# Geometry Potpourri

In this workbook, we're going to walk through a number of common geometric tasks such as finding intersections, computing distances, performing rotations, etc. For each task, we'll showcase not only the Klein code that would produce the desired result, but also show mathematically how it can be computed by hand. Pen and paper is not only encouraged, it is likely a requirement to get the most out of this workbook.

Lack of rigor ahead!

In this article, I make very little effort to explain why an operation is defined a certain way or explain some of the deeper underlying mathematics. This article was written because it's often helpful to see how computation is done first, and perform some calculations by hande as well gain some intuition before taking a closer look at the theory.

## Multivectors

Info

Klein API: kln::entity

In this section, we look at all the various operations and elements that show up in $$\PGA$$. A common question that people run into when getting exposed to GA for the first time is, "why are their so many operations?" The best answer is that, well, geometry has a lot of operations! More than can be captured easily with just the dot product and cross product you're likely familiar with. This section just gets you acquainted with the mechanics of $$\PGA$$, but to glean more geometric intuition, all the sections afteward will focus on concrete examples (plane distance to point, meeting to points to make a line, projecting a line onto a plane, rotating a point about a line, reflecting a point through a plane, etc). You'll find that for a bit of upfront complexity presented here, what's gained is a satisfying degree of uniformity in all the geometry that comes later.

### PGA Basis and Geometric Product

Info

Klein API: operator*

Written as two elements adjacent to each other ($$ab$$).

Let's get used to some basis elements and operations first. In $$\mathbf{P}(\mathbb{R}^*_{3, 0, 1})$$ (also known as PGA), we have four grade-1 vector basis elements, $$\ee_0$$, $$\ee_1$$, $$\ee_2$$, and $$\ee_3$$. Let's define the square of these vectors in the following manner. The first element squares to $$0$$ ($$\ee_0 \ee_0 = 0$$) while the latter 3 elements square to $$1$$ ($$\ee_1\ee_1 = \ee_2\ee_2 = \ee_3\ee_3 = 1$$). As we now immediately have a means to create $$1$$, we can generate all the numbers this way, so let's declare $$1$$ as our grade-0 basis element.

Very non-mathematically, the "grade" of an element is the number of subscripts it possesses. Soon, you should get a good feel of what grades result from what operations, and what information the grades present in a multivector quantity conveys.

When we write elements adjacent to one another as in $$\ee_1\ee_1$$, the operation being represented here is the geometric product. In code, the geometric product is expressed as the multiplication (*) operator. So far, we've written down what the geometric product does when the elements multiply against themselves. Next, we need to describe what happens when we take the geometric product between different elements.

Mixed products of the basis vectors produce the basis bivectors, which can themselves be used in linear combinations to produce general bivectors. Bivectors that can be written as the product of two vectors may be referred to as blades or simple bivectors. The products between basis vectors are all blades by definition. As shorthand, we combine subscripts as shown below:

\begin{aligned} \ee_1\ee_0 &= \ee_{10} \\ \ee_1\ee_2 &= \ee_{12} \\ \ee_3\ee_0 &= \ee_{30} \\ &\dots \end{aligned}

At the moment, we would have $$4\times 3 = 12$$ basis bivectors if defined this way, but we impose a relation on them that relates basis bivectors that contain the same subscripts. Namely, the product of two vectors is minus the product of the two vectors with the order interchanged.

\begin{aligned} \ee_{01} &= -\ee_{10} \\ \ee_{02} &= -\ee_{20} \\ \ee_{03} &= -\ee_{30} \\ \ee_{12} &= -\ee_{21} \\ \ee_{31} &= -\ee_{13} \\ \ee_{23} &= -\ee_{32} \tag{1} \end{aligned}

For reasons that will become clear in a future writeup, we'll select the blades on the left hand side of $$(1)$$ to be the canonical 2-blades in PGA. There is some choice index order selection as will be explained later, but the order here is used to match existing literature and so that computations later are made more convenient.

Now that we know we can swap the order of the multiplication by basis vectors by introducing a sign, we can compute things like $$\ee_{01}\ee_0$$.

\begin{aligned} \ee_{01}\ee_0 &= \ee_0\ee_1\ee_0 \\ &= -\ee_0\ee_0\ee_1 \\ &= 0 \end{aligned}

Another example (that doesn't vanish):

\begin{aligned} \ee_{20}\ee_2 &= \ee_2\ee_0\ee_2 \\ &= -\ee_2\ee_2\ee_0 \\ &= -\ee_0 \end{aligned}

