Chapter 3: The Dot Product

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This chapter is about a powerful tool called the dot product. It is one of the essential building blocks in computer graphics, and in Interactive Illustration 3.1, there is a computer graphics program called a ray tracer. The idea of a ray tracer is to generate an image of a set of geometrical objects (in the case below, there are only spheres). These are lit by a number of light sources, located at three-dimensional positions. The user must also set up a virtual camera, i.e., a camera position and field of view, and the camera's direction (i.e., where does it look). The ray tracer then traces rays from the camera position in the camera direction, through a set of pixels in the image plane of the camera. The program then finds the closest geometric object, and determines whether any light reaches that point directly from the light sources (otherwise, it will be in shadow). Reflection rays may also be traced in order to create reflective objects (such as the middle sphere in Interactive Illustration 3.1). Try out the ray tracing program by pressing Start below the illustration. In the ray tracing program above, the dot product was used to calculate the intersection between a ray and a sphere, and the dot product was also used to measure the length to the intersection points. In addition, the law of reflection, which is used to calculate reflective objects, was implemented using dot products. Both these topics will be explained again in Section 3.7, when the reader knows how the dot products works.

In general, the dot product is really about metrics, i.e., how to measure angles and lengths of vectors. Two short sections on angles and length follow, and then comes the major section in this chapter, which defines and motivates the dot product, and also includes, for example, rules and properties of the dot product in Section 3.2.3. Section 3.3 introduces the concept of an orthonormal basis and in Section 3.4, a set of inequalities are presented that often are useful. Section 3.5 shows some examples on how to use the dot product. Then follows a section about lines and planes, and finally, there is a follow-up section on ray tracing.

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The smallest angle between one vector, $\vc{u}$, and another vector, $\vc{v}$, is denoted by $[\vc{u},\vc{v}]$. To the right, the smallest angles between pairs of vectors are illustrated with green arcs, and in one case, the angle is illustrated with a green square. In this latter case, the angle makes $90^\circ$ or $\pi/2$ radians, i.e., $[\vc{u},\vc{v}]=\pi/2$. This is also denoted $\vc{u} \perp \vc{v}$, which illustrates that the vectors are orthogonal, which is the same as perpendicular. Note that the angle is $0$ in the lower left corner, and $\pi$ radians in the middle in the bottom row. In both these cases, one can say that the vectors are collinear since they lie on a shared line, and they are in fact also parallel. In the lower left corner, the vectors are parallel and have the same direction, while in the middle in the bottom row, the vectors are parallel but have opposite directions. When two vectors are parallel, this is denoted $\vc{u}\, || \,\vc{v}$. Also note that if $\vc{u}$ and $\vc{v}$ are parallel, then it must hold that $\vc{v} = k \vc{u}$ for some value of $k$.

We are now ready for the definition of the dot product itself:

Recall from Chapter 2 that $||\vc{v}||$ denotes the length of the vector $\vc{v}$. Since the length of a vector is always positive, the dot product will be positive if and only if $\cos[\vc{u},\vc{v}]$ is positive. From this we can deduce the following rules about the dot product

\begin{align} \vc{u} \cdot \vc{v}>0 \ \ \ &\Longleftrightarrow\ \ \ \ 0 < [\vc{u},\vc{v}] < \pi/2,\\ \vc{u} \cdot \vc{v}<0 \ \ \ &\Longleftrightarrow\ \ \ \ \pi/2 < [\vc{u},\vc{v}] \leq \pi, \\ \vc{u} \cdot \vc{v}=0 \ \ \ &\Longleftrightarrow\ \ \ \vc{u} \perp \vc{v}, \mathrm{\ i.e.,\ } [\vc{u},\vc{v}]=\pi/2 \mathrm{\ or\ } \vc{u} = 0 \mathrm{\ or\ } \vc{v} = 0. \\ \end{align}

(3.2)

Especially the last one is important: if the two vectors $\vc{u}$ and $\vc{v}$ are orthogonal (perpendicular) to each other, then $\vc{u}\cdot \vc{v} = 0$. As we will see later in this chapter, orthogonality is a useful feature which often simplifies calculations.

