Lecture 4: Planes

Overview

A plane is defined by a point $p_1$ and a normal $\hat n$. The plane consists of all points for which $(p-p_1)$ is perpendicular to $\hat n$.

Or, using the dot product:

$$f(p) = (p - p_1) \cdot \hat n = 0$$

Recall our equation for a ray:

$$p(t) = p_0 + t * \hat d$$

Plugging in for $p$ gives us:

$$f(p) = (p_0 + t * \hat d - p_1) \cdot \hat n = 0$$

Derivation

Solve for $t$:

$$((p_0 - p_1) + \hat d * t) \cdot \hat n = 0$$ $$(p_0 - p_1) \cdot \hat n + d * t \cdot \hat n = 0$$ $$(p_0 -p_1) \cdot \hat n = -t * (\hat d \cdot \hat n)$$ $$t = \frac{(p_1 - p_0) \cdot \hat n}{\hat d \cdot \hat n}$$

But now… $p_0$ is the origin of our ray but we need a $p1$

Povray gives us planes in this format:

plane {
    <A, B, C>, D
}

Where <A, B, C> is the normal of the plane and D is the “distance” of the plane, or how far from the origin the plane is along the normal. Therefore, $\hat n * D$ is a point on the plane.

$$t = \frac{(\hat n * D - p_0) \cdot \hat n}{\hat d \cdot \hat n}$$

The dot product of a normalized vector with itself is 1, so we really have:

$$t = \frac{D * (\hat n \cdot \hat n) - (p_0 \cdot \hat n)}{\hat d \cdot \hat n}$$ $$t = \frac{D - (p_0 \cdot \hat n)}{\hat d \cdot \hat n}$$

This gives us $t$ in terms of known quantities.

Other Plane Equation

It’s important to note that there is another commonly used equation for a plane.

$$A*x + B*y + C*z + D = 0$$

It looks like we can use this equation with the four parameters given by the Povray plane specification. However, for “simplicity”, Povray actually gives us -D for this equation. i.e. in Povray, the intended equation is:

$$A*x + B*y + C*z - D = 0$$ $$A*x + B*y + C*z = D$$

You can also use this equation to find a $p_1$ for the plane.