The same algorithm can be used to produce the 4 trivectors:

\begin{aligned} \ee_{123} &= \ee_{12}\ee_3 = -\ee_{132} = \ee_{312} \\ \ee_{021} &= \ee_{02}\ee_1 = -\ee_{201} = \ee_{210} \\ \ee_{013} &= \ee_{01}\ee_3 = -\ee_{031} = \ee_{301} \\ \ee_{032} &= \ee_{03}\ee_2 = -\ee_{302} = \ee_{320} \end{aligned}

Not all permutations were listed, but just a few so you can gain familiarity with swapping indices and introducing signs or removing them as ncessary. The last element we haven't touched on is the pseudoscalar $$\ee_{0123}$$. The pseudoscalar is so named because while it does not have grade zero, it is uniquely identified within its grade to within a permutation of its indices (compared to, say, the grade-2 elements of which there are six distinct members). Note that there is no way in which we can introduce an element with grade higher than this, and so our $$16$$ algebra elements (1 scalar, 4 vector, 6 bivector, 4 trivector, and 1 pseudoscalar) are fully described. Linear combinations can be made from all of these elements to produce a general multivector. To take the geometric product between general multivectors, we leverage the fact that the geometric product obeys the distributive law.

\begin{aligned} (4\ee_1 - 2\ee_{32})(3\ee_{012} + \ee_{10}) &= 4\ee_1(3\ee_{012} + \ee_{10}) - 2\ee_{32}(3\ee_{012} + \ee_{10}) \\ &= 12\ee_1\ee_{012} + 4\ee_1\ee_{10} - 6\ee_{32}\ee_{012} - 2\ee_{32}\ee_{10} \\ &= -12\ee_{02} + 4\ee_0 - 6\ee_{301} - 2\ee_{3210} \\ &= 4\ee_0 - 12\ee_{02} - 6\ee_{013} - 2\ee_{0123} \end{aligned}

In the last step, we simply took the time to sort the resulting terms by grade and also performed swaps as necessary to write terms with the canonical subscript ordering.

Exercises

1. Write down two vectors whose product is $$-\ee_{01} + 2\ee_{13}$$.
2. The geometric product is neither commutative, nor anticommutative. Find examples that demonstrate this fact.
3. Not every bivector can be written down as the product of two vectors. Find an example that demonstrates this fact.

### The Symmetric Inner Product

Info

Klein API: operator|

Written as $$a \cdot b$$.

The symmetric inner product (inner product for short) is a bilinear binary operation like the geometric product except it is always grade decreasing such that the final grade is the absolute value of the difference of the operand grades. If the grade of the element resulting from a geometric product would have been greater than this difference, the inner product extinguishes it to zero. In other words, all indices must contract for the inner product to produce a non-zero value (note that if the degenerate element contracts, a zero is produced anyways). Before, we saw that the geometric product can decrease the grade in the event that indices match (which contracts the result to $$1$$ or $$0$$ depending). The easy way to perform a symmetric product is to perform the geometric product as normal but discard terms produced that produce the undesired grade. For example, $$\ee_1\cdot \ee_1 = 1$$ but $$\ee_1\cdot \ee_2 = 0$$. For a more nuanced example, $$\ee_{12}\cdot\ee_{02}$$ is also $$0$$ because the $$1$$ index did not contract. Note that we write the symmetric inner product as $$\cdot$$ in equation form, but in code, it is expressed via the pipe (|) operator. As with the geometric product, the symmetric inner product obeys the distributive law. Here's an example mirroring an example above but using the inner product instead of the geometric product:

\begin{aligned} (4\ee_1 - 2\ee_{32})\cdot (3\ee_{012} + \ee_{10}) &= 4\ee_1\cdot(3\ee_{012} + \ee_{10}) - 2\ee_{32}\cdot (3\ee_{012} + \ee_{10}) \\ &= -12 \ee_{02} + 4\ee_0 \\ &= 4\ee_0 - 12\ee_{02} \end{aligned}

In addition, because of our "rule" that the inner product can only decrease grade, the inner product between any quantity and a scalar quantity must be zero.

Exercises

1. Can the symmetric inner product ever produce the pseudoscalar? Why or why not?
2. Explain why $$\ee_0 \cdot \ee_{012}$$ is $$0$$ and not $$\ee_{12}$$.
3. The symmetric inner product not always commutative! Provide a counterexample for the inner product between other grades. When is it commutative?

### The Exterior Product

Info

Klein API: operator^

Written as $$a\wedge b$$.