Note that the dot product produces a scalar value, and therefore, it is sometimes called the scalar product.

3.2.1 Unit Vectors and Normalization

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A unit vector is a vector whose length is 1, that is, $\vc{v}$ is a unit vector if $\ln{\vc{v}}=1$. From every non-zero vector, $\vc{v}$, a unit vector can be created. This is called to normalize the vector, and the process is called normalization. A normalized vector $\vc{n}$ is created from $\vc{v}$ by dividing $\vc{v}$ by its length, $\ln{\vc{v}}$, that is

\begin{equation} \vc{n} = \frac{1}{\ln{\vc{v}}}\vc{v}. \end{equation}

(3.4)

Next, we need to show that $\vc{n}$ is indeed a unit vector, i.e., $\ln{\vc{n}}=1$. Let us denote $l=1/\ln{\vc{v}}$, and simplify the expression for the length of $\vc{n}$ as $\ln{\vc{n}} = \ln{l\vc{v}} = \abs{l}\,\ln{\vc{v}}=\frac{1}{\ln{\vc{v}}}\ln{\vc{v}}=1$.

Note that if both vectors are normalized, i.e., $\ln{\vc{u}}=\ln{\vc{v}}=1$, the dot product simplifies to $\vc{u} \cdot \vc{v} = \cos[\vc{u},\vc{v}]$. This fact is very useful in shading computations for computer graphics, where the cosine for the angle between two vectors are often needed. In fact, normalized vectors were used extensively in Interactive Illustration 3.1 for the tracing of rays and for the shading there.

3.2.2 Projection

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From trigonometry, it is known that in a right triangle, the cosine of one of the smaller angles is related to the hypotenuse and the length of one of the shorter sides. More exactly, this is expressed as

\begin{equation} \cos \theta= \frac{a}{c}, \end{equation}

(3.5)

where $c$ is the length of the hypotenuse, $a$ is the length of the shorter side that makes an angle, $\theta$, with the hypotenuse. This is also illustrated to the right. So in these very specific situations, it is possible to calculate an angle given that the two lengths, $a$ and $c$, are known. Alternatively, one side length may be calculated given that the angle and the other side length are known.

Assume that one vector, $\vc{u}$, shall be projected orthogonally onto another vector, $\vc{v}$, in order to create a new vector, $\vc{w}$. This is illustrated to the right. Since $\vc{u}$ and $\vc{w}$ make up a triangle with a right angle, the following must hold: $\cos [\vc{u},\vc{v}] = \ln{\vc{w}} / \ln{\vc{u}}$, which is simply an application of Equation (3.5) above. This in turn means that the length of $\vc{w}$ is $\ln{\vc{w}} = \ln{\vc{u}}\cos [\vc{u},\vc{v}]$. If $\vc{v}$ has the length one, i.e., $\ln{\vc{v}}=1$, then the projected vector can be computed as

\begin{equation} \vc{w} = \ln{\vc{w}}\ \vc{v} = \ln{\vc{u}} \cos [\vc{u},\vc{v}]\ \vc{v}. \end{equation}

(3.6)

To convince yourself that $\ln{\vc{w}}\vc{v}$ really is equal to $\vc{w}$, note that it both has the right direction (the direction of $\vc{v}$, since $\ln{\vc{w}}$ is just a scalar) and it has the right length (namely $\ln{\vc{w}}$, since the length of $\vc{v}$ equals one). However, it would be nice to handle cases where $\vc{v}$ does not have length one as well. This can be done by normalizing the vector $\vc{v}$, i.e., multiplying it with $\frac{1}{\ln{\vc{v}}}$, which will give it length one. If $\vc{v}$ is replaced by $\frac{1}{\ln{\vc{v}}}\vc{v}$ in Equation (3.6) above, the following is obtained

\begin{equation} \vc{w} = \frac{\ln{\vc{u}} \cos [\vc{u},\vc{v}]}{\ln{\vc{v}}} \vc{v}. \end{equation}

(3.7)

