title: Advanced Mathematics
subtitle: DLMDSAM01
Author: Dr. Robert Graf
publisher: IU International University of Applied Sciences
year: 2023

Unit 3 Vector Algebra

Think of a vector like one piece of information, but broken down into its most fundamental components. Focusing on 2 and 3 dimensional space, R2\mathbb{R}^2 and R3\mathbb{R}^3, we will probably look at the Cartesian plane and space. That is, with typical xx, yy, and zz axes.

Applications of vector operations lends to information science, machine learning, and the wider computer science.

Suggested reads are:

3.1 Scalars and Vectors

A vector is an element of a mathematical structure called a vector space. It can be thought of as a mathematical object with magnitude and direction. However, for machine learning and information science, vectors might not have anything to do with physical space.

We analyse vectors by breaking them into their component parts.

2-Dimensional Vectors

Let’s consider initial point P=(p1,p2)P=(p_1, p_2) in the x,yx,y coordinate plane. It has a terminal point Q=(q1,q2)Q=(q_1, q_2). The vector representing travel from PP to QQ is PQ→\overrightarrow{PQ}. The magnitude is the length of the line segment, which we obtain from the distance formula. In our case:

∣PQ→∣=(q1−p1)2+(q2−p2)2|\overrightarrow{PQ}|=\sqrt{ (q_1-p_1)^2+(q_2-p_2)^2 }

A vector is standard position has an initial point at the origin, (0,0)(0,0), and is called a position vector. Additional notation is described for nn-dimensions.

Fundamental Definitions:

Trigonometry can play a big role in separating vectors into their component values in physical space.

Suppose we have a golfer who just hit a golf ball. We are going to ignore most of the other forces and things in this space as it is a quick and general demo of vector analysis. The ball is hit with 100 Newtons of force at 50o∡50^o\measuredangle. Can we separate into x,yx,y coordinates?

The book as a different example. Sine will give you the y value, and cosine should be x. Since the angle is greater than 45o45^o, the yy values should be greater than xx. (64.3,76.6)(64.3 ,76.6).

I didn’t believe it at first but

64.32+76.62≈100264.3^2+76.6^2 \approx 100^2

Similarly, if you are given starting and ending points, you can use inverse (or arc) trig to get the angle.

3.2 Addition and Subtraction of Vectors

Properties of Addition and Subtraction of Vectors

A mathematical field is a set of which addition, multiplication, subtraction, and division are defined and behave the same way as for rational and real numbers. What do we mean by a set? Yes, like the set of real numbers R\mathbb{R}, and/or complex numbers C\mathbb{C}.

Addition and subtractions of vectors works by adding and subtracting their components. In Physics, it works the same when adding forces together if they collide.

Letv⃗=(v1,...,vn)Letu⃗=(u1,...,un)Thenv⃗+u⃗=(v1+u1,...,vn+un)Andv⃗−u⃗=(v1−u1,...,vn−un)\begin{align*} \text{Let} \quad \vec{v} &= (v_1, ..., v_n)\\ \text{Let} \quad \vec{u} &= (u_1, ..., u_n)\\ \text{Then} \quad \vec{v} + \vec{u} &= (v_1 + u_1, ..., v_n + u_n)\\ \text{And} \quad \vec{v} - \vec{u} &= (v_1 - u_1, ..., v_n - u_n) \end{align*}

Components of the vectors are elements of the field of real numbers, or complex numbers, addition is commutative (ie a+b=b+aa+b=b+a).

Vector addition is also associative, there exists an additive identity 0, and every element vv has an additive inverse −v⃗-\vec{v}. The book as a wonderful table of properties on pp. 71-72.

Vectors in Two Dimensions

By R2\mathbb{R}^2, we mean the two-dimensional space of real numbers. The book also goes over a graphical way of adding and subtracting to make sense. Draw one vector, and then at the tip of that (as the origin), draw the next to add them. It’s the same as component-wise addition.