The exterior product (also known as the wedge product or the outer product) is similar to the geometric product except it always extinguishes results upon contraction. Thus, the result of an exterior product either has grade equal to the sum of the grades of its operands, or it is exactly zero. The exterior product is represented in equations with a $$\wedge$$ symbol, and in code, is represented with a caret (^) operator. The exterior product obeys the laws of distributivity as with the other products, so let's return to our familiar example:

\begin{aligned} (4\ee_1 - 2\ee_{32})\wedge (3\ee_{012} + \ee_{10}) &= 4\ee_1\wedge(3\ee_{012} + \ee_{10}) - 2\ee_{32}\wedge (3\ee_{012} + \ee_{10}) \\ &= -2\ee_{3210} \\ &= -2\ee_{0123} \end{aligned}

Exercises

1. The geometric product between vectors (grade-1) is the sum of the inner product and the exterior product. Prove this by explicitly expanding out the geometric product between two general vectors.
2. The geometric product is not generally the sum of the inner product and the exterior product. Why is that?
3. The exterior product isn't generally anti-commutative! Find counterexamples demonstrating this fact.

### The Poincaré Dual Map

Info

Klein API: operator!

Written as $$\JJ(a)$$.

To perform the regressive product in the next section, we need a map that is grade-reversing. That is, we need a map that takes multivectors $$v$$ to multivectors $$v^*$$ such that the grade of $$v$$ is equal to $$4$$ minus the grade of $$v^*$$ (and vice versa). For example, vectors should be mapped to trivectors, bivectors to bivectors, and the scalar to the pseudoscalar. In addition, this map needs to be an involution (two application of the map is the identity). Here, a map is provided but note that several such maps are possible. The nice property of the map provided here is that the application of the map is just a coordinate tuple reversal without any sign changes. The mechanism for how this basis was chosen is elegant but will be the subject of a different article. By convention, the dual map is written as $$\JJ$$ in expressions.

\begin{aligned} \JJ(1) = \ee_{0123}&,\quad \JJ(\ee_{0123}) = 1 \\ \JJ(\ee_0) = \ee_{123}&,\quad \JJ(\ee_{123}) = \ee_0 \\ \JJ(\ee_1) = \ee_{032}&,\quad \JJ(\ee_{032}) = \ee_1 \\ \JJ(\ee_2) = \ee_{013}&,\quad \JJ(\ee_{013}) = \ee_2 \\ \JJ(\ee_3) = \ee_{021}&,\quad \JJ(\ee_{021}) = \ee_3 \\ \JJ(\ee_{01}) = \ee_{23}&,\quad \JJ(\ee_{23}) = \ee_{01} \\ \JJ(\ee_{02}) = \ee_{31}&,\quad \JJ(\ee_{31}) = \ee_{02} \\ \JJ(\ee_{03}) = \ee_{12}&,\quad \JJ(\ee_{12}) = \ee_{03} \end{aligned}

Exercises

1. Double check that all elements in the algebra are accounted for above.
2. The dual map is a unary operator that can be distributed across each term in a multivector sum. For example, $$\JJ(\ee_1 + 2\ee_{02}) = \ee_{032} + 2\ee_{31}$$. Verify that the map is still an involution (that is, two applications of $$\JJ$$ should map back to the original argument).

### Regressive Product

Info

Klein API: operator&

Written as $$a\vee b$$.

The regressive product (written as $$\vee$$ and expressed as the & operator in code) is defined in terms of the dual map and the exterior product like so:

$a \vee b = \JJ(\JJ(a)\wedge\JJ(b))$

Here's a worked example of the regressive product, again using a familiar operands:

\begin{aligned} (4\ee_1 - 2\ee_{32})\vee (3\ee_{012} + \ee_{10}) &= \JJ(\JJ(4\ee_1 - 2\ee_{32})\wedge \JJ(3\ee_{012} + \ee_{10})) \\ &= \JJ((4\ee_{032} + 2\ee_{01}) \wedge (-3\ee_3 - \ee_{23})) \\ &= \JJ(-6\ee_{013} - 2\ee_{0123}) \\ &= -2 - 6\ee_2 \end{aligned}

If you're working out the example above yourself and finding some disagreement in the signs, remember that to use the dual map given above, the indices must match exactly. Thus, in this example, the dual $$\JJ(3\ee_{012})$$ is $$-\JJ(3\ee_{021}) = -3\ee_3$$.

Exercises

1. If the exterior product is zero, will the regressive product be zero? Why or why not?
2. By the same token, if the regressive product is zero, will the exterior product be zero?

## Construction

### Planes

Info

kln::plane

A plane $$p$$ is the manifestation of a reflection. It is often helpful not to think of a plane as a "set of points" in PGA as will be evident when we look at rotations and translations later. Let's look at some simple examples first.