Here, we have used the fact that the normalization does not change the direction, and we can hence continue using $\cos[\vc{u},\vc{v}]$ instead of $\cos[\vc{u}, \frac{\vc{v}}{\ln{v}}]$. Next, multiply both the numerator and the denominator by $\ln{\vc{v}}$, which leads to the general orthogonal projection formula

\begin{equation} \vc{w} = \frac{\ln{\vc{u}}\ \ln{\vc{v}} \cos [\vc{u},\vc{v}] }{ \ln{\vc{v}}^2 } \vc{v}. \end{equation}

(3.8)

The reason for this last step is that the numerator, $\ln{\vc{u}}\ \ln{\vc{v}} \cos[\vc{u},\vc{v}]$, now becomes equal to the definition of the dot product. Thus we can write the formula even shorter as

\begin{equation} \vc{w} = \frac{\vc{u} \cdot \vc{v} }{ \ln{\vc{v}}^2 } \vc{v}. \end{equation}

(3.9)

We have thus proved the following formula:

3.2.3 Rules and Properties

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With the help of the definition and the projection formula, it is now possible to deduce the following rules:

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The rules in Theorem 3.1 are intuitive, since they are the same as for scalar addition and scalar multiplication. In Section 3.5, several examples will be presented on how to use these rules.

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This section will describe a simple way of calculating the dot product. Assume we have two three-dimensional vectors $\vc{u} = (u_1, u_2, u_3)$ and $\vc{v} = (v_1, v_2, v_3)$ expressed in the same basis

\begin{align} \vc{u} & = u_1 \vc{e}_1 + u_2 \vc{e}_2 + u_3 \vc{e}_3,\\ \vc{v} & = v_1 \vc{e}_1 + v_2 \vc{e}_2 + v_3 \vc{e}_3. \end{align}

(3.20)

The dot product $\vc{u} \cdot \vc{v}$ can now be written as

\begin{align} \vc{u} \cdot \vc{v} & = (u_1 \vc{e}_1 + u_2 \vc{e}_2 + u_3 \vc{e}_3) \cdot (v_1 \vc{e}_1 + v_2 \vc{e}_2 + v_3 \vc{e}_3),\\ \end{align}

(3.21)

which can be expanded to

\begin{align} \vc{u} \cdot \vc{v} & = u_1 v_1 \vc{e}_1 \cdot \vc{e}_1 + u_1 v_2 \vc{e}_1 \cdot \vc{e}_2 + u_1 v_3 \vc{e}_1 \cdot \vc{e}_3 \\ & + u_2 v_1 \vc{e}_2 \cdot \vc{e}_1 + u_2 v_2 \vc{e}_2 \cdot \vc{e}_2 + u_2 v_3 \vc{e}_2 \cdot \vc{e}_3 \\ & + u_3 v_1 \vc{e}_3 \cdot \vc{e}_1 + u_3 v_2 \vc{e}_3 \cdot \vc{e}_2 + u_3 v_3 \vc{e}_3 \cdot \vc{e}_3. \end{align}

(3.22)

This looks long and complicated. However, assume that $\vc{e}_i \cdot \vc{e}_j$ is $0$ when $i \neq j$. This is equivalent to saying that each basis axis is orthogonal to every other basis axes. Then all but three terms would go away. Assume further that $\vc{e}_i \cdot \vc{e}_i = 1$ for all $i$. This is equivalent to saying that the length of each basis vector should be one. The remaining terms could then be simplified too, yielding

\begin{equation} \vc{u} \cdot \vc{v} = u_1 v_1 + u_2 v_2 + u_3 v_3. \end{equation}

(3.23)

Now we are ready to define orthonormal basis in any dimension.

We also generalize the simplified dot product to vectors of any dimensionality:

Note that the two different ways of indexing the components of a vectors have been used above, e.g., recall that $\vc{v}=(v_1, v_2, v_3) = (v_x, v_y, v_z)$.

