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━ Maths - Vectors ━​

Introduction

WIP Intro

Hi, you're early! I haven't ported this page over from my draft just yet but you can preview it at the link below for early-access:
​tinyurl.com/gamedevStudySheet

Introducing Vectors

Vector Notation
A vector is a data structure that holds numbers. When typed, a variable that refers to a vector will always be bolded. For example: v. A vector is like a matrix but a vector is always only one column or row. In that sense, a vector is like a very tall or very wide matrix. Vectors are typed in bold and lowercase, matrices are typed in bold and uppercase (like M). Matrices are described as mxn "m by n" where m is the row count and n is the column count. When typed, a vector (like v) always refers to specifically a tall, vertical, one-column vector. In terms of mxn this means n = 1. If we want to refer to a wide vector ( m = 1 ) then we write v^T, this looks like "v to the power of T" but is simply "v transpose". A transpose is a counter-clockwise rotation of a vector. 
          [ 1  ]
If v = [ 2 ] , then v^T = [ 1 , 2, 3 ].
          [ 3 ]
If I write v = [ 1, 2, 3 ], then I am still describing a vertical vector that looks exactly like the v shown on the line above but I am just writing it horizontally for practicality. If a vector is actually horizontal (wide) then I will always use the transpose notation. Now let's explore the significance of a vector with respect to lines which may be a structure you feel more familiar with...

Rays, Lines, Line Segment, and Vector Formulas
A line is a 2D structure that has zero width, zero height, and infinite length.  In school you may have learned the line equation y = mx + b. We will not use this equation, it is a very constrained 2D representation of a line that will not be useful for game dev where most of our applications are in 3D. Instead, I want you to think of a line as a container in 3D space. Our line formula is an equation that defines which points in the 3D line within our line. In that regard, it is also a test to see if any given point is on our line. A line contains an infinite amount of points, but two points are sufficient to define the line. If we have two points v1 and v2 we can subtract them to get the direction between them, v (this is a vector). Our line formula is defined using this v in addition to our of the points on the line which we call P (it can be v1, v2, or any other point on the line).
L = P + ( t * v )
The formula above defines line L, starting from point P, into the direction v which is scaled by scalar t. Because a line extends in two directions, t can be negative. Because a line is infinite, t can be any value from -infinity to +infinity. Now here is the cool part... this is also the equation for a direction vector and a line segment. The only difference is that the data type defines the range of our scalar, t. Below are the parametric equations for each of these structures, so-called "parametric" because they all rely on that input parameter t.
​
L = P + ( t * v ) is a line if t = [ -infinity, +infinity ]
L = P + ( t * v ) is a ray if t = [ 0, +infinity ]
L = P + ( t * v ) is a line segment if t = [ 0, n ]
L = t * v is a vector t = [ -infinity, +infinity ]  ( vectors have a magnitude and a direction but no position )

The realization that we can think of direction vectors and line segments as just lines with bounds is a critical insight. Most of the problems we will face with these data types are much easier if we instead solve the problem with lines. For example, we will later explore the problem: "does line segment L intersect plane P?", the solution is basically to first solve "does line L' intersect plane P?" and if it does, then we just need to check if the intersection between L' and P is on line segment L.
Practice Question: ​What is the equation of a line that contains these two points: ( 11, 13, 17 ) and ( 10, 11, 14 )?
Answer:
​v =  ( 11, 13, 17 ) - ( 10, 11, 14 ) = [ 1, 2, 3 ]
L = P + ( t * v ) --> L = ( 11, 13, 17 ) + ( t *  [ 1, 2, 3 ] ) 
3D Shapes from 2D Vectors
Above we discussed how the line's equation is something like a test to see if a point is in a 2D line. Abstracting that idea a bit... a group of vectors when considered together are sort of like a formula themselves.  For example, a plane is 2D structure that can be defined by two vectors. Any two vectors will work as long as they are not coincident aka collinear aka in the same direction. We'll explore planes in far greater depth later, but the key insight here is that vectors can be thought of as equations of points and so too can planes be though of as equations of vectors. Now let's take that one step further into 3D...

Basis Vectors
A basis is a collection of vectors that can be combined in order to reach any point within a space. Here's anexample:
av + bu = P
If 2D vectors v and u can be combined and scaled to express any 2D point P, then v and u form a basis for R2 (which means 2D space). 
  • For example let's say P = [ 1, 10], v = [1, 0], and u = [0, 1]. Then the equation can be completed with a = 1 and b = 10.
  • But if P = [ 1, 10], v = [1, 0], and u = [2, 0]. Then the equation can never be completed because the second term of P, 10, can never be reached by the equation a*0 + b*0 = 10. There are no values for a or b that can overcome the zero scalar to reach 10.
The key insight here is that the only thing that matters is for you to have the same number of vectors as the dimension and for them to all be disjoint aka non-colinear aka not in the same direction. I like to imagine basis vectors like rigid segments of a robotic arm. It doesn't help if two segments go the same direction, in order for the robot to be able to reach any point, the segments need to extend in different directions.
Practice Question: Which of the following are a basis in their respective dimensions? 
  • 3 disjoint vectors in R4
    • These vectors are disjoint, that's important! But we need 4 to form a basis in R4. 
  • 2 collinear vectors in R3
    • These vectors are on the same line, they alone cannot form a basis. Additionally we would need at least three vectors to form a basis in R3.
  • [ 0, 1 ], [ 0, 2 ], and [ 5, 1 ].
    • Here we have three vectors in R2. The first two are colinear but the third is not. So if we take the third and either of the first two then we have two disjoint vectors in R2 which is sufficient to form a basis. 