#### Planes through the origin

The plane represented implicitly by $$x = 0$$ appears in PGA simply as the vector $$\ee_1$$. Similarly, the planes $$y = 0$$ and $$3z = 0$$ correspond to $$\ee_2$$ and $$3\ee_3$$ respectively. It's easy to see that planes like $$x + 3y - z = 0$$ can be represented by taking linear combinations of $$\ee_1$$, $$\ee_2$$ and $$\ee_3$$. In this case, the plane we just created would be represented as $$\ee_1 + 3\ee_2 - \ee_3$$.

// Plane x = 0
kln::plane p1{1.f, 0.f, 0.f, 0.f};

// Plane y = 0
kln::plane p2{0.f, 1.f, 0.f, 0.f};

// Plane 3z = 0
kln::plane p3{0.f, 0.f, 3.f, 0.f};

// Plane x + 3y - z = 0
kln::plane p4{1.f, 3.f, -1.f, 0.f};

// Equivalent to p4
kln::plane p5 = p1 + 3.f * p2 - p3 / 3.f;


#### Planes away from the origin

To shift a plane from the origin, we can add a multiple of the degenerate vector $$\ee_0$$. This vector is different from the other vectors in a very important way as we will see soon.

The plane $$2x + 1 = 0$$ is represented as $$\ee_0 + 2\ee_1$$.

// Plane 2x + 1 = 0
kln::plane p6{2.f, 0.f, 0.f, 1.f};


Other planes like to the one above can be constructed similarly.

#### Angles between planes

The angle between planes is given by the inner product which produces the cosine of the angle between them as you would expect. The planes must be normalized first for this to work, and the norm of a plane $$p$$ can be calculated as $$\sqrt{p\cdot p}$$.

Suppose we have two planes $$p_1 = 3\ee_0 + \ee_1 + \ee_3$$ and $$p_2 = \ee_0 + \ee_3$$. The first plane we recognize as $$x + z - 3 = 0$$, a plane parallel to the y-axis that passes through the points $$(0, y, 3)$$ and $$(3, y, 0)$$ for any value of $$y$$. The second plane the plane $$z + 1 = 0$$ which is parallel to the $$xy$$-plane and intercepts the $$z$$-axis one unit below the origin. We expect the angle between these planes to be $$\frac{\pi}{4}$$ radians, so let's quickly verify this with the inner product. First though, we must normalize them by dividing by the norm. This gives:

\begin{aligned} p_1 &= \frac{3\sqrt{2}}{2}\ee_0 + \frac{\sqrt{2}}{2}\ee_1 + \frac{\sqrt{2}}{2}\ee_3 \\ p_2 &= \ee_0 + \ee_3 \end{aligned}

Remember that when computing the norm, because $$\ee_0^2 = 0$$, the $$\ee_0$$ component does not participate in the computaiton. The angle between them is then computed as:

\begin{aligned} p_1 \cdot p_2 &= \left(\frac{3\sqrt{2}}{2}\ee_0 + \frac{\sqrt{2}}{2}\ee_1 + \frac{\sqrt{2}}{2}\ee_3\right) \cdot (\ee_0 + \ee_3) \\ &= \frac{\sqrt{2}}{2}\ee_3\cdot\ee_3 \\ &= \frac{\sqrt{2}}{2} \end{aligned}

indicating that the angle between $$p_1$$ and $$p_2$$ is $$\cos^{-1}{\frac{\sqrt{2}}{2}} = \frac{\pi}{4}$$ radians. The Klein code to compute the angle between planes is shown below:

float plane_angle_rad(kln::plane p1, kln::plane p2)
{
p1.normalize(); // Normalizes p1 in place
p2.normalize(); // Normalizes p2 in place
return std::acos((p1 | p2).scalar());
}


Exercises

1. The angle calculation above is a great example to demonstrate why the degenerate element $$\ee_0$$ is so important. Consider what would have happened if instead $$\ee_0 \cdot \ee_0 \neq 0$$.
2. Now, suppose we made the "choice" of representing planes with trivector coordinates instead of vector coordinates. Repeat the above calculation using the duals of each plane. Does the computation still work?

### Lines

Info

kln::line

#### Intersecting planes to create lines

The "meet" operation in $$\PGA$$ is defined in terms of the exterior product ($$\wedge$$). Given two planes, $$p_1 = 3\ee_0 + \ee_1 + \ee_3$$ and $$p_2 = \ee_0 + \ee_3$$, we can compute the exterior product like so:

\begin{aligned} p_1\wedge p_2 &= (3\ee_0 + \ee_1 + \ee_3) \wedge (\ee_0 + \ee_3) \\ &= 3\ee_{03} + \ee_{10} + \ee_{13} + \ee_{30} \\ &= -\ee_{01} + 2\ee_{03} - \ee_{31} \end{aligned}