For vectors in $\R^1$, $\R^2$, and $\R^3$, there is a natural notion of what an angle is. We use this notion to define the scalar product according to (Definition 3.1). If the basis is orthonormal, we obtain the above simple formula $\vc{u} \cdot \vc{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 $ for calculating the scalar product. For vectors in higher dimensions, there is no notion of what an angle is. In the case of an orthonormal basis, the solution here is to use the simple formula for the scalar product from Definition 3.4 and infer the notion of angle from the scalar product. The angle between two non-zero vectors $\vc{u}$ and $\vc{v}$ then becomes as follows.

Now, let us illustrate how the simple dot product evaluation (Definition 3.4) works with a simple example:

To get a more intuitive feeling for what the dot product provides, the reader is encouraged to play with Interactive Illustration 3.8 below.

3.3.1 Vector Length in Orthonormal Basis

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As we have already seen, having the vectors in an orthonormal basis simplifies the calculation of their dot product. In this section, it will become clear that it also simplifies the calculation of the length, which is denoted $\ln{\vc{v}}$, of a vector. This is also called magnitude and norm.

Recall rule $(iv)$ in Theorem 3.1, which says $\vc{v}\cdot\vc{v} = \ln{\vc{v}}^2$. If the vector has coordinates $(v_x, v_y)$ in an orthonormal basis, we can use the simple formula (Definition 3.4) for the dot product and we have $\ln{\vc{v}}^2 = v_x^2 + v_y^2$. In the top figure to the right, we have drawn the vector $\vc{v}$. Since the values $v_x$ and $v_y$ are the coordinates of $\vc{v}$, they are also equal to the lengths of the dashed lines. Note how this figure is analogous with the triangle below, where $c = \ln{\vc{v}}$, $a = v_x$ and $b = v_y$. Hence the expression $\ln{\vc{v}}^2 = v_x^2 + v_y^2$ is a proof of Pythagorean theorem, which states that $c^2 = a^2 + b^2$.

Therefore, the length of a vector in an orthonormal basis can be calculated as

\begin{equation} \ln{\vc{v}} = \sqrt{v_x^2 + v_y^2}. \end{equation}

(3.32)

Likewise, for a three-dimensional vector, $\vc{v} = (v_x, v_y, v_z)$, in an orthonormal basis, the dot product $\vc{v} \cdot \vc{v} = \ln{\vc{v}}^2$ can be calculated as $v_x^2 + v_y^2 + v_z^2$, and hence the vector length is calculated as

\begin{equation} \ln{\vc{v}} = \sqrt{v_x^2 + v_y^2 + v_z^2}. \end{equation}

(3.33)

Also in this case, it is possible to use the Pythagorean theorem to get an geometrical explanation why this formula is correct, as is shown in Interactive Illustration 3.10. Note that it is often convenient to know what happens to the length of a vector when it is scaled by a factor, $k$. Using the definition of the dot product, we get

\begin{equation} (k\vc{v})\cdot(k\vc{v}) = \ln{k\vc{v}}\,\ln{k\vc{v}}\cos[k\vc{v},k\vc{v}] = \ln{k\vc{v}}^2. \end{equation}

(3.34)

One many also use rule $(ii)$ twice in order to obtain

\begin{equation} (k\vc{v})\cdot(k\vc{v}) = k ( \vc{v}\cdot(k\vc{v})) = k(k(\vc{v}\cdot\vc{v})) = k^2 \ln{\vc{v}}^2. \end{equation}

(3.35)

Hence, we have

\begin{equation} \ln{k\vc{v}}^2 = k^2 \ln{\vc{v}}^2, \end{equation}

(3.36)

which gives

\begin{equation} \ln{k\vc{v}} = \abs{k}\, \ln{\vc{v}}. \end{equation}

(3.37)

Note that only the zero vector has zero length, i.e., $\ln{\vc{0}} = 0$, and $\ln{\vc{v}}>0$ if $\vc{v}\neq\vc{0}$.

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The following are some very useful inequalities in mathematics. They are part of the dot product chapter since they are easy to prove using the definition of the dot product.

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Another related inequality is the triangle inequality shown below.