Bases are really important in game development. An object's "transform" is a basis that defines its orientation. Have you ever seen an image like the one below? This is commonly how transforms are visualized. As you can see, there are three disjoint vectors shown. All of a transform's basis vectors are perpendicular.
Vectors for Distance
In addition to defining 3D objects, vectors are commonly used to define the space between objects. In an earlier example we discussed how if you take two points on a line: v1 and v2, then you can subtract them to derive the direction of the line. The order of the points in the subtraction ( "v1 - v2" vs  "v2 - v1" ) doesn't matter for a line because it is infinite in both directions, but it matters a lot when directionality is relevant. For example if S is a race's start position and E is the race's position, how do you derive the direction from the start to the end of the race? It's E - S. You always put the end point as the first argument. A pneumonic for this can be "keep your eye on the price" or "put the destination first". It's easy to derive in a 1D case, just consider if you want to get from value 0 to value 10 do you need "10 - 0 = +10" or "0 - 10 = -10" of course we need +10 to get from 0 to 10, as -10 would send us in the wrong direction into negative numbers. 

Vector Norms
The magnitude of a vector is the total distance it represents. The vector magnitude is also known as the "L2-norm", "Euclidean norm", or simply "the norm". For 1D vectors, the magnitude is just the value's absolute value. And just like an absolute value for an integer is written with vertical bars, a vector norm is also written this way. An absolute values are only relevant in 1D (basic numbers), but a vector norm is like an absolute value formula that can be applied for higher dimensions. To acknowledge this, some authors will use one line on the sides for absolute value "|x|" and two lines for the vector norm "||x||", but I am not going to do that. For higher dimensional vectors we have a formula to help us out:
|v| = sqrt( ( v.x * v.x ) + ( v.y * v.y ) + ... )
This norm (the Euclidean norm) is the most common kind of norm which is why people sometimes just call it "the norm". But there are actually many kinds of norms. One kind of norm is called the p-norm which takes a variable "p" into an equation. The L2-norm is simply the p-norm with p = 2. You probably won't come across the other norms in game dev.
Practice Question: ​What is the magnitude of vector v, where v = [ 2, 3, 4 ]?
​|v| = sqrt( ( v.x * v.x ) + ( v.y * v.y ) + ( v.z * v.z ) )
​|v| = sqrt( ( 2 * 2 ) + ( 3 * 3 ) + ( 4 * 4 ) )
​|v| = sqrt( ( 4) + ( 9 ) + ( 16 ) )
​|v| = sqrt( 29 )

Normalizing Vectors
Often, we don't care at all about the magnitude of a vector and instead would prefer to normalize aka "unitize" it which means scale it to be a length of 1. We do this when we want a vector only for it's direction or if we want to simplify a formula that requires us to use the vector's magnitude as a factor (if we magnitude is 1 then the factor can be removed). To normalize a vector, divide each term by the vector's magnitude. If a vector is normalized, then it will be italicized.
Syntax Key:  x<--scalar    v<--vector   u<--normalized vector   M<--matrix   P<--point   PS<--line (or line segment) defined by points P and S
Practice Question: ​Convert vector v, where v = [ 2, 3, 4 ], into a unit vector u.
|v| = sqrt( 29 ) <-- see previous problem
u = v / |v| 
u = [ 2, 3, 4 ] / sqrt( 29 )
u = [ 2  / sqrt( 29 ), 3  / sqrt( 29 ), 4 / sqrt( 29 ) ]

Dot Product

Congratulations, you have made it to the most important section of this entire guide. The dot product aka "the inner product" is absolutely the most common interview topic out of all possible game programming questions. I have never experienced a gameplay engineer interview that has not touched upon the dot product. It is imperative that you learn this topic inside-and-out, upside-down, and sideways. It is ESSENTIAL!

Dot Product: “the product of two vector magnitudes multiplied by the cosine of the angle between the vectors.”

Its primary uses are to find the angle between two vectors and to perform vector projections. To remember that cosine needs to be used for the dot product, I suggest the pneumonic of observing that the ‘o’ and ‘c’ in cosine looks like a dot. 

The dot product requires O(n) FLOPs to compute which is pretty fast compared to the cross product.

The dot product can be expressed in many ways, each formula has it's own uses. Here are some common expressions followed by explanations and greater context on their significance. However it does require a horizontal add-an operation that SIMD units cannot do, this makes them repeated dot products slow. One optimization we can do, if you have to do multiple dot products with the same vector v, is to put the unique vectors (for example: x, y, z) into the rows of a matrix and multiply the vector v times the matrix. This yields the three results at the cost of only one horizontal add. I've never done this optimization personally, but I want you to be familiar with some of the tricks like this that we will use in practice for optimization.

a • b = |a| |b| cos( θ ) 
The most common dot product expression. You must have this memorized. We can use the arccos() function on both sides to pull out the theta angle which represents the angle between vectors a and b. If a and b are of unit length, aka "unit vectors", then their magnitudes are 1 and those factors (|a| and |b| ) can be removed to simplify the formula to acos( a • b ) = θ. The theta we receive from a dot product formula is unsigned meaning it does not have any directionality. So if you have two vectors, the dot product can tell you how much to rotate the first for it to align with the second but you won't know which direction to turn. We will later explore the cross product which does provide that directionality.

a • b = cos( θ ) 
If both a and b are normal vectors (that means they are of unit length) then the formula can be simplified to the form above. I have a riddle for you: if  a • b in this formula result in a negative number, then what does this tell us about the angle θ? Think about the unit circle. Cos is the x value of the angle. And it is only negative if the angle is in the top left or bottom left quadrant of the unity circle. Therefore a negative number on the left hand side of this equation means that that θ is obtuse. Now what if the angle is acute? We then the left hand side value will always be positive. But unlike in the negative case, a positive left hand side value does not mean the angle is always acute since a very obtuse angle in the bottom right quadrant of the unit circle will have a positive Cos. In the next formula we will abstract these observations for any vectors, not just normalized ones.