How can we verify that this is correct? Well, let's first try to calculate the equation of this line using more traditional means. In classical Euclidean geometry, the six degrees of freedom of a line are often represented using Plücker coordinates. Let's express our planes using the implicit equation form:

\begin{aligned} 0 &= 3 + x + z \\ 0 &= 1 + z \end{aligned}

The Plücker displacement is determined by the cross product of the plane normals, which in this case is the cross product $$(\mathbf{i} + \mathbf{k})\times\mathbf{k} = \mathbf{i}\times\mathbf{k} = -\mathbf{j}$$. Meanwhile, the Plücker moment of the line is given as $$3\mathbf{k} - (\mathbf{i} + \mathbf{k}) = -\mathbf{i} + 2\mathbf{k}$$. Putting it together, the Plücker coordinate tuple of our line is $$(d_1:d_2:d_3:m_1:m_2:m_3) = (0: -1: 0:-1:0:2)$$, in exact accordance with the calculation above.

Observe that the coordinates associated with $$\ee_{21}$$, $$\ee_{13}$$, and $$\ee_{32}$$ capture the information about the orientation of the line, while the coordinates associated with $$\ee_{01}$$, $$\ee_{02}$$, and $$\ee_{03}$$ work in tandem to impart a translational element as well.

What happens if the planes are parallel? Let's try our meet between planes $$-2\ee_0 + 2\ee_1$$ and $$\ee_1$$ to see what happens.

$(-2\ee_0 + 2\ee_1) \wedge \ee_1 = -2\ee_{01}$

We ended up with a bivector all the same! Quantities that have no direction components like the above are referred to as ideal lines, also known as lines at infinity. What's important about concepts like lines and points at infinity is that their existence allows our algebra to have closure without needing to introduce vague notions of $$\infty$$. Another way to word this is to say that planes always intersect, even when parallel, and the algebra does not need to make any special exceptions for them. Furthermore, these entities at infinity have uses! We'll see later that rotations about ideal lines can be used to generate translations.

The code that produces the line above is the following snippet:

// plane 1: 3 + x + z = 0;
kln::plane p1{1.f, 0.f, 1.f, 3.f};

// plane 2: 1 + z = 0;
kln::plane p2{0.f, 0.f, 1.f, 1.f};

// line intersection of planes 1 and 2
kln::line intersection = p1 ^ p2;


Exercises

1. Match up the Plücker coordinates above with the coefficients of the meet operation above. Do you see how the element indices matter in determining the sign?
2. Lines at infinity have 3 degrees of freedom. Is there an analogous plane at infinity? How many degrees of freedom does it have?
3. What happens when you evaluate $$p_2\wedge p_1$$ instead of $$p_1\wedge p_2$$? Is the change justified?

### Points

Info

kln::point

#### Meet a line and a plane

With the same meet operation ($$\wedge$$) we used to intersect two planes to construct a line, we can intersect a line and a plane to construct a point. Let's use the line $$\ell = -\ee_{01} + 2\ee_{03} - \ee_{31}$$ from the previous example and the plane $$\ee_1 + \ee_2$$.

\begin{aligned} (-\ee_{01} + 2\ee_{03} - \ee_{31}) \wedge (\ee_1 + \ee_2) &= -\ee_{012} + 2\ee_{031} + 2\ee_{032} - \ee_{312}\\ &= -\ee_{123} + 2\ee_{032} + 2\ee_{031} + \ee_{021} \\ &\rightarrow \;\; \ee_{123} - 2\ee_{032} + 2\ee_{013} - \ee_{021} \end{aligned}

In the last step, we divided the expression by $$-1$$ so that the weight of $$\ee_{123}$$ is exactly one. For points, this weight is the homogeneous coordinate, and when it is unity, there is a direct association between the point's Cartesian coordinates and the other trivector weights. For $$\PGA$$, given a normalized point with $$\ee_{123}$$ weight $$1$$, the $$x$$ coordinate is the weight of the $$\ee_{032}$$ trivector, the $$y$$ coordinate is the weight of the $$\ee_{013}$$ trivector, and the $$z$$ coordinate is the weight of the $$\ee_{021}$$ trivector.

Projective equivalence

In projective geometry, all geometric entities (planes, lines, points) exhibit a property known as projective equivalence. For any such entity $$X$$, the entity $$aX$$ represents the same entity for any non-zero real scalar $$a$$. To compare weights from one entity to another meaningfully however, we tend to use this projective equivalence to keep things normalized in the same way. For example, it doesn't make sense to compare the $$\ee_{021}$$ weight (the $$z$$-coordinate) between two points unless they are both normalized in the same manner.