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Note that the triangle inequality is easy to understand geometrically, as shown in the figure to the right. It is clear from the figure that the sum of the lengths of $\vc{u}$ and $\vc{v}$ must be longer or equal to the length of $\vc{u}+\vc{v}$. In fact, equality can only happen when when $\vc{u}$ and $\vc{v}$ are parallel and in the same direction. The reader is encouraged to move the $\vc{u}$ vector so that happens.

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In this section, some useful examples will be shown, where we point out which rules are used to reach the result. The rule is put on top of the equal sign, for example,

\begin{equation} \vc{a} \cdot (\vc{a}+\vc{b}) \overset{(iii)}{=} \vc{a} \cdot \vc{a} + \vc{a} \cdot \vc{b}, \end{equation}

(3.48)

implies that rule $(iii)$ (Theorem 3.1) was used to arrive at the right hand side of the equal sign. Next follows an example where the law of parallelograms is derived.

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Lines and planes are very common and important geometrical entities useful in many situations, such as when determining whether a ray (i.e., a line) from a virtual eye looking through a pixel center hits a geometrical object, such as a sphere. More broadly, lines and planes are often used in computational geometry, computer vision, computer graphics, computer-aided design (CAD), etc.

3.6.1 Lines

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Throughout this book, straight lines will mostly be used, and therefore, the shorter term line is often used instead. A straight line can be described with a starting point, $S$, and a direction, $\vc{d}$, as illustrated to the right. To describe a line, it may be convenient to have a representation that can find all possible points on the line. First, start with the point, $P$, to the right, and create a vector from $S$ to $P$, i.e., $\overrightarrow{SP}$. If $P$ is located on the line, described by $S$ and $\vc{d}$, then $\overrightarrow{SP}$ must be parallel to $\vc{d}$. In fact, a scalar, $t_1$, must exist which makes $t_1\vc{d}$ exactly as long as $\overrightarrow{SP}$, which is expressed as

\begin{equation} \overrightarrow{SP} = t_1 \vc{d}. \end{equation}

(3.55)

Since $P$ is located in the direction of $\vc{d}$, we also know that $t_1>0$. For $Q$, which also lies on the line, there is another scalar, $t_2$, which satisfies: $\overrightarrow{SQ} = t_2 \vc{d}$. Since $Q$ is located in the opposite direction of $\vc{d}$, it is clear that $t_2<0$. For $R$, on the other hand, there is no scalar, $t_3$, that satisfies a similar relation. That is, $\overrightarrow{SR} \neq t_3 \vc{d}$ for all values of $t_3$. The only difference between $P$ and $Q$ is that they use different scalars, $t_1$ and $t_2$. Hence, it makes sense that any point on the line can be described by using a particular scalar, $t$. Therefore, we use $P(t)$ to denote a function of a scalar, $t$, which returns a point, $P$. That is, for different values of $t$, different points, $P$, will be generated. This expression can be rewritten as

\begin{gather} \overrightarrow{SP(t)} = t\vc{d} \\ \Longleftrightarrow \\ P(t) - S = t\vc{d} \\ \Longleftrightarrow \\ P(t) = S + t\vc{d}. \end{gather}

(3.56)

Note that all points, $P(t)$, on the line are described by starting at $S$ and then adding a scaled direction vector, $t\vc{d}$, to reach $P(t)$. Some examples: $P(0) = S$, $P(1) = S + \vc{d}$, and $P(-2.5) = S -2.5\vc{d}$. This type of parameterized line is summarized in the following definition.

One often says that the line above is on explicit form, which simply means that the points, $P(t)$, on the line can be generated directly from the expression. Lines on explicit form in one, two, and three dimensions can be found in Interactive Illustration 3.15 below. Note that taking the length of both sides of a line, $P(t)-S = t\vc{d}$, gives $\ln{P(t)-S} = \ln{t\vc{d}}$. When the line direction is normalized, i.e., $\ln{\vc{d}}=1$, then $\abs{t} = \ln{P(t)-S}$. This can be very useful when computing intersections between, for example, a line and a sphere, as will be seen in Section 3.7.