Introduction to Bases

A Spanning Set is a set that allows us to express every vector as a linear combination of elements in said set. What do we mean by linear combination?

A linear combination of elements is a sum of those elements. They may, or may not, be multiplied by a scalar.

A basis for a vector space is the minimal spanning set of vectors in that vector space. By minimal, we mean that if any vectors were removed, the set would no longer span the space. Some vector(s) would not be able to be formed as a linear combination of the remaining elements in the set.

Consider the following… a two-dimensional space can be spanned by vectors i⃗\vec{i} and j⃗\vec{j}, pointing along the xx and yy axes. Any element in this 2-D space (or plane) can be expressed by a linear combination (adding and possible scalar multiplication) of these two vectors. For example, the vector v⃗=2i⃗+3j⃗\vec{v}=2\vec{i} + 3\vec{j}. However, if we removed a vector from the space, say i⃗\vec{i}, we no longer span the space and could not create v⃗\vec{v} as a linear combination of elements in this set.

The above is more of an abstract understanding to assist in problem solving. Bases are fundamental to the study of vector spaces. We typically express the spanning set of vectors with unit vectors, discussed more below.

Moving from one basis to another is an important question.

Unit Vectors

A unit vector is a vector with magnitude of one. Each unit vector should point in a particular direction. To find a unit vector u⃗\vec{u},

u⃗=v⃗∣v⃗∣=1∣v⃗∣v⃗\vec{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{|\vec{v}|}\vec{v}

Recall that ∣v⃗∣|\vec{v}| is the magnitude. Standard unit vectors are unit vectors that run parallel to the xx and yy axes. That is;

i⃗=(1,0);j⃗=(0,1)\vec{i} = (1,0); \quad \vec{j} = (0,1)

Any vector in 2-D space can be written as a linear combination of these two vectors. We can write a vector a⃗=(3,−4)\vec{a} = (3,-4) as the following linear combination:

a⃗=3i⃗−4j⃗\vec{a} = 3\vec{i} - 4\vec{j}

Another basis is of polar coordinates, with distance rr from the origin and angle θ\theta with respect to the positive xx axis. Interesting to note, any two vectors that are not scalar multiple of one another will form a basis in 2-D space. This is guaranteed in 2-D but not in higher dimensions. Food for thought.

Vectors in Three-Dimensional Space

Moving right along, we can express unit vectors in 3-dimensional space like, the set is {i⃗, j⃗, k⃗}\{\vec{i},\, \vec{j},\, \vec{k}\}.

i⃗=(1,0,0);  j⃗=(0,1,0);  k⃗=(0,0,1);\vec{i}=(1,0,0);\;\vec{j}=(0,1,0);\;\vec{k}=(0,0,1);

They of course are directed along the axes. The magnitude can be found as ∣r⃗∣=x2+y2+z2|\vec{r}|=\sqrt{x^2+y^2+z^2}.

Collinear Vectors

Two vectors, a⃗ and b⃗\vec{a} \text{ and } \vec{b}, are collinear (generalization of parallel) if there exists a real valued constant cc such that a⃗=cb⃗\vec{a} = c\vec{b}. The notation is a⃗∥b⃗\vec{a}\Vert \vec{b}. As you can see, the point in the same direction, but just have different magnitudes. Like tires on a car, or rockets and a spaceship.

3.3 Multiplication of Vectors: Dot Product and Scalar Product

Multiplication of vectors is not the same as with normal numbers.

Scalar or Dot Product

The dot product (AKA scalar product, but no one calls it this) produces a scalar output. Here it is… Let a⃗=(a1,...,an)\vec{a} = (a_1, ..., a_n) and b⃗=(b1,...,bn)\vec{b} = (b_1, ..., b_n). The dot product is:

a⃗⋅b⃗=a1b1+a2b2+...+anbn=∑i=1nai+bi=∑i=0n−1ai+bi\begin{align*} \vec{a} \cdot \vec{b} &= a_1b_1 + a_2b_2 + ... + a_nb_n\\ &= \sum_{i=1}^n a_i+b_i\\ &= \sum_{i=0}^{n-1} a_i+b_i \end{align*}

. The last two parts are just additional notation, the latter aimed at programming with a 0 index. They solidify the fact the dot product returns a scalar value.