In this case, the trivector is associated with the point at $$(-2, 2, -1)$$. How can we verify that this point is the intersection we're seeking? Let's represent our plane and line as a system of equations. Recalling that $$\ell$$ was constructed as the intersection from two planes from before, this means that we can form the following system of three equations:

\begin{aligned} 0 &= 3 + x + z \\ 0 &= 1 + z \\ 0 &= x + y \end{aligned}

Subtracting the second equation from the first yields $$x = -2$$. Substituting in the third equation yields $$y = 2$$. Finally, the second equation immediately gives $$z = -1$$.

In code, the intersection of the plane and the line to construct a point can be done as follows:

kln::line l(-1.f, 0.f, 2.f, 0.f, -1.f, 0.f);
kln::plane p{0.f, 1.f, 1.f, 0.f};
kln::point intersection = l ^ p;


Exercises

1. As mentioned previously, the exterior product is only anticommutative in certain situations. This isn't one of them! In fact, it would be somewhat concerning if the point intersection depended on the order $$\ell \wedge p$$ vs $$p \wedge \ell$$. Evaluate $$p \wedge \ell$$ for the above example and verify that you get the same result.
2. Seeing points as trivectors might feel a bit confusing at first. This isn't so much an exercise, but a request for you, the reader, to reserve any doubts that the representation makes sense until we look at the geometric product in view of symmetric actions.
3. Here, we met a line and a plane to get the point of intersection point. Three planes can be met with the same operator $$\wedge$$ to produce the intersection point as well. Can you see why?

### Joins

In the constructions above, we used the exterior product to meet entities to elegantly construct intersections. The $$\wedge$$ operator is a grade-increasing operation that allowed us to go from planes to lines to points. What about going the other direction? Well, there's a handy tool for that! Earlier, we learned about the Regressive Product, defined in terms of the exterior product on the duals of the operands. As a result, it should be evident that the regressive product is a grade decreasing operation that allows us to join entities to go in the other direction, from points to lines to planes.

Sure enough, this works exactly as you'd expect. Two points can be joined to construct a line, and a line and a point (or equivalently three points) can be joined to create a plane. Here's a code snippet doing exactly this.

kln::point p1{x1, y1, z1};
kln::point p2{x2, y2, z2};

kln::line p1_to_p2 = p1 & p2;

kln::point p3{x3, y3, z3};

/// Equivalent to p1 & p2 & p3;
kln::plane p1_p2_p3 = p1_to_p2 & p3;


Exercises

1. Try to perform the computation above by hand given sufficiently simple initialization values of the three points. Do the results agree with what you expect?
2. Explain why the Poincaré dual map we introduced earlier needed to be an involution for our join to work so conveniently.

## Projection, Rejection, Containment

A common question is how to compute whether a point lies on a line or plane, or whether a line lies on a plane. When examining the implicit forms of lines and planes, the answer is obvious. For example, given a plane $$0 = ax + by + cz + d$$, points that lie on the plane are simply coordinate tuples that satisfy the implicit equation. In GA, coordinates are "substituted" in this manner, and we need a way to express projection/rejection/containment in terms of the algebraic operations available to us. The key is to use the symmetric inner product to perform metric "measurements". Let's consider the various cases separately.

### Point to plane

Let's consider a point $$P = \ee_{123} + 2\ee_{032} + \ee_{013}$$ and a plane $$p = 3\ee_0 + \ee_1$$. The inner product between them is computed as:

\begin{aligned} P \cdot p &= (\ee_{123} + 2\ee_{032} + \ee_{013}) \cdot (3\ee_0 + \ee_1) \\ &= \ee_{23} - \ee_{03} \end{aligned}

This is the line perpendicular to the plane through the point. A way to intuitively see why this is so is to observe the fact that the inner product contracts indices, leaving only the parts that are not in common between the two. As a result, not only does the inner product produce a bivector, each corresponding term in the bivector produced contains indices not present in the plane. Can we now compute the distance between the point and the plane?

The answer is no. Notice how in the above expression, the weight of $$\ee_0$$ of the plane didn't participate in the calculation at all. This means that translating the plane towards or away from the origin doesn't effect its inner product with a point, indicating that the line cannot be used to compute the distance between a point and a plane. Note that whether or not the plane was normalized made no difference.

Before continuing, let's quickly see what would have happened if we computed $$p \cdot P$$ instead of $$P \cdot p$$.

\begin{aligned} p \cdot P &= (3\ee_0 + \ee_1) \cdot (\ee_{123} + 2\ee_{032} + \ee_{013}) \\ &= \ee_{23} - \ee_{03} \end{aligned}

Conveniently, the results match!

Exercises

1. Can you justify the equality between $$p\cdot P$$ and $$P \cdot p$$?
2. In the example above, the line produced by the inner product has a translational component. Where did that component come from?
3. Can you see how the translational component of the produced line changes when we shift the position of the point?