Recall that a two-dimensional point, $S$, has two scalar components $(s_x, s_y)$, and similar, a two-dimensional vector, $\vc{d}$ has two scalar components, $(d_x,d_y)$. Now, note that a two-dimensional line, $P(t)=S + t\vc{d}$, can be expressed in terms of its scalar components of the vectors and points as

\begin{equation} P(t) = S + t\vc{d} \ \ \ \Longleftrightarrow \begin{cases} p_x(t) = s_x + td_x,\\ p_y(t) = s_y + td_y. \end{cases} \end{equation}

(3.58)

The following gives a taste of Chapter 5 on Gaussian elimination. Take the top row above and multiply by $d_y$ and multiply the bottom row by $d_x$. This leads to the following, where the parameter $t$ has been dropped from $p_x$ and $p_y$ for clarity purposes,

\begin{equation} \begin{cases} p_x d_y = s_x d_y + d_x d_y t,\\ p_y d_x = s_y d_x + d_x d_y t. \end{cases} \end{equation}

(3.59)

From algebra, it is known that a term can be subtracted from an expression as long as the same term is subtracted on both sides. Hence, it is legal to subtract the bottom row from the top row, which leads to

\begin{equation} p_x d_y - p_y d_x = s_x d_y - s_y d_x + d_x d_y t - d_x d_y t = s_x d_y - s_y d_x \\ \Longleftrightarrow \\ d_yp_x - d_x p_y + s_y d_x - s_x d_y =0, \end{equation}

(3.60)

and as can be seen, the $t$-terms have disappeared from the expression. This can further be rewritten as

\begin{equation} d_y p_x - d_x p_y + s_y d_x - s_x d_y =0 \\ \Longleftrightarrow \\ a p_x + b p_y + c = 0, \end{equation}

(3.61)

where $a=d_y$, $b=-d_x$, and $c=s_y d_x - s_x d_y$. This equation may be familiar to some, and especially, if we use $x=p_x$ and $y=p_y$, which gives

\begin{equation} a x + b y + c = 0. \end{equation}

(3.62)

For $b\neq 0$, one can further rewrite the expression above as $y =(-ax -c)/b=$ $kx+m$, where $k=-a/b$ and $m=-c/b$. This expression of a line is definitely familiar to most people, where $k$ describes how much $y$ changes when $x$ is increased by 1, and $m$ is the $y$-value at $x=0$. Note, however, $a x + b y + c = 0$, is a more general description of a two-dimensional line, since it makes it possible to describe vertical lines as well, which is not possible with $y=kx+m$.

It is important to note that the implicit form of the line, $a p_x + b p_y + c = 0$, and the explicit form of the line, $P(t) = S + t\vc{d}$, describe the same line exactly, since the former expression was derived from the latter. Next, we will show that if the basis is orthonormal, the explicit form can be rewritten using dot products. For that purpose, we introduce $\vc{n} = (n_x,n_y) = (a,b) = (d_y,-d_x)$, which leads to

\begin{equation} d_y p_x - d_x p_y + s_y d_x - s_x d_y =0 \\ \Longleftrightarrow \\ \vc{n} \cdot (P - S) = 0, \end{equation}

(3.63)

where $P=(p_x, p_y)$ and $S=(s_x, s_y)$. Here we have used the fact that $(d_y, -d_x) \cdot (p_x, p_y) = d_y p_x + (-d_x) p_y$ when vectors ard described in an orthonormal base, as described in Theorem 3.4. As can be seen, if we take the vector from $S$ to any point, $P$, on the line, then its dot product with $\vc{n}$ must be zero if $P$ is to be on the line. Interestingly, we see that $\vc{n}\cdot\vc{d} =$ $(d_y,-d_x)\cdot (d_x,d_y) = $ $d_y d_x - d_x d_y = $ $0$, i.e., $\vc{n}$ is orthogonal to the line direction, $\vc{d}$. Therefore, $\vc{n}$ is often called the normal of the line. Note also, that $\vc{n} \cdot (P - S) = 0$ is said to be in implicit form, since it does not easily generate points on the line. However, it is straightforward to test whether a point, $P$, lies on the line.

This leads to the following definition of a two-dimensional line on implicit form.

