First round incorporating Trevor's feedback

Author: hollasch

books/RayTracingTheNextWeek.html

@@ -2266,7 +2266,8 @@
 -----------------------
 Spheres are generally the first ray tracing primitive because their implicit formula makes it so
 easy to solve for ray intersection. Like spheres, planes also have an implicit formula, and
-ray-plane intersection is even _easier_ to solve.
+we can use their implicit formula to produce an algorithm that solves for ray-plane intersection.
+Indeed, ray-plane intersection is even _easier_ to solve than ray-sphere intersection.
 
 You may already know this implicit formula for a plane:
 
@@ -2279,12 +2280,15 @@
 
 (We didn't flip the sign of D because it's just some constant that we'll figure out later.)
 
-Here's another way to think of this formula: given the plane perpendicular to the normal vector
+Here's a good way to think of this formula: given the plane perpendicular to the normal vector
 $\mathbf{n} = (A,B,C)$, and the position vector $\mathbf{v} = (x,y,z)$ (that is, the vector from the
-origin to any point on the plane), then
+origin to any point on the plane), then we can use the dot product to solve for $D$:
 
   $$ \mathbf{n} \cdot \mathbf{v} = D $$
 
+for any position on the plane. This is an equivalent formulation of the $Ax + By + Cz = D$ formula
+given above, only now in terms of our 3D values.
+
 Now to find the intersection with some ray $\mathbf{R}(t) = \mathbf{P} + t\mathbf{d}$. Plugging in
 the ray equation, we get
 
@@ -2298,12 +2302,16 @@
 
   $$ t = \frac{D - \mathbf{n} \cdot \mathbf{P}}{\mathbf{n} \cdot \mathbf{d}} $$
 
-This gives us $t$, which we can plug into the ray equation to find the point of intersection.
+This gives us $t$, which we can plug into the ray equation to find the point of intersection. Note
+that the denominator $\mathbf{n} \cdot \mathbf{d}$ will be zero if the ray is parallel to the plane.
+In this case, we can immediately record a miss between the ray and the plane. As for other
+primitives, if the ray $t$ parameter is less than the minimum acceptable value, we also record a
+miss.
 
-Ok, we can find the point of intersection between a ray and a plane. This will help us determine
-whether a ray hits a given quadrilateral, using the plane that contains the (planar) quadrilateral.
-In fact, we can use this approach to define any planar primitive, like triangles and disks (more
-later).
+All right, we can find the point of intersection between a ray and a plane. This will help us
+determine whether a ray hits a given quadrilateral, using the plane that contains the (planar)
+quadrilateral. In fact, we can use this approach to define any planar primitive, like triangles and
+disks (more later).
 
 We now have three questions we must answer:
 
@@ -2333,8 +2341,8 @@
 
 Finding the Plane That Contains a Given Quadrilateral
 ------------------------------------------------------
-We have $O$, $u$, and $v$, and want the corresponding equation of the plane containing all three of
-these values (and thus the quad itself).
+We have $\mathbf{O}$, $u$, and $v$, and want the corresponding equation of the plane containing all
+three of these values (and thus the quad itself).
 
 Fortunately, this is very simple. Recall that in $Ax + By + Cz = D$, $(A,B,C)$ represents the
 normal vector. To get this, we just use the cross product of the two side vectors $u$ and $v$:
@@ -2365,7 +2373,7 @@
 Since a plane is a 2D construct, we just need a plane origin point $\mathbf{O}$ and _two_ basis
 vectors: $\mathbf{u}$ and $\mathbf{v}$. Now normally, you see axes that are perpendicular to each
 other, so they all meet at 90° angles. However, this doesn't need to be the case in order to span
-the entire space -- you just can't have two axes that are parallel to each other.
+the entire space -- you just need two axes that are not parallel to each other.
 
   ![Figure [ray-plane]: Ray-plane intersection](../images/fig-2.06-ray-plane.jpg)
 
@@ -2381,8 +2389,9 @@
 
 If $\mathbf{u}$ and $\mathbf{v}$ were guaranteed to be orthogonal to each other (forming a 90°
 angle between them), then this would be a simple matter of projecting $\mathbf{P}$ onto the two
-basis vectors $\mathbf{u}$ and $\mathbf{v}$. However, since they can form any angle (other than
-zero), the math's a little bit tricker.
+basis vectors $\mathbf{u}$ and $\mathbf{v}$. However, since we are not restricting $u$ and $v$ to be
+perpendicular, and allowing them to form any angle (other than zero), the math's a little bit
+trickier.
 
 $$ \mathbf{P} = \mathbf{O} + \alpha \mathbf{u} + \beta \mathbf{v}$$
 
@@ -2563,8 +2572,7 @@
 
 We'll leave these additional 2D primitives as an exercise to the reader, depending on your desire to
 explore. Consider triangles, disks, and rings. You could even create cut-out stencils based on the
-texture map! We've included the solution for triangle, ellipse, and annulus in the commit history
-(search the history for "annulus"), so you can take a peek if you'd like to see our solution.
+texture map!
 
 
 
@@ -2847,7 +2855,7 @@
 Instances
 ====================================================================================================
 The Cornell Box usually has two blocks in it. These are rotated relative to the walls. First, let’s
-make an axis-aligned block primitive that holds 6 rectangles:
+create a function that returns a box, by creating a `hittable_list` of six rectangles:
 
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
     ...