To compute the distance from the point to the plane, we need to use an operator that doesn't annihilate the translational component of the plane. The answer is to use the either the join ($$\vee$$) or meet ($$\wedge$$) operator to construct the metric-free quantity from which we can extract the distance by inspecting the magnitudes. Let's compute both to see what we come up with:

\begin{aligned} P \wedge p &= (\ee_{123} + 2\ee_{032} + \ee_{013}) \wedge (3\ee_0 + \ee_1) \\ &= -3\ee_{0123} - 2\ee_{0123} \\ &= -5\ee_{0123} \end{aligned}
\begin{aligned} P \vee p &= (\ee_{123} + 2\ee_{032} + \ee_{013}) \vee (3\ee_0 + \ee_1) \\ &= \JJ\left((\ee_0 + 2\ee_1 + \ee_2) \wedge (3\ee_{123} + \ee_{032})\right) \\ &= \JJ\left(3\ee_{0123} + 2\ee_{0123}\right) \\ &= \JJ\left(5 \ee_{0123}\right) \\ &= 5 \end{aligned}

The ideal norm (norm computed with only elements containing subscript $$0$$) of the meet matches the norm of the join which encodes the distance from the point to the plane. As an equation, we'd express the ideal norm as $$||P\wedge p||_\infty$$. Let's check if this is right! The plane $$p$$ corresponds to $$0 = x + 3$$ and the point $$P$$ corresponds to the coordinate tuple $$(2, 1, 0)$$. The point-to-plane distance formula tells us that the distance between $$P$$ and $$p$$ is, in fact, $$5$$, in agreement with our point-plane meet and join calculation. To understand why this works, pay attention to which terms survived the meet and the join. The point's homogeneous component paired with the plane's translational component, and all the plane's direction components paired with the point's positional components. Geometrically, we've projected the point onto the normal direction of the plane and subsequently taken the norm which included the plane's translation from the origin.

To compute this distance in Klein, given a point P and plane p, we'd simply do std::abs((p ^ P).e0123()). Alternatively, we could do std::abs((p & P).scalar()). Of course, you must normalize the plane with p.normalize() for this computation to work.

Exercises

1. The above computation worked because the plane $$p$$ was normalized such that $$p^2 = 1$$. Change the normalization factor or change $$p$$ itself and verify that normalization is required for the distance measure above to work.
2. Considering the meet and join operators, does it make sense that the meet and join of a point and plane produced the pseudoscalar and scalar respectively?

What about the line perpendicular to the plane through the point $$P\cdot p$$? It can't be used to measure the distance to the plane as we pointed out before, but it can be used to project the point to the plane. The succinct formula for this projection is $$(p \cdot P)P$$.

\begin{aligned} (p\cdot P)P &= (\ee_{23} - \ee_{03})(3\ee_0 + \ee_1) \\ &= 3\ee_{023} + \ee_{123} - \ee_{031} \\ &= \ee_{123} - 3\ee_{032} + \ee_{013} \end{aligned}

which corresponds to the point $$(-3, 1, 0)$$. That this point lies on the plane is immediate. Also, this point is $$5$$ units away from $$P$$ and the line between them is parallel to the $$x$$ axis, indicating that we have indeed projected the point to the plane. Klein code that computes this projection is shown below:

kln::plane p{1, 0, 0, 3};
kln::point P{2, 0, 1};
kln::point P_on_p{(p|P) * p}; // Equivalent to (P|p) * p


### Line to plane

Another potential projection we should consider is projecting a line to a plane. Without proof, the equation projecting $$\ell$$ onto $$p$$ is $$(p\cdot \ell)p$$. Before trying this out, let's check the dimensionality first. The expression $$p\cdot\ell$$ will produce a vector quantity, but the geometric product of two vectors doesn't necessarily produce a bivector (there could be scalar components). The "trick" is to realize that because the inner product contracts all subscripts, the vector produced by $$p\cdot \ell$$ will be completely orthogonal to both $$\ell$$ and $$p$$, so we are justified in claiming that the final result $$(p\cdot \ell)p$$ is a bivector, meaning that this formula does, in fact, produce a line.