It should be noted that Definition 3.7 still holds in the case of a non-orthogonal basis. That is, the line can still be written as $\vc{n} \cdot (P-S) = 0$ where $\vc{n}$ is a normal vector to the line. However, it is no longer straightforward to calculate the coordinates of $\vc{n}$, since, in general, it is no longer equal to $\vc{n} = (d_y, -d_x)$.

As seen above, there are two different types of mathematical representations for two-dimensional lines. The same cannot be done for a one-dimensional or three-dimensional line. However, as will be seen in Section 3.6.2, a three-dimensional plane equation has two similar representations, both an implicit and an explicit.

Returning to two-dimensional lines on the form: $\vc{n} \cdot (P - S) = 0$, which says that all points $P$ that lie on the line, represented by $S$ and $\vc{d}$, fulfil the expression above, i.e., the dot product is equal to zero. What happens if $P$ is not on the line? Well, the dot product will not be zero, of course, but can something else be read into that result? It turns out it can be quite useful. To demonstrate this, a scalar function, $e$, of $P$ is created as

\begin{equation} e(P) = \vc{n} \cdot (P - S). \end{equation}

(3.65)

This function is sometimes called edge equation (in computer graphics) or signed distance function (this is explained later in this section). Since, $e(P)$ is defined using a dot product, it is known from Section 3.4, that $e(P)$ is positive when the angle $[\vc{n}, P-S] < \pi/2$, that $e(P)$ is negative when $[\vc{n}, P-S] > \pi/2$, and that $e(P)=0$ only when $[\vc{n}, P-S] = \pi/2$. However, $e(P)=0$ also means that $P$ lies on the line defined by $S$ and $\vc{d}$. Therefore, when $e(P)>0$, we say that $P$ is in the positive half-space of the line, and when $e(P)<0$, $P$ is in the negative half-space, i.e., the line divides the entire two-dimensional plane into two half-spaces. This is shown in Interactive Illustration 3.16. With help from the Definition 3.1 (dot product), Equation (3.65) can be rewritten as

\begin{equation} e(P) = \vc{n} \cdot (P - S) = \ln{\vc{n}}\, \ln{P - S} \cos [\vc{n}, P-S]. \end{equation}

(3.66)

Note that $\ln{P - S} \cos [\vc{n}, P-S]$ is actually the orthogonal distance from $P$ to the line, with the caveat that the "distance" is signed. This means that if $P$ is in the positive half-space, then signed distance is positive, and if $P$ is in the negative half-space, the signed distance is negative. Also note that if $\vc{n}$ is normalized, then $e(P)$ is exactly the signed orthogonal projection distance function, i.e., $e(P) = \ln{P - S} \cos [\vc{n}, P-S]$.

3.6.2 Planes

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Planes in three dimensions (and higher) are similar to lines in two dimensions in that they both split their domain into two half-spaces. In Section 3.6.1, we saw that a two-dimensional line splits the $xy$-plane into a positive half-space and a negative half-space. There are also two types of representations (one implicit and one explicit) for planes, similar to lines in two dimensions.

To define a plane on explicit form, i.e., similar to $P(t) = S + t\vc{d}$ for lines, a starting point, $S$, and two direction vectors, $\vc{d}_1$ and $\vc{d}_2$, are needed. The direction vectors may not be collinear, i.e., $\vc{d}_1 \neq k \vc{d}_2$ for all values of $k$. Another way to put it is that the direction vectors may not be parallel, not even parallel but with opposite directions. The direction vectors both lie in the plane. Hence, if a point, $P$, is to lie in the plane, it must hold that

\begin{equation} \overrightarrow{SP} = t_1 \vc{d}_1 + t_2\vc{d}_2, \end{equation}

(3.70)

for some values of the scalars $t_1$ and $t_2$. Similar to lines, this expression is rewritten as below, where the two scalars, $t_1$ and $t_2$, have been set as parameters to $P$, i.e.,

\begin{gather} \overrightarrow{SP(t_1,t_2)} = t_1 \vc{d}_1 + t_2\vc{d}_2 \\ \Longleftrightarrow \\ P(t_1, t_2) - S = t_1 \vc{d}_1 + t_2\vc{d}_2 \\ \Longleftrightarrow \\ P(t_1, t_2) = S + t_1 \vc{d}_1 + t_2\vc{d}_2. \\ \end{gather}

(3.71)

This form is illustrated to the right as well. As can be seen, this is very similar to vector addition. The direction vectors, $\vc{d}_1$ and $\vc{d}_2$, are scaled by $t_1$ and $t_2$, respectively. The resulting vectors are added together with the starting point, $S$. This leads to the following definition.