x • y = 0 if x ⟂ y
When you complete a dot product, if the result is zero then that means the vectors are perpendicular aka orthogonal (regardless of if the vectors are normalized). This is the easiest way to check if vectors are perpendicular. I'd like to combine this insight with the previous equations insights to present you with the following statements which we can use to interpret the result of any dot product.

a • b = |a| |b| if they’re exactly same direction
a • b > 0 separated by an acute angle
a • b = 0 if perpendicular
a • b < 0 separated by an obtuse angle
a • b = -1 |a| |b| if they’re exactly opposite direction
a • b = a b^T = ( a.x *  b.x ) + ( a.y b.y ) 
This formula show that we compute the dot product via the summation of pairwise multiplication of two vectors' components. Vectors are mx1 matrices as as we discussed in the previous chapter, mxn matrices cannot be multiplies unless the n of the first factor equals the n of the second factor. So you cannot multiply two vectors because you can't multiply an mx1 by mx1, when m != 1. However if we transpose the second vector, it becomes 1xm and now we can multiply an mx1 x 1xm because the 1's meet in the middle. This is effectively what the dot product does: it transposes the second vector and then represents basic matrix multiplication. You may notice that in the expanded expression, on the far right, that this looks very similar to the vector norm formula, except there is no square root and we have two distinct vectors. We will explore that similarity in the next formula.

a • a = |a| |a| = sqrt( ( a.x * a.x ) + (a.y * a.y) )^2 = ( a.x * a.x ) + (a.y * a.y) 
This formula shows that a vector dotted with itself is equal to it's magnitude squared. This basically amounts to the vector norm formula but the square cancels out the square root.
Checkout that cool triangle formed with side lengths of x, y, and z. Triangles have the interesting property that |x + y| <= |x| + |y|. This is called the Triangle Inequality and it means that any one side length of the triangle cannot be greater than the sum of two other side lengths. Think about why this is true, if the red side length was greater than x + y then x and y could not be combine to reach the end of the red arrow and the triangle could not be closed. Now imagine that the sides of the triangle are each vectors. If x, y, and x + y are vectors, then we we can express the following inequality: | v • w | ≤ |v| |w|. This is known as the “Cauchy-Schwarz” inequality and it is a special case of the Triangle Inequality.​
Practice Question: ​Here is a practical guided question on the dot product: 
​Answer: https://www.youtube.com/watch?v=-OKPCMHr7Qc&ab_channel=MatthewVentures
Practice Question: ​You are given the forward vectors of the player and an enemy: p and e; and the positions of the player and enemy: P and E. Determine if the enemy is facing the player.
​Answer: This question is a little ambiguous because it doesn't clarify the angle threshold within which two things are considered "facing each other". I'm going to consider the angle between two vectors: the enemy's forward vector and the vector from the enemy to the player. If the angle between those vectors is less than 90 degrees then I would consider them facing each other. We can test this easily with the dot product e • EP = r. If r is negative then we have an angle that is less than 90 degrees and the enemy is therefore facing the player.

Vector Projections

Vector projection is the process of isolating a portion of a vector. Take a look at the image below which shows four vectors: a, b, x, and y. The vector a can actually be broken into two different vectors: x and y. For example y might represent the portion of a that goes in the direction of the y axis and x might represent the portion of a that goes in the direction of the x axis. If I asked for the "projection of a onto b" then that would be the same as asking for "the portion of a that goes in the direction of b". In this case that would simply be x.
Introduct projections and then relate them to the Cauchy shears inequality. TODO
In this next example, I have changed b. Now the projection of a onto b is z. One way that may help you to conceptualize this operation is to think of a sun projecting a shadow of a onto b. Can you imagine how z in the example below and x in the example above are shadows of a? If not, don't worry about it, the main insight is to think of z as a part of a.
Practice Question: ​In the example above, z is described as a part of a. Derive w, where w is the remaining part of a.
Hint: that means that z + w = a. Just like x + y = a.
Answer: Since w is the remaining part of a, that must means that a = z + w. Therefore w = a - z. But here's something really cool... in this 2D example, if z is a's component in b's direction, then that means w is a's component in the direction perpendicular to b. So just like we were able to project a onto b to get z. If we found some vector perpendicular to b, say c, then the projection of a onto c would be w. This is a 2D case, so a only has two components. But if we were in 3D then we could extrapolate this method further to find a final vector d that would be perpendicular to both b and c and projecting a onto d would give us our third component. Projections are like chopping up our vector into it's composite parts.

​Formula: Projection of a onto b
Below is the formula for calculating the projection of "a onto b".
I've written the formula in two ways because I think both have unique utility.
  • The one on the left shows that the magnitude of b is used to scale two terms: a • b and b. This is important to realize because if b is a unit vector (meaning that you are projecting onto a unit vector), then the projection of "a onto b" simplifies to simply ( a • b ) b. This form is nice because the left term is a scalar that completely accounts for the magnitude of the resulting vector (including the sign) and the right term is a unit vector that completely accounts for the direction.
  • The second formula is nice because it combines the two magnitude into a simple denominator dot product then groups all the scalars together on the left. This is a great way to show that when projecting onto b, we are really just trying to calculate how much to scale b. All of our computation simply amounts to a scaling factor to apply to b.

Projections must be onto the origin
I want to address a major caveat that caused me to fail an interview some years ago. If projecting onto a structure with position, then that structure needs to contain the origin for the projection to work. This is somewhat complicated, but critical to understand. For example, if you are projecting onto a line, line-segment, or ray, then you need to first translate it to the origin before projecting onto it. Vectors have no position and do not require this step but things like lines do have position so this step is imperative.

If line segment AB does not go through the origin, here are the steps to projecting vector p onto AB: 
1 ) Subtract A from p to obtain p’
2 ) Then project p’ onto AB for partial solution h. 
3) Finally, now we add A to h (undoing our offset earlier) for the final answer. 