Let's consider the line $$\ell = \ee_{32} + \ee_{13}$$ (you should recognize this line as a line through the origin between the $$x$$ and $$y$$ axes). The projection onto the plane $$p = 3\ee_0 + \ee_1$$ should clearly be the $$y$$ axis shifted $$3$$ units so let's verify that this is the case:

\begin{aligned} (p \cdot \ell)p &= \left((3\ee_0 + \ee_1)\cdot(\ee_{32} + \ee_{13})\right) p \\ &= \ee_{3}(3\ee_0 + \ee_1) \\ &= -3\ee_{03} - \ee_{13} \end{aligned}

At first glance, this doesn't seem like the line that we want, but we must remember that projectively, the lines $$3\ee_{03} + \ee_{13}$$ and $$-3\ee_{03} - \ee_{13}$$ are projectively equivalent. In Klein, we can produce this projection as follows:

kln::plane p{1, 0, 0, 3};
kln::branch l{1, 1, 0};
// A branch is a line through the origin
// This calculation works the same with kln::line l{0, 0, 0, 1, 1, 0} but
// is more efficient using a branch since a branch uses half the storage.
kln::line l_on_p{(p | l) * p};


Exercies

1. Repeat the projection above using $$(\ell \cdot p)p$$ instead of $$(p\cdot \ell)p$$. You should find that you get a negated result that remains projectively equivalent.

### Point to line

With the point and a line, two reasonable questions to ask are, as with the plane, how to compute the distance to the line, and how to project the point onto the line. Before continuing, guess what the formulae should be!

Guess! Then click to reveal spoilers

If you thought the distance between the point and the line is $$||P\vee \ell||$$ and the projection is $$(P\cdot \ell)\ell$$, you'd be right! The uniformity between this equation and all the others we've seen should be quite satisfying. Note that the point must be normalized so that the weight of $$\ee_{123} = 1$$ and the line must be normalized so that the square norm of the directional components is unity for these formulae to work.

In Klein, given a point P and line l, the projection is kln::point P_on_l{(P|l) * l} as you would expect.

The projection is somewhat straightforward, but perhaps the distance formula may take some additional processing. First, let's consider the join $$P\vee\ell$$. What type of quantity does this produce? Well, from the section detailing the regressive product above, we know that his is a plane containing both the point and the line. It turns out that the norm of this plane will encode the distance between the point and the line. Let's see how this works for the line $$\ell = \ee_{02} + \ee_{23}$$ and point $$P = \ee_{123} + \ee_{013} + \ee_{021}$$ (we expect the distance to be $$1$$).

\begin{aligned} ||P\vee \ell|| &= ||(\ee_{123} + \ee_{013} + \ee_{021})\vee (\ee_{02} + \ee_{23})|| \\ &= ||\JJ\left((\ee_0 + \ee_2 + \ee_3)\wedge(\ee_{31} + \ee_{01})\right)|| \\ &= ||\JJ\left(-\ee_{013} -\ee_{021} + \ee_{123} + \ee_{013}\right)|| \\ &= ||\JJ\left(-\ee_{021} + \ee_{123}\right)|| \\ &= ||-\ee_3 + \ee_0|| \\ &= 1 \end{aligned}

If you have trouble remembering how lines are defined, remember that terms containing a $$0$$ in the subscript are translational components (the moment of the line) and the remaining terms determine the line's direction. The line $$\ee_{02} + \ee_{23}$$ can be decomposed as the meet of two planes $$\ee_0 - \ee_3$$ and $$\ee_2$$ so we can intuit that the line is the $$x$$ axis shifted a unit in the $$+z$$ direction. The norm of a point is just the homogeneous coordinate. Code computing the distance between a point and a line looks like:

kln::line l{0, 1, 0, 1, 0, 0};
kln::point p{0, 1, 1};
float distance = kln::plane{l & p}.norm();


For completeness, we can project the point on the line with the following snippet:

kln::line l{0, 1, 0, 1, 0, 0};
kln::point p{0, 1, 1};

// Note that for this calculation to work, if the line l and point p were
// not normalized, we would need to invoke l.normalize() and p.normalize()
// first before proceeding.

kln::point p_on_l{(p | l) * p};


Exercises

1. The weight of $$\ee_{013}$$ in the point above should correspond directly with its distance from the line. Verify that this is so.
2. Try shifting the point in a different direction (say along $$x$$). Does the distance stay unaffected as you'd expect? What about shifting the point in the $$z$$ direction?

## Reflections

So far, we've seen how the exterior product and regressive product can be used to construct planes, lines, and points. We also saw how the symmetric inner product can be used to measure angles between planes. Furthermore, we used the ideal norm in conjunction with the exterior and regressive products to compute distances between entities. Already, $$\PGA$$ has yielded many fruitful results which has consequently led to a compact and elegant API. However, we haven't yet encountered a primary usage of the geometric product, and this is where we'll fully appreciate the decision of using the representation we have.

### Reflection through a plane

A reflection of an entity $$X$$ through a plane $$p$$ is given by $$pXp$$. This is true regardless of whether $$X$$ is another plane, a line, or a point!

ARTICLE CURRENTLY BEING DRAFTED