Here are some properties. We say that for all real vectors u⃗  ,  v⃗  and  w⃗  in  Rn\vec{u}\;,\;\vec{v}\;\text{and}\;\vec{w}\;\text{in}\;\mathbb{R}^n, the following properties hold:

u⃗⋅v⃗=v⃗⋅u⃗u⃗⋅(v⃗+w⃗)=u⃗⋅v⃗+u⃗⋅u⃗u⃗⋅u⃗=∣u⃗∣2u⃗⋅(cv⃗)=c(u⃗⋅v⃗)=c(u⃗)⋅v⃗\begin{align} \vec{u}\cdot\vec{v} &= \vec{v}\cdot\vec{u}\\ \vec{u}\cdot(\vec{v}+\vec{w}) &= \vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{u}\\ \vec{u}\cdot\vec{u} &= |\vec{u}|^2\\ \vec{u}\cdot(c\vec{v}) &= c(\vec{u}\cdot\vec{v})= c(\vec{u})\cdot\vec{v}\\ \end{align}

The dot product can be extended to handle vectors in Cn\mathbb{C}^n with complex numbers as well. We let u⃗∗\vec{u}^* indicate the complex conjugate. It has the same properties except that the squared magnitude ends up being real.

Angle between two vectors

In 2-D and 3-D space, the dot product can be interpreted geometrically as the angle between 2 vectors.

In two dimensional space, R2\mathbb{R}^2, let θ\theta denote the angle between vectors a⃗=(a1,a2)\vec{a}=(a_1, a_2) and b⃗\vec{b}. Then, the cosine of the angle between the vectors is given by:

cos⁡(θ)=a⃗⋅b⃗∣a⃗∣∣b⃗∣=a1b1+a2b2∣a⃗∣∣b⃗∣\begin{align*} \cos(\theta) &= \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\\ &= \frac{a_1b_1+a_2b_2}{|\vec{a}||\vec{b}|}\\ \end{align*}

equation 3.1

Without proof, this also works in 3-dimensions. It implies:

θ=arccos⁡(a⃗⋅b⃗∣a⃗∣∣b⃗∣)\theta = \arccos \left( \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} \right)

There’s a proof on pp. 81-82. The cool think apparently is that we now also have an easy test for orthogonality in the case that u⃗⋅v⃗=0\vec{u} \cdot \vec{v} =0.

Vector Projection

This is like breaking forces into their spacial coordinates. You are projecting a vector onto others, I think like the bases. The books suggests thinking of it like a shadow. The length of the shadow is scalar projection. The length and direction of the shadow is vector projection.

The projection of a⃗\vec{a} onto b⃗\vec{b} is:

Projb⃗(a⃗)=b⃗⋅a⃗∣b⃗∣2b⃗\text{Proj}_{\vec{b}}(\vec{a}) = \frac{\vec{b}\cdot\vec{a}}{|\vec{b}|^2} \vec{b}

The book on pp. 83-84 has a little proof based on logical thinking.

Inner Product

The inner product is a generalization of the dot product, and assigns a real, or complex, number to two vectors. We generally assume we will be working with complex numbers when using the inner product though.

Definition - Inner Product: The inner product between vectors a⃗\vec{a} and b⃗\vec{b} is written as

⟨a⃗∣b⃗⟩OR⟨a⃗, b⃗⟩\begin{gather*} \langle{ \vec{a} | \vec{b} }\rangle\\ \text{OR}\\ \langle{ \vec{a},\, \vec{b} }\rangle\\ \end{gather*}

The following properties hold:

⟨u⃗∣v⃗⟩=⟨v⃗∣u⃗⟩∗⟨u⃗∣(cv⃗+dw⃗)⟩=c⟨u⃗∣v⃗⟩+d⟨u⃗∣w⃗⟩⟨cu⃗+dv⃗∣w⃗⟩=c∗⟨u⃗∣w⃗⟩+d∗⟨v⃗∣w⃗⟩⟨cu⃗∣dv⃗⟩=c∗d⟨u⃗∣v⃗⟩\begin{align*} \langle \vec{u} | \vec{v} \rangle &= \langle \vec{v} | \vec{u} \rangle^*\\ \langle \vec{u} | (c\vec{v} +d\vec{w}) \rangle &= c\langle \vec{u} | \vec{v} \rangle + d\langle \vec{u} | \vec{w} \rangle\\ \langle c\vec{u} + d\vec{v} | \vec{w} \rangle &= c^*\langle \vec{u} | \vec{w} \rangle + d^*\langle \vec{v} | \vec{w} \rangle\\ \langle c \vec{u} | d \vec{v} \rangle &= c^*d\langle \vec{u} | \vec{v} \rangle \end{align*}

And for orthogonal vectors, ⟨u⃗∣v⃗⟩=0\langle \vec{u} | \vec{v} \rangle = 0. It’s very interesting to see that there’s a difference in results based on the order things happen within the inner product. It is more rigid than a regular dot product.

I believe it to be the same operation. The norm of a vector is ∥u⃗∥=⟨u⃗∣u⃗⟩\|\vec{u}\|=\sqrt{\langle \vec{u} | \vec{u} \rangle}.

Cross Product

The cross product is is perpendicular to both vectors and normal to the plane that contains the vectors. Cross product is exclusive to vectors. There’s also the linear algebra concept of matrix multiplication which is different.

In three-dimensional space, let a⃗=a1i⃗+a2j⃗+a3k⃗\vec{a} = a_1\vec{i} + a_2 \vec{j} + a_3 \vec{k} and b⃗=b1i⃗+b2j⃗+b3k⃗\vec{b} = b_1\vec{i} + b_2 \vec{j} + b_3 \vec{k}. The cross product is perpendicular to both given vectors and defined as:

a⃗×b⃗=(a2b3−b2a3)i⃗−(a1b3−b1a3)j⃗+(a1b2−b1a2)k⃗\vec{a}\times \vec{b} = (a_2b_3-b_2a_3)\vec{i} - (a_1b_3-b_1a_3) \vec{j} + (a_1b_2-b_1a_2)\vec{k}

Each component is made up exclusively of only the other components. Additionally, any component crossing itself is 0 since something cannot be perpendicular to itself, in a manner of speaking.

Algebraic Properties of the Cross Product

For vectors u⃗\vec{u} and v⃗\vec{v} in R3\mathbb{R}^3, the following properties hold:

u⃗×v⃗=−(v⃗×u⃗)u⃗×(v⃗+w⃗)=(u⃗×v⃗)+(u⃗×w⃗)u⃗×u⃗=0⃗u⃗⋅(v⃗×w⃗)=(u⃗×v⃗)⋅w⃗\begin{gather*} \vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})\\ \vec{u} \times (\vec{v}+\vec{w}) = (\vec{u} \times \vec{v})+(\vec{u} \times \vec{w})\\ \vec{u} \times \vec{u} = \vec{0}\\ \vec{u} \cdot(\vec{v} \times \vec{w}) = (\vec{u} \times \vec{v}) \cdot \vec{w} \end{gather*}

Note that the cross product is not associative, that is a(b×c)≠(a×b)ca(b\times c) \neq (a \times b)c.

Geometric Properties of Cross Product

I know, it doesn’t end…

  1. If the vectors are scalar multiples of each other, then the cross product is 0. Again, if they are along the same access, you can’t get perpendicular.
  2. The cross product is orthogonal to both vectors. Think of the vectors forming a plane and the cross product shooting out of it into the 3rd dimension.
  3. ∣u⃗×v⃗∣=∣u⃗∣∣v⃗∣sin⁡(θ)|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin(\theta) is the magnitude. It’s the area of the parallelogram formed by the vector with adjacent sides.

