How to calculate the intersection point of a vector and a plane defined as a point and normal vector? I've read through some of the questions on line-plane intersection but I'm struggling because many define the plane in a format like:  
$4x-2y+5z-9=0$,  
but my plane is defined as a point ($x_p, y_p, z_p$) and the unit vector normal to the plane ($i_p, j_p, k_p$).  
My line is defined by a point ($x_v, y_v, z_v$) and a unit vector in the direction of the line ($i_v, j_v, k_v$),
I can't figure out how to apply what I've read in other questions to my scenario, which I will need to implement in some generic VB.NET code.
 A: Let $P_p=(x_p,y_p,z_p)$, $P_v=(x_v,y_v,z_v)$, $\vec{v_v}=(i_v,j_v,k_v)$ and $\vec{v_p}=(i_p,j_p,k_p)$. Then
$$P_{intersection} = P_v + \frac{(P_p-P_v)\cdot\vec{v_p}}{\vec{v_v}\cdot\vec{v_p}} \ \vec{v_v}$$
I assume you already know the basics of operations with vectors and points...

EDIT: This formula can be easily derived from the vectorial definition of planes and lines. A point $P$ belongs to the plane defined by $P_p$ and $\vec{v_p}$ if $(P-P_p)\cdot\vec{v_p}=0$. Moreover, points of the line defined by $P_v$ and $\vec{v_v}$ are of the form $P_v+\lambda \vec{v_v}$. Note that the intersection point has to satisfy both conditions, so it is enouh to plug in the line form into the plane equation and solve:
$$(P_v+\lambda \vec{v_v}-P_p)\cdot\vec{v_p}=0 \iff \lambda =\frac{(P_p-P_v)\cdot\vec{v_p}}{\vec{v_v}\cdot\vec{v_p}}$$
Of course, if $\vec{v_v}\cdot\vec{v_p}=0$, both elements would be parallel, so there would not be any intersection point.

Edit: the originally stated formula was incorrect. This has now been fixed.
A: If $\vec P=(x,y,z)$ is some point of the plane that is given by point $\vec A$ and a normal $\vec n$, then by definition of normal the vector $\vec{AP}=\vec P -\vec A$ is perpendicular to $\vec n$, so the scalar product $\vec{AP}\cdot\vec n=0$.
By writing this in coordinate form, you will get the plane equation:
$$
(x-x_v)i_v+(y-y_v)j_v+(z-z_v)k_v = 0,\\
i_vx+j_vy+k_vz-(i_vx_v+j_vy_v+k_vz_v)=0
$$