To explain step 1: we needed to subtract any point of AB from p, so A was an easy choice. This may seem like a very strange step but basically it is just a walk to the origin. For example, if A = [ 0, 0, 10 ], then it is 10 units up the z-axis away from the origin. By subtracting A from p we are undoing that 10 unit difference.

Perpendicular Projection
In the projection examples we have thus far discussed, the projection of a onto b resulted in a vector representing the component of a that was parallel to b. In other words, the result of the projection of a onto b is always a vector parallel to b; this has led to some folks referring to this projection as the "parallel projection". In contrast, you might come across something called a "perpendicular projection" aka "orthogonal projection" in which we actually want the component of a that is perpendicular to b. In R3, this is easy to compute. To find the perpendicular projection of a onto b, simply subtract the parallel projection of a onto b from a. This process removes any part of a that was parallel to b, leaving only the perpendicular parts.

3D Cross Product

The 3D cross product aka "outer product" or simply "vector product" returns a vector perpendicular to two input vectors, multiplied by the product of their magnitudes, multiplied by the sine of the angle between them. The resulting vector's magnitude is equal to the area of the parallelogram formed by the input vectors. A pneumonic to remember to use the sine function is to think that the ‘s’ in sine looks like two crossed curves.

The 3D cross product formula: a ⨯ b = |a| |b| sin(θ) n
  • n is a unit vector perpendicular to both a and b.
  • θ is the angle between a and b.
  • If a and b are parallel, then the resulting vector will be the zero vector, 0.
​
Hey check it out, this formula returns the angle between the two vectors just like the dot product formula, but here it is the sine of the single instead of the cosine of the angle. As you will soon see, a cross product takes a lot of computation, about O(nm) FLOPs, which is much more than the dot product so if you're interested in the angle it makes more sense to just use the dot product. But the directionality of n, can tell us if the angle between input vectors is positive or negative which is not something the dot product can tell us.

Unlike the dot product, which returned a scalar (a number), the cross product returns a vector. In the example below, the purple vector a⨯b is the cross product of vectors a and b. Observe how it is perpendicular to both of them. 
Cross Product Magnitude
The above diagram also shows that the volume of the parallelogram formed by a and b is equal to the magnitude of a⨯b. When taking the magnitude of a cross product, remember that n is a unit vector so it can basically be removed from the formula: |a⨯b| = |a| |b| sin(θ).
Practice Question: ​If you take the cross product of two unit vectors, is the resulting vector going to be unit length?
Answer: the magnitude of a cross product of two vectors is equal to the area of a parallelogram formed by two vectors. In which case would a parallelogram has an area of 1? Think of of the formula: area = base * height. For the area to be 1, the base and height would both need to be one as well. The two unit vectors are of length 1 so the answer is yes.
SIKE! Did I fool you? Yes, it is true that area = base * height. But two sides of a parallelogram are not necessarily their base and height. Take a look at the image below, in this case a is not the height, h is! So we need to ask, when is a the same length as h? The answer is only when the parallelogram is a rectangle. So the answer to this question is "the resulting vector will only be unit length if the angle between the input vectors is a right angle".
Cross Product Magnitude Summary
The cross product magnitude |a⨯b| equals...
  • 0 if the input vectors are parallel. Can you think of why this is? Two colinear lines would result in a parallelogram with zero height! And therefore zero area.
  • |a||b| if a ⟂ b, basically this is the same insight from the practice question above. If the two sides of the parallelogram are perpendicular then it will be a rectangle and have area = base * height.
​
​Cross Product Properties
The cross product is anticommutative:  a⨯b = -b⨯a 
So basically, when you switch the order of the vectors, one has their sign flipped. If the cross product is going to return the sign of the resulting vector than this makes sense as a required attribute. That's because if you had two input vectors and order did not matter, then crossing both of them would provide you with the the number of degrees between them, but no indication of which direction you would need to turn. The dot product is commutative so we can re-order terms but it alone cannot determine the sign of the angle between them. This trait is one of the things that makes the cross product really useful.

Following up on the cross product's anticommutative-ness, the right-hand rule can be used to determine the direction of the resulting vector.  Its a neat little trick where if you use your orient your pointer and middle finger towards each of the input vectors and then your thumb will point in the direction of the cross product result. In the left example below, I have plotted some vectors using an online visualizer.
The blue vector is [-1, -1, 0], the red is [-1, -1, 0], and finally the orange is [0, 0, 2]. The orange vector should be the result of blue vectors crossed with the red vector. In other words, [-1, -1, 0]⨯[-1, -1, 0] = [0, 0, 2]. And check it out, if you aim your pointer finger in the blue direction (to the left), and your middle finger in the red direction (towards you), then your thumb naturally points upwards in the orange direction. See the middle image below.
To demonstrate the anticommutative property, you can now try to switch the order of red and blue. First aim your pointer finger in the red direction (towards you) and now this time aim your middle finger in the blue direction (to the left). In order for your right hand's middle finger to point left, you will need to invert your hand upside down and the result is your thumb will point down instead of up. 
​The cross product is additively distributive ( a ⨯ (b + c)) = ( a ⨯ b ) + ( a ⨯ c)
Remember that the cross product's result vector is perpendicular to the two input vectors. So here we are saying that the cross product of two inputs is going to be the same vector that we would get by combining cross products of one of those input vectors with each of the components of the other input vector.

