Reading: Geometry

A fair amount of 3D geometry is involved in automatically setting up the camera. (Even more is involved in using the mouse to drag the camera to a different viewpoint, but we will defer that for now.) We also need to understand some geometry of vectors to do the computations for material and lighting.

So, we're going to attack that now. This reading explains the relevant math. The most important part of the following is understanding the dot product. The dot product is used in several ways in the lighting model, so the more comfortable you are with it, the better. We won't use cross-products in material and lighting, but we will later in the semester, and it fits reasonably well here, but if you get lost in understanding the cross-product, don't worry.

Vector Math

Intuitively and geometrically, a vector is an arrow. It has length and direction. In a coordinate system, we can represent a vector as a triple of numbers $(\Delta x, \Delta y, \Delta z)$.

Of course, a triple of numbers is also used to represent points. Is there any difference? Yes. A point has location but not length, and a vector has length but not location. You can get one from the other, though. If we have two points, A and B, we can find the vector from A to B using subtraction: \[ \vecIII{\Delta x}{\Delta y}{\Delta z} = \vec{v} = B - A = \vecIII{b_x}{b_y}{b_z} - \vecIII{a_x}{a_y}{a_z} = \vecIII{b_x-a_x}{b_y-a_y}{b_z-a_z} \]

Vectors and Points in Homogeneous Coordinates

Interestingly, we can tell the difference between vectors and points in homogeneous coordinates, which we learned about in the context of affine transformations. Points have $w=1$, and (as you can see from the subtraction above) vectors have $w=0$: \[ \vecIV{\Delta x}{\Delta y}{\Delta z}{0} = \vec{v} = B - A = \vecIV{b_x}{b_y}{b_z}{1} - \vecIV{a_x}{a_y}{a_z}{1} \]

Dot Product

• $v\cdot w\in\Re$. That is, the dot product of two vectors is just a real number, a scalar (not a vector).
• $\forall\ v, v\cdot v=0 \mathrm{~iff~} v=\vec{0}$. That is, the dot product of a vector with itself is zero if and only if the vector is the zero vector. In 3D, the zero vector is (0,0,0).
• (commutativity) $v \cdot w = w \cdot v$. That is, the order of the vectors doesn't matter.
• (linearity) $v \cdot (au+bw) = a(v\cdot u) + b(v \cdot w)$. That is, the dot product of a vector $v$ with the linear combination of two other vectors, $u$ and $w$, is the same as the linear combination of the two dot products.

But in this course, we're going to be very concrete. We will define the dot product as the sum of products: \[ v \cdot w = \sum_{i=0}^{n} v_i w_i \]

In fact, since we are only using 3 dimensions, $x$, $y$ and $z$: \[ v \cdot w = v_x w_x + v_y w_y + v_z w_z \]

You can easily check that this definition of a dot product satisfies the required properties.

Note that calculating one element of a matrix multiplication is the same thing as a dot product: the result element is the dot product of a row from the first matrix with a column from the second matrix.

Example

Suppose we have the following two vectors: \[ \begin{eqnarray*} v &=& (1,2,3) \\ w &=& (6,5,4) \\ \end{eqnarray*} \]

The dot product of these vectors is: \[ v \cdot w = 1\ast6+2\ast5+3\ast4 = 6+10+12 = 28 \]

Note that the dot product of two vectors is a scalar: a real number, not a vector.

Length

Given the dot product, we can define the length of a vector: \[ |v| = \sqrt{v\cdot v} \]

This is just our old friend the Pythagorean theorem in disguise! Why? Because the dot product of a vector with itself ends up squaring each element and adding them up.

Note that we can rescale (shrink or grow) any vector (except the zero vector) to have unit length. This is called normalizing. \[ w = \frac{v}{|v|} \] \[ |w| = 1 \]

The result is called a unit vector. We often will use unit vectors in computer graphics. For example, normal vectors (vectors perpendicular to a surface) are usually specified as unit vectors. See the last section, below.

