If the ray is defined by a position and direction vector, and the plane is defined by a position and a normal vector, how can I find out the vector position of intersection?
For the case I'm interested in, the direction vector of the ray is the same as the opposite of the normal of the plane.
On
Given a ray $r$ defined by position $p$ and direction $d$, we have the parameterisation $$r(t) = p + td$$ A plane $\Pi$ defined by position $a$ and normal $n$ has a defining equation $$x · n = a· n$$ that is, $\Pi = \{ x : x · n = a· n\}$ An intersection means that for some parameter $t$, $r(t) ∈ \Pi$. This means $r(t) · n = a· n$. Using the definition of $r(t)$, we obtain the linear equation $$ td· n = (a-p)· n$$ If $d· n = 0$ then the point on the line $p$ needs to also be in the plane for a solution; in this case the ray $r$ is contained in the plane so any $t$ is a solution. If $d· n ≠ 0$ then the parameter is the unique solution given by $$ t = \frac{(a-p)· n}{d· n}$$
Which corresponds to the point $$r(t) = p + \frac{(a-p)· n}{d· n}d$$
In your special case with $d = -n$, we have the final answer $$ r(t) = p + ((a-p)· n)n $$
On
The plane equation is given by $\overrightarrow{y}.s.t.\overrightarrow{n}\dot{} (\overrightarrow{x}-\overrightarrow{y})=0$, where $\overrightarrow{n}$ is normal vector and $\overrightarrow{x}$ the position one. With the notation $\overrightarrow{y}=(x,y,z)$ Then you have to replace a general vector of the ray in the plane equation
A plane can be defined as the set of all vectors $\vec{x}$ satisfying $\vec{n}_{\rm Plane}\cdot(\vec{x} - \vec{x}_{\rm Plane}) = 0$ where $\vec{x}_{\rm Plane}$ is any point in the plane and $\vec{n}_{\rm Plane}$ is a normal vector.
A ray can be parameterized as $\vec{x}(t) = \vec{x}_{\rm Ray} + t\vec{D}_{\rm Ray}$ where $\vec{x}_{\rm Ray}$ is a point on the ray, $\vec{D}_{\rm Ray}$ is the direction vector and $t$ ranges over all real numbers from $-\infty$ to $\infty$.
To find the intersection point we simply substitute the equation for the ray into the equation for the plane and solve for $t$ to find
$$t_{\rm Intersection} = \frac{\vec{n}_{\rm Plane}\cdot (\vec{x}_{\rm Plane}-\vec{x}_{\rm Ray})}{\vec{n}_{\rm Plane}\cdot\vec{D}_{\rm Ray}}$$
Using this value for $t$ in the ray equation gives you the intersection point
$$\vec{x}_{\rm Intersection} = \vec{x}_{\rm Ray} + \frac{\vec{n}_{\rm Plane}\cdot (\vec{x}_{\rm Plane}-\vec{x}_{\rm Ray})}{\vec{n}_{\rm Plane}\cdot\vec{D}_{\rm Ray}}\vec{D}_{\rm Ray}$$
Note that if $\vec{n}_{\rm Plane}\cdot\vec{D}_{\rm Ray} = 0$ then there is no intersection point (or the whole ray lies in the plane).