Do you remember how the cross product magnitude is equal to the area of the parallelogram formed by the two input vectors? Well I want to show you something like an abstraction of that into 3D. There's this really cool product called the triple product. It's formed by using the dot product AND the cross product, crazy I know! It looks like this: a • ( b ⨯ c ). And so here is the really cool part, this expression evaluates to the signed volume of the parallelepiped formed by a, b, and c. The below image is an example of such a volume.
Cross Product Computation
So far we have discussed at length what cross product represents and what it equals but not how to actually calculate it. In order to calculate it we will need to use the determinant formula covered in the previous chapter. And we will need a new equation. The equation below shows the cross product expressed as a determinant. In this formula, the unit vector variables i, j, and k represent the three components of a 3D vector. When we solve the determinant it simplify to the form: a⨯b = ei + fj + gk. Where i, j, and k are unit vectors and e, f, and g are scalar factors.
Below is the fully expanded derivation. As you can see, the 3x3 determinant was broken up into three smaller 2x2 determinants. Then the 2x2 determinant formula was used to express a⨯b into the form of a⨯b = ei + fj + gk. Pardon me, I don't have the ability to use subscripts in my text so I will use brackets, hopefully you will understand that in this case e = a[2]*b[3] - a[3]b[2].
Here is the equivalent expansion written in code:
void crossProduct( int v_A[], int v_B[], int c_P[] )
{ 
   c_P[0] = v_A[1] * v_B[2] - v_A[2] * v_B[1]; 
   c_P[1] = -(v_A[0] * v_B[2] - v_A[2] * v_B[0]);
   c_P[2] = v_A[0] * v_B[1] - v_A[1] * v_B[0];
}
As you can see, the code is zero-indexed so the index numbers are one less than in the algebraic representation. For example the first multiplication we see is v_A[1] * v_B[2] instead of a[2]*b[3]. As you can see, the cross product requires quite a lot of computation!​
Practice Question: ​If given a player's forward vector f and the vector from a player to a target t, explain how you would derive the signed magnitude of the angle the player must turn to that that face the target. Basically, you need to determine how much we need to rotate f so that it is collinear with t and in the same direction as t. 
Answer: The first part of the problem is to get the angle θ between f and t by using the dot product. That angle is unsigned so we now just need to determine if the rotation is clockwise or counter-clockwise. This second part requires two steps. First we determine the plane we are rotating within. This plane will be defined by the normal vector perpendicular to both f and t. To find that vector we use the cross product: Vector3 planeNormal = Vector3.Cross( characterForward, characterToTarget )
Now to see if our rotation is clockwise from the player's perspective we just need to test if the plane normal is in the same direction as our character's up direction. 

bool shouldTurnClockwise = Vector3.Dot( planeNormal, character.up ) > 0;
​This line works because the dot product returns a positive value only when the input vectors are separated by an acute angle. And if the rotation plane's normal and our player's up are seperated by an acute angle, then that means they point in the same general direction and that means we are rotating clockwise.

Our final answer is that we need to rotate around the plane defined by planeNormal for angle θ in the direction dictated by shouldTurnClockwise.

2D Cross Product

In the previous section we learned about the 3D cross product, and you might think that the 2D cross product is the same thing just in 2D. But you'd be wrong! It is true that the 2D cross product takes 2D vectors instead of 3D vectors, but this product returns a scalar instead of a vector. However the 2D cross product has a similar role to the 3D cross product in that it's result will tell us if the first input vector is on the left or right side of the second input vector, just like the 3D cross product can be used to determine the direction of an angle between input vectors.

2D Cross product formulas:
a⨯b = ( a.x * b.y ) - ( b.x * a.y )
AB⨯CD = ( B.x - A.x )( B.y - A.y ) - ( D.x - C.x )( D.y - C.y )
The first formula is for vectors, the second formula is for line segments.

The magnitude of a⨯b is...
  • Positive if the a is on the left hand side of b
  • Negative if a is on the right hand side of b
  • Zero if a and b are collinear
Practice Question: ​You are given two points: A and B, and a line segment defined by a start and end points: S and E. Determine if A and B are on the same side of line segment SE.
Answer: We can use the 2D cross product to determine what side of a line a point is on. First we test point A. AS ⨯ AE = a. If a < 0, then the point lies on one side of the line, and if a > 0, then it lies on the other side. If a=0, then the point lies exactly on the line. Now we can repeat this step for the other point: B. BS ⨯ BE = b. Now we can compare the results, a and b, to see if they are the same polarity (aka both negative or both positive) which will tell us if A and B are on the same side of SE.
Practice Question: ​Imagine you are making a 2D game. Given a vector, g, representing the ground plane and a vector, b, pointing to the position of a ball, determine if the ball is above or below the ground.
Answer: https://www.youtube.com/watch?v=-n_C7tD55_A&ab_channel=PothOnProgramming
Practice Question: You are given a triangle formed by points ABC. And a separate point, P. Determine if P lies within the triangle.
Answer: Basically, to test which side of AB the point P is on, we do AP x AB. If P is on the same side of all the triangle sides then it is in the triangle, so the solution is to just repeat this check for the remaining sides. This approach actually works for all polygons! It just requires you to go through the points in order. Remember the order matters a lot because if you compare a point to AB vs BA those two line segments are literally going in the opposite directions! So make sure to pick a direction (like clockwise) and stick to it such as AB, BC, CA.
Code: https://stackoverflow.com/questions/2049582/how-to-determine-if-a-point-is-in-a-2d-triangle

Thankfully, this is a very common problem and alternative approaches have various benefits/drawbacks. We will discuss a Barycentric coordinate approach in the next chapter. But for completeness I will provide a link to that derivation here in case you're interested: https://www.youtube.com/watch?v=EZXz-uPyCyA&list=PLW3Zl3wyJwWN6V7IEb2BojFYOlgpryp1-&index=7&ab_channel=JorgeRodriguez