Interestingly, there is also an implicit form of the plane equation. First, let us write out the plane equation on component form, as shown below, where $d_{1,y}$ means the $y$-component of $\vc{d}_1$, and so on.

\begin{equation} P(t_1, t_2) = S + t_1 \vc{d}_1 + t_2\vc{d}_2 \ \ \ \Longleftrightarrow \ \ \ \begin{cases} p_x(t_1,t_2) = s_x + t_1 d_{1,x} + t_2 d_{2,x},\\ p_y(t_1,t_2) = s_y + t_1 d_{1,y} + t_2 d_{2,y},\\ p_z(t_1,t_2) = s_z + t_1 d_{1,z} + t_2 d_{2,z}. \end{cases} \end{equation}

(3.73)

Recall that for two-dimensional lines, the parameterized line: $P(t)=S+t\vc{d}$ could be transformed into a line on implicit form: $\vc{n}\cdot (P-S)$. This was done by eliminating $t$, and the same can be done for Equation (3.73) above, where both $t_1$ and $t_2$ can be eliminated. This can be done using Gaussian elimination, and there is an example in Chapter 5 on this. Here, we simply state that Equation (3.73) above can be transformed into

\begin{equation} \vc{n} \cdot (P - S) = 0, \end{equation}

(3.74)

and refer to Chapter 5 for more details on how the elimination of $t_1$ and $t_2$ was done. Note that $\vc{n}$ is the normal of the plane, i.e., it is perpendicular to any vector that lies in the plane. There is also a relationship between $\vc{n}$ and $\vc{d}_1$ & $\vc{d}_2$. This is the cross product, which is the topic of Chapter 4.

Exactly as for lines, we can also create a signed distance function as

\begin{equation} e(P) = \vc{n} \cdot (P - S), \end{equation}

(3.76)

where $e(P)=0$ when $P$ lies in the plane. Also, $e(P)>0$ when $P$ lies on the same side of the plane as the point $S+\vc{n}$, and that is called the positive half-space. Similarly, if $e(P)<0$ then $P$ lies on the same side of the plane as the point $S-\vc{n}$. That other part of the space is called the negative half-space.

Next follows an example, where both line equations and plane equations are used.

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In the introduction of this chapter (Section 3.1), a graphics program, called a ray tracer, is shown in Interactive Illustration 3.1. Such a program is rather straightforward to write, once some knowledge about linear algebra has been obtained. At the core of a ray tracer, there is a visibility function that determines which object a ray can "see". An example is shown to the right. In a ray tracer, one creates a virtual viewer, which has a position, and is looking in a certain direction. The ray tracer then computes an image from that position in that direction. The view position is the starting point of the blue ray to the right. A set of rays is then created. In the simplest case, one ray per pixel in the image plane is created. It is then up to the ray tracing program to examine the relevant objects in the scene, and compute whether a ray through a pixel hits an object, and also to find the closest object. For the ray in Figure 3.21, we can see that the ray hits two of the three circles. However, since the yellow circle is closer, the pixel is colored yellow. For the pixel above, however, the corresponding ray that goes through that pixel will hit the green circle, and the pixel is therefore colored green. To generate images with shadows, reflections, and refraction, more rays may be shot from the first intersection point between the yellow circle and the ray.

In this section, two examples, which relate to ray tracing, will be shown. The first shows how to compute the intersections between a three-dimensional ray and a three-dimensional sphere. The second example, shows how a vector can be reflected in a surface, whose normal is known. Both of these examples use the dot product as a major tool.