntersectionPoint}\big([c,\hat{n}],[a,\hat{v}]\big)$.
Solution does not exist, i.e. the line does not intersect the plane, whenever $(\hat{v}\cdot\hat{n}) = 0$. This means that that the vector $\hat{v}$ is orthogonal to $\hat{n}$ and is therefore parallel to the plane $P$. Consequently, this happens only when the line is parallel to the plane.
A: $(\mathbf{p}_s-\mathbf{c})\cdot\hat{\mathbf{n}}=0$ is the standard equation of a plane: $\mathbf{p}_s$ is the generic point on the plane, $\mathbf{c}$ is a fixed point on the plane and $\hat{\mathbf{n}}$ is a unit vector perpendicular to the plane. You have to find the intersection point between that plane and a generic line. That means that you must find the value of $t$ in the parametric equation of the line $\mathbf{p}=\mathbf{a}+t\mathbf{b}$ (with $\mathbf{a}$ and $\mathbf{b}$ generic vectors), such that $\mathbf{p}$ belongs to the plane.