Vector Maths Practice

Are you ready for the gauntlet of vector maths review questions? Good luck!
Practice Question: Given vectors a and b, determine if they are ∥ or ⟂.
Answer: 
​A • B = 0 if ⟂ 
A x B = 0 if ∥ (this is the same as checking if |A x B| = 0 )
Practice Question: Given vectors a and b, determine if they are ∥ or ⟂.
Answer: 
​A • B = 0 if ⟂ 
A x B = 0 if ∥ (this is the same as checking if |A x B| = 0 )
Practice Question: Given line L, find any other line that is perpendicular to L.
Answer: This question is deceptively difficult. Since L is a line, we have theoretically probably have at least one position on the line and also it's slope. A perpendicular line will be any line with a "negative reciprocal" slope when compared to L. In a 2D line this is usually as easy as taking the first lines slope "x/y" and converting it to "-y/x" but... since any of L's component could be zero, this approach takes a few checks. Here is the approach I suggest:

If L has two non-zero values, we use the “negative reciprocal” approach to swap the two nonzero elements
if xyNonZero return (y, -x, 0)
if yzNonZero return (0, z, -y)
if xzNonZero return (z, 0, -x)

Otherwise, if only 1 value is nonzero, just pick another axis to be 1 and leave the rest zero 
if xNonZero return (0, 1, 0)
if yNonZero return (1, 0, 0)
if zNonZero return (1, 0, 0)
​
If no values are nonzero then L is an edge case and we have no valid answer. We can probably just return L or something like that.
Practice Question: ​You are given a player's and enemy's position represented by the points P and E. You are also given the player's normalized forward vector, f. Given that both the player and enemy are on the same plane (the XY plane), determine the amount of degrees the player needs to turn in order to face the enemy. Indicate if this rotation is positive or negative.
Answer: The 2D cross product can be used to determine the side of a line that a point is on. It may seem like the right fit here as we could 2D cross product E with f  to determine if the E is on the right or left side of the player. However, that would not tell us how many degrees we need to turn.
The dot product is another useful vector operation that can be used to determine the angle between two vectors. We could compare f and PE in this manner to see how much rotation is necessary for the player to turn to face E. However, this result will provide us with an unsigned result so we won't know which way to turn. 
You could combine the dot product approach (which gives us the magnitude of the angle) and the 2D cross product approach (which gives us the sign of the angle).
Alternatively, since we know the player and enemy are on the XY plane, we know that the Z axis vector, z, will be perpendicular to the line segment PE. You could get the 3D cross product of f and PE to get a vector perpendicular to both of those vectors, v. Then you could calculate the dot product of v and the z  to determine if they both point in the same direction.  
Practice Question: Find the distance between a line segment, BC, and point, A.
Answer: Classic question, this is something you need to feel comfortable solving. I absolutely recommend drawing a diagram for questions like these.​

The first approach I would like to show you uses the parallelogram area formula. ​Here is a diagram for you to reference: ​https://i.gyazo.com/30720d2a714569900b61b552af92a11f.png We consider the parallelogram which is partially defined by the three points we are provided. Since we know that the magnitude of the cross product is equal to the area of the parallelogram defined by the input vectors, we can determine that area with |BA x BC|. Then since we know that the area of a parallelogram is also it's base * height, we can use this area with the base length, |BC| to derive the height length. And the cool part is that our height length is actually the distance to A. We know this because the shortest distance between a point and a line is always going to be along a line perpendicular to our starting line. And that also happens to be the height of our parallelogram.

An alternative approach to this problem is to use vector projection. Here is a diagram for your to reference: https://i.gyazo.com/e4a300ff9834697fe5c56d97592bfc90.png If we project BA onto BC then we will get a vector which points from B to D. D is a point on the line that is closest to A. We can use the vector we derived to get from B to D and then calculate the magnitude of the vector DA which is the distance between these two points. We know that the shortest distance between a point and a line is always going to be a straight line from the input line to the point and we know that DA is this 'straight line' because it will be perpendicular to BD. So to start this approach we are going to need to first move our problem to the origin. This is always a necessary step when projecting onto a structure with position data, without this step the projection result may not be valid. I have copied the below steps from earlier in the guide while replacing the variables in the steps to correspond to this problem. For this step I will refer to the vector from B to A as v.
"...here are the steps to projecting vector v onto BC:
1 ) Subtract B from v to obtain v’
2 ) Then project v’ onto BC for partial solution h. 
3) Finally, now we add B to h (undoing our offset earlier) for the final answer."
The result, h, is the vector from B to a point D which is the closest point on line BC to A. Let's derive D via D = B + v.
Okay, now that we have D, there is only one more step: determine the distance between D and A. This can be calculated using the earlier formula "|k| = sqrt( ( k.x * k.x ) + ( k.y * k.y ) + ... )" we just need to convert DA into a vector which is easy enough, k = A-D. Then we plug k into the magnitude formula and we are done. There is a psuedo code write-up of this approach here: ​https://stackoverflow.com/questions/3120357/get-closest-point-to-a-line
Practice Question: Consider a bullet that is bouncing off a wall defined by it's normal vector, n. The bullet starts at position S with velocity v. Determine g, the directional vector of the ball following the collision with the wall. 
Answer: I created a detailed walkthrough of this problem here: ​https://youtu.be/IIVcGCdkMbA?si=hOjEhas6rw6UVTMz

An alternative approach is to project v onto n to get p, where p is basically just a vector along n but in the opposite direction of n. And then you can use the dot product between v and p, to determine the angle of the bullet going into the wall. And you can rotate p that same angle just away from S. You would need to rotate around the axis defined by the p x n, but I think this approach would work just as well.
Practice Question: You are given a line defined by points AB and a line defined by points PQ. Determine how far apart the lines are from each other at their closest point of intersection.

​Answer: This question is medium-to-hard difficulty. I don't think it would come up on a beginner's interview but it is an excellent question and a good test for mid-level to senior candidates. The key insight to solving this problem is to understand that the shortest distance between two points is always going to be a straight line. If you tried the earlier problem that determines the distance between an point and a line then you should remember that insight means we need to find a perpendicular vector to our input line. In this case, we need to determine the a line segment which is perpendicular to both of the input lines and the length of that line segment will be our solution.

