\( \newcommand{\vecIII}[3]{\left[\begin{array}{c} #1\\#2\\#3 \end{array}\right]} \newcommand{\vecIV}[4]{\left[\begin{array}{c} #1\\#2\\#3\\#4 \end{array}\right]} \newcommand{\Choose}[2]{ { { #1 }\choose{ #2 } } } \newcommand{\vecII}[2]{\left[\begin{array}{c} #1\\#2 \end{array}\right]} \newcommand{\vecIII}[3]{\left[\begin{array}{c} #1\\#2\\#3 \end{array}\right]} \newcommand{\vecIV}[4]{\left[\begin{array}{c} #1\\#2\\#3\\#4 \end{array}\right]} \newcommand{\matIIxII}[4]{\left[ \begin{array}{cc} #1 & #2 \\ #3 & #4 \end{array}\right]} \newcommand{\matIIIxIII}[9]{\left[ \begin{array}{ccc} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \end{array}\right]} \)
  1. Can you briefly go over Distance from a Point to a Plane?

    Sure. We need

    • Q the point off the plane
    • N, the surface normal of the plane.
    • P, the point on the plane
    • W, the vector from P on the plane to Q, which is just Q-P
    • $\theta$ the angle between N and W.

    We want to find D, the (scalar) distance. \[ \begin{eqnarray*} d &=& \cos\theta |W| \\ d &=& \frac{N\cdot W}{|N||W|} |W| \\ d &=& \frac{N\cdot W}{|N|} \\ d &=& \frac{N\cdot (P-Q)}{|N|} \\ \end{eqnarray*} \]

  2. Can you go over Intersecting a Ray with a Plane? / Math behind intersection of ray and plane?

    Yes, let's do that. The key idea is:

    The dot product of the surface normal with every point on the plane is the same
    \[ N \cdot (x,y,z) = N \cdot P \]
  3. In the "better way" of intersecting a ray with a plane, what do the variables P and N represent exactly (I know the plane is defined by them though)?