Example

What is the length of the vector $v=(2,3,6)$? \[ |v| = \sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7 \]

If we normalize $v$, what do we get? Let $w=v/|v|$, we get \[ w = (2,3,6)/7 = (2/7,3/7,6/7) \approx (0.28,0.43,0.86) \]

You can easily check that $|w|=1$. The vector $w$ is in exactly the same direction as $v$, but it's one unit long.

Angles Between Vectors

The angle between two vectors $v$ and $w$ can be found using the following formula: \[ \cos(\theta) = \frac{v\cdot w}{|v||w|} \]

which we can also think of as: \[ \cos(\theta) = \frac{v}{|v|} \cdot \frac{w}{|w|} \]

In other words, it's just the dot product of the two unit vectors. That is, if both of two vectors $v$ and $w$ are normalized (a length of one unit), the formula for the cosine of the angle between them becomes the following, which is remarkably simple: \[ \cos(\theta) = v\cdot w \]

This is good news, because a dot product is very quick to compute: in 3D, it's just 3 multiplies and 2 adds. It turns out that the cosine of an angle is often desired in computer graphics, so being able to compute it so simply is enormously valuable. In fact, there are occasions when the tail wags the dog: the cosine of the angle is used because it's so quick to compute. It's used a lot in lighting calculations.

Example 1

Given the following vectors, what is the cosine of the angle between them? \begin{eqnarray*} v &=& (2,3,6)\\ w &=& (9,6,2)\\ \end{eqnarray*}

Since we already know the length of $v$, let's first find the length of $w$: \[ |w| = \sqrt{w\cdot w} = \sqrt{9^2+6^2+2^2} = \sqrt{81+36+4} = \sqrt{121} = 11 \]

Therefore, the cosine of the angle between $v$ and $w$ is: \[ \cos(\theta) = \frac{v\cdot w}{7\ast 11} = \frac{2\ast9+3\ast6+6\ast2}{77} = \frac{18+18+12}{77} = \frac{48}{77} \approx 0.62 \]

If, for some reason, we needed the actual angle, we could compute the arc cosine, but we'll almost never need that. (This angle is about 51 degrees.)

Example 2

Suppose we have the vectors $x=(1,0,0)$ and $y=(0,1,0)$. That is, these are unit vectors that point in the direction of the X and Y axes. It's easy to see that the dot product between these is zero.

What does this mean? If the cosine of an angle is zero, that means the angle is a right angle (ninety degrees). That is, these two vectors are perpendicular. (We're not surprised that they're perpendicular, but it's nice to see that the calculation confirms it.) In linear algebra, the term orthogonal is used for vectors that are perpendicular. In geek-speak, we say that two things are "orthogonal" if one doesn't affect the other. This makes sense because we can translate an object by multiples of the vector $x$ without affecting its Y coordinate, and translate it by multiples of the vector $y$ without affecting its X coordinate.

Example 3

What can we say about the angle, $\theta$, between the vectors: \[ \begin{eqnarray*} v &=& (2,3,6)\\ u &=& (-6,10,-3)\\ \end{eqnarray*} \]

The cosine of the angle is found by: \[ \cos(\theta) = \frac{(2,3,6)\cdot(-6,10,-3)}{|v||u|} = \frac{-12+30-18}{|v||u|} = \frac{0}{|v||u|} = 0 \]

So, these vectors are perpendicular (orthogonal).

Applying the Law of Cosines

This section proves that the dot product of normalized vectors is the angle between them. You can skip it if you're not interested.