First, let's get the edge case out of the way. If these lines are parallel, then any point on the lines is going to be the closest distance from the other line. We can take an arbitrary point on AB, for example let's just use A. Then we can solve the sub-problem "What is the closest distance from line PQ to point A" for our answer. We solved this sub-problem in an earlier practice problem.

Okay so now that we know the lines are not parallel, we can find a vector that is perpendicular to both. Can you think of a vector operation that takes in two vectors and returns a vector perpendicular to both of them? It's the cross product! Use the cross product to get vector n which is the shortest direction between the lines. However, though it is in the right direction, it is not necessarily the correct distance. If we take any line segment that connects a point from each of the lines, we can project it onto n for the correct distance. For example project line segment AP onto n for vector s, and then our solution is the magnitude of s. As a tip, I would recommend you normalize n into n before projecting onto it, because projecting onto a normalized vector simplifies the formula considerably.

Here are some resources on this approach: https://www.youtube.com/watch?v=6vpXM1dskmY&ab_channel=AnviDalal
--------------------------------
Now I'd like to share with you a much more complicated alternative approach. This approach may be more complicated but it avoids the expensive cross product calculation so it is much faster. Additionally this approach works in 2D while the first approach does not. Even if you prefer the earlier approach I think it's important to understand this approach because we will use these strategies for more complex problems coming up in later chapters. This approach relies on a math proof to determine the solution points on AB and PQ that are closest together, then you can determine the magnitude of the vector between these solution points. 

This solution comes from a book called Geometry Algorithms that I can sadly no longer find online but I do have screenshots of the relevant pages if you would like to read the original presentation which spans 3 pages. Please mind that the variables used in the book are totally different.
- Page 79: https://i.gyazo.com/75ff6e449e26293fc6f8500c75ddfd1c.png
- Page 80: https://i.gyazo.com/903a29b06993097a37029aefe0819ac7.png
- Page 81: https://i.gyazo.com/eab91dccfd4befd927c534779cd6b97a.png
- My explanation of this approach: https://i.gyazo.com/0244400b70f404faeb98f99c034dab85.png

Let us refer to the closest point on AB to PQ as C; and the closest point on PQ to AB as R. We know that the line segment RC that connects these solution points will be perpendicular to both AB and PQ therefore the dot product of PQ with both of these line segments is zero. In the first three lines of our proof we express that fact and then, since both equal zero, we set the expressions equal to each other.
AB ⊥ RC  
PQ ⊥ RC  
AB • RC = 0
PQ • RC =  0
AB • RC =  PQ • RC
Here is a diagram for your reference: https://i.gyazo.com/5c5229802d674511cf300c0703b4a2cc.png

Now the next step of this approach is to write the line segments in terms of their parametric equations. This requires us to have a term for the direction vector of each line segment. I will call the vector from A towards B "a", the vector from P towards Q "p", and lastly the vector from R to C "r". Since we know that C lies somewhere on AB, we can express C as "A plus some amount of a". To express "some amount" we introduce a new scalar variable. In a parametric equation, you use a parametric scalar to represent this "some amount" concept. For AB this parameter can be d. And for PQ this parameter can be s. By the way, I am deliberately trying to use letters for each line segment that are groups on different sides of the alphabet to hopefully make this easier for you to differentiate. The final parametric form for expressing C and R are as follows:
C = A + a * d
R = P + p * s

Now we can return to our earlier dot product expressions "AB • RC = 0" and expand RC into the expression C-R. Then we substitute our new parametric equations in for both C and R.
AB • RC = 0
AB • [(C ) - (R)] = 0
AB • [(A + a * d) - (P + p * s)] = 0

We can do the same replacements for our dot product expression "PQ • RC =  0".
PQ • RC = 0
PQ • [(C ) - (R)] = 0
PQ • [(A + a * d) - (P + p * s)] = 0

Using the two equations we derived, we can solve for the parameters s and d in terms of the a, p, and r.
The full derivation is a few pages long but you can see it in full here: 
​https://www.youtube.com/watch?v=ELQG5OvmAE8&ab_channel=EgoMoose

The resulting equations are...
d = [(a • p)(p • r) - (p • p)(a • r)] / [(a • a)(p • p) - (a • p)(a • p)]
s = [(a • a)(p • r) - (a • p)(a • r)] / [(a • a)(p • p) - (a • p)(a • p)]

You may be concerned about the denominator term "[(a • a)(p • p) - (a • p)(a • p)]" being zero. If it is zero, a divide-by-zero could crash our program. But this is only true when AB and PQ are parallel so if we check that first then we are not at risk.

We can now use these expressions to find C and R, then determine the distance between them for our final answer.
Practice Question: ​You are developing an AI for a video game. The AI will shoot a projectile at the player but since the projectile is slow we need to design that AI to lead it's target. For example, if you are travelling towards the right, we need to calculate how far off to your right we need to fire the projectile so that it will hit you by the time it reaches you. At least, this is one setup for the following iconic problem.
Given an entity (for example a Player) at position P with velocity vector p; and another entity (for example a Bullet) at position B with velocity vector b, determine if these entities will collide. If they will collide, then determine when. At this point, the problem is the same as the previous problem, basically the intersection point of two lines. But since we are discussing collision here, each of our entities are going to have a collision sphere radii. The first entity (the Player) has radius p. The second entity (the bullet) has radius b. So now you have six variables: p, P, p, B, b, and b. I personally find it easier to keep using the same letters so I can easily associate which belong to each entity but of course you can use your own variable symbols. I can't type subscripts in this format but ideally p would be r-subscript-p. 

Answer:
We can imagine two rays, one from P in direction p and another from B in direction b. We will have an intersection at points where the distance between the two rays is equal to the sum of the entities radii. These points are probably not the points closest to each other on each ray. If that is confusing, imagine two long train cars colliding at an intersection: we need to find out when the heads of the trains collide, not when their centers of mass will collide (because the heads of the trains will meet before their centers of mass meet). So in pursuit of these points on each ray, let's define each ray in parametric form then see if we can solve the equation. Here are the parametric equations for an arbitrary point "P2" on the first ray,  and an arbitrary point "B2" on the second ray. 
P2 = P + p * t
B2 = B + b * t

You can see that I have introduced a scalar parameter "t" as well, but interestingly it is the same "t" in both equations because they are moving at the same time. The distance between these two points can be expressed as follows:
|P2 - B2| = distance between points

We want the distance when it is equal to the sum of the radii, in other words we need to solve:
p + b = |P2 - B2|

Let's expand out the right hand side's vector magnitude formula:
p + b = sqrt( [P2.x - B2.x]^2 + [P2.y - B2.y]^2 )

You'll notice that I am solving this for a 2D case, and I have not included the ".z" values but the formula will result in the same expression for 3D cases. Our next step is to square both sides which will remove the sqrt.
(p + b)^2 = [P2.x - B2.x]^2 + [P2.y - B2.y]^2

I'm going to call our equation above the "pre-substitution equation" because we are going to take a little break from this equation and substitute the base terms of each exponent expression. If you're interested in why we do a substitution here, it's so that we can convert the pre-substitution equation into a quadradic form which we can solve. But I am getting ahead of myself, first let's just perform our substitution.  We will convert "P2.x - B2.x" into "x1 + t * x2". For each of these steps, we are basically only considering the x aspect of the points and vectors which I denote with ".x".
P2.x - B2.x 
( P2.x ) - ( B2.x )
( P.x + p.x * t ) - ( B.x + b.x * t )
( P.x + p.x * t )  - B.x - b.x * t )
( P.x - B.x ) + ( p.x * t  - b.x * t )
( P.x - B.x )  + t ( p.x - b.x )

Now we can replace some of these terms because they are known. For example we know the value of P and B so we can substitute the term ( P.x - B.x ) with whatever it's actual value is, we will call it x1. The same is true of ( p.x - b.x ) for which I will use x2. I will use the three bar equals sign ( ≡ ) which means "is defined as".
( P.x - B.x ) ≡ x1
( p.x - b.x ) ≡ x2

We can apply the same steps to simplify the "(P2.y - B2.y)" term of our pre-substitution equation.
( P.y - B.y ) ≡ y1
( p.y - b.y ) ≡ y2

Now, for the moment we have all been waiting for, we plug in our substitutions into the pre-substitution equation.
(p + b)^2 = [P2.x - B2.x]^2 + [P2.y - B2.y]^2
(p + b)^2 = [( P.x - B.x )  + t ( p.x - b.x )]^2 + [( P.y - B.y )  + t ( p.y - b.y )]^2
(p + b)^2 = [ x1 + t * x2 ]^2 + [ y1  + t * y2]^2

Check it out, our right hand side is now completely in the form of an expression using x1, x2, y1, y2 and t. Only t is a variable here. All the other terms are constants which we know. We can expand out the right hand side to convert our expression into a quadratic form.
(p + b)^2 = [ (x2)^2  + (x2)^2 ] * t^2     +      2t * [ x1 * x2 + y1 * y2]     +     x1^2 + y1^2

We are almost in the form of a quadratic equation, we just need to move the left hand side over so that we have an expression equal to zero. Remember that p and b are also constants so we only have one variable in this entire equation and it's the parametric scalar t which has been carried along from the very start.

0 =  t^2 * [ (x2)^2  + (x2)^2 ]     +      t * 2 * [ x1 * x2 + y1 * y2]     +     x1^2 + y1^2 - (p + b)^2
                  \_____________/                    \_______________/              \________________/
                             A                                                    B                                           C

Okay so hopefully the form above looks familiar.  It's a quadratic equation where t is our only variable. We can now plug the terms into a quadratic formula for our solutions. That is a bit too much work for me to type out here but it shouldn't be too difficult to follow that part. I want to note that the quadratic equation gives two results. As explained in the link below, these two results are the two times that the entities are the desired distance away from each other. One time is before the centers of mass collide and one time is after. For this problem we want the smaller t value.

Here is a pseudo code implementation of stack overflow:
https://stackoverflow.com/questions/33140999/at-what-delta-time-will-two-objects-collide​​

And now... for the grand finale vector maths question....
Practice Question: ​You recently noticed an issue on a 3D Spider-Man game you are working on. You push the move joystick forward (shown by green arrow below) but Spider-Man moves in the red direction. It seems that this forward input is moving him with resect to his current forward axis. You want him to respect the player input regardless of which way he is currently facing. Given the green directional input, instead of going in the red direction he should always go in the purple direction regardless of which way he is currently facing.
You are given the following vectors as input:
  • j, the vector of joystick input. It's a 2D vector and both component have values ranging from -1 to 1. Indicating how much the joystick is pushed forward, back, left, and right. The first component has the x data (left vs right) and the second component has the y data (up vs down).
  • c, the forward vector of the camera. It's a 3D vector.
  • u, Spider-Man's up vector. It's a 3D Vector. This vector always points up and implicitly defines his walking plane.
Your output should be a 3D vector m which represents the direction Spider-Man should move given the input parameters. Your answer should be in the format of pseudo code.

Answer: 
  • Normalize the joystick input as a vector with components joyX and joyY
  • Project camera forward onto the avatar’s walking plane, then normalize the resulting vector, call it F
  • Now you can multiply joyY * F to get the forward vector
  • We can find R by doing Cross(F, walkingPlanesNormal), and then calculate joyX * R for the sideways vector​