According to the law of cosines, from trigonometry: \[ |v-w|^2 = |v|^2 + |w|^2 - 2 |v| |w| \cos(\theta) \]

where $\theta$ is the angle between $v$ and $w$. With some algebra, we get: \[ \cos(\theta) = \frac{|v|^2 + |w|^2 - |v-w|^2}{2 |v| |w|} \]

All the terms in $v-w$ are of the form \[(v_0-w_0)^2\] multiplying out to \[v_0^2-2 v_0 w_0+w_0^2\] That means that all the squared terms in the numerator disappear, the $-2$ cancels with the denominator, and we are left only with the cross terms: \[ \cos(\theta) = \frac{\sum_{i=0}^n v_i w_i}{|v| |w|} \]

but that numerator is just the dot product of $v$ and $w$, so \[ \cos(\theta) = \frac{v\cdot w}{|v| |w|} \]

Cross Product

As we know, the dot product of two vectors produces a scalar. The cross product produces a vector: one that is perpendicular to both of them. In 3D, the cross product is: \[ \begin{eqnarray*} \vec{n} &=& \vec{v}\times\vec{w} \\ &=& \vecIII{v_x}{v_y}{v_z}\times\vecIII{w_x}{w_y}{w_z} \\ &=& \vecIII{v_yw_z-v_zw_y}{v_zw_x-v_xw_z}{v_xw_y-v_yw_x} \end{eqnarray*} \]

Note that order matters: $v \times w = - w \times v$. Notice also that in computing $n_x$, you use only terms from the Y and Z components, and similarly for $n_y$ and $n_z$. There's a nice graphical way to remember the way to compute this. Looking at the following figures, multiply the terms connected by lines, and subtract one line from another. For the X and Z, you subtract the major diagonal (sloping down to the right) from the minor diagonal, and for Y, you subtract them in the opposite order (or just negate the result).

Direction of the Cross Product

In what direction does the vector point? For that, we need the right-hand rule: point the fingers of your right hand in the direction of the first vector, sweep them towards the second, and your thumb points in the direction of the cross product. Alternatively, point your thumb in the direction of the first vector, your index in the direction of the second, and your middle finger in the direction of the cross-product.

Interestingly, $z=x\times y$, $x = y \times z$ and $y = z \times x$. That is, a normalized vector parallel to the $z$ axis, (0,0,1) is the cross product of vectors for the $x$ axis, (1,0,0), and the $y$ axis, (0,1,0).

Length of the Cross Product

The length of the cross-product divided by the lengths of the two vectors gives the sine of the angle between the two vectors: \[ \sin(\theta) = \frac{|v\times w|}{|v| |w|} \]

Cross-Product Example

Find the cross product of the following two vectors: \[ \begin{eqnarray*} v &=& (2,3,6)\\ w &=& (9,6,2)\\ \end{eqnarray*} \]

We'll write the vectors vertically in order to find the result. \[ \begin{eqnarray*} \vec{n} &=& \vec{v}\times\vec{w} \\ &=& \vecIII{2}{3}{6}\times\vecIII{9}{6}{2} \\ &=& \vecIII{3\ast2-6\ast6}{9\ast6-2\ast2}{2\ast6-9\ast3} \\ &=& \vecIII{6-36}{54-4}{12-27} \\ &=& \vecIII{-30}{50}{-15} \\ \end{eqnarray*} \]

We already know from previous examples that $|v|=7$ and $|w|=11$. The sine of the angle between $v$ and $w$ is \[ \sin(\theta) = \frac{|n|}{|v||w|} = \frac{\sqrt{900+2500+225}}{77} = \frac{\sqrt{3625}}{77} \approx 0.78 \]

Earlier, in the example above, we found that the cosine of the angle between $v$ and $w$ is 48/77. Therefore, we should be able to confirm that: \[ 1 = \cos^2(\theta)+\sin^2(\theta) = (48/77)^2+(\sqrt{3625}/77)^2 = (2304 + 3625)/77^2 = 5929/5929 \]

Whew!

Normal Vectors

A vector that is perpendicular to a surface, such as a plane or a sphere, is said to be normal to it. It's therefore called a normal vector. (Other vectors are not abnormal.)

Normal vectors for surfaces are crucial in lighting calculations, because the normal vector tells us the orientation of the surface, and the orientation is important to how light falls on the surface and where it reflects to.

A common way to find a normal vector for a plane is to find two vectors that lie in the plane and take their cross product. To find a normal vector for a surface, find a plane that is tangent at the desired point, and find the normal vector for that plane. We'll see examples of this later in the course.