□\Box.

title: Single and Multivariable Calculus
subtitle: Early Transcendentals
Author: David Guichard
publisher: ?
year: 2022-12-01

Single and Multivariable Calculus

Ch. 12 Three Dimensions

12.1 The Coordinate System

p. 297

Explaining Cartesian coordinates z=f(x,y)z=f(x,y).

Distance between two points can be derived from the 2-D Pythagorean case. Let point 1 be P1(x1,y1.z1)P_1(x_1, y_1. z_1), and point 2 be P2(x2,y2.z2)P_2(x_2, y_2. z_2). We then have the distance between points written as:

c=d(P1, P2)=(x1−x2)2+(y1−y2)2+(z1−z2)2c = d(P_1,\,P_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

Funny enough, the distance formula in 2 dimensions is the equation of a circle. In three dimensions, this is the equation for a sphere, or:

r2=(x−h)2+(y−k)2+(z−l)2r^2 = (x-h)^2+(y-k)^2+(z-l)^2

Where h,k,lh,k,l represents the centre of the sphere.

Choose some points like (2,2,3)(2,2,3), (8,6,5)(8,6,5) and (−1,0,2)(-1,0,2) and find the distance between them. Would they form a triangle?

12.2 Vectors

p. 300

A vector is “a quantity consisting of a non-negative magnitude and a direction.” One representation can be with polar coordinates, (m, θ)(m,\,\theta).

A displacement vector represents a distance travelled. You can also have a speed vector, or just velocity, indicating rate of movement in a particular direction. You can continue this trend for acceleration and force as well.

Definition - Scalar Multiplication: The scalar multiplication of vector A⃗=(m, θ)\vec{A}=(m,\,\theta) and scalar a≥0  ∀  a∈Ra \geq 0\; \forall \; a \in \mathbb{R} is:

aA⃗=(am, θ)a\vec{A} = (am,\,\theta)

Notice it points in the same direction, but has a different magnitude. The scalar can be a<0a \lt 0. The magnitude is always non-negative, but the direction will be inverted.

for  a<0aA⃗=(∣a∣m, θ+π)\text{for} \;a \lt 0 \quad a\vec{A} = (|a|m,\,\theta+\pi)

However, typically polar coordinates are not used. In general:

⟨v1⃗, w1⃗⟩+⟨v2⃗, w2⃗⟩=⟨v1⃗+v2⃗,  w1⃗+w2⃗⟩\langle \vec{v_1},\,\vec{w_1} \rangle + \langle \vec{v_2},\,\vec{w_2} \rangle = \langle \vec{v_1}+\vec{v_2},\;\vec{w_1}+\vec{w_2} \rangle

This is the sum of vectors. It just means that we sum the vector components that are pointing along the same axis, or in the same direction. Scalar multiplication is just multiplying through the vector. And the magnitude of the vector is:

∣⟨v,w⟩∣=v2+w2|\langle v,w \rangle| = \sqrt{v^2 + w^2}

Same concepts extend into three dimensions.

Unit vectors have magnitude of one and mainly denote a direction. In 3 dimensions we use i=(1,0,0), j=(0,1,0), k=(0,0,1)i = (1,0,0),\, j=(0,1,0),\,k=(0,0,1).

Theorem 12.2.2 on pp 304-305 lists properties already outlined in the course book. However, we are also provided a proof.

The Dot Product

p. 306

The question is that if two vectors begin at the same point, what is the angle between them? We can use trigonometry and the law of cosines here.

Let A⃗=(a1,a2,a3)\vec{A} = (a_1, a_2, a_3) and B⃗=(b1,b2,b3)\vec{B} = (b_1, b_2, b_3). Then,

∣A−B∣2=∣A∣2+∣B∣2−2∣A∣∣B∣cos⁡(θ)2∣A∣∣B∣cos⁡(θ)=∣A∣2+∣B∣2−∣A−B∣2=a12+a22+a32+b12+b22+b32−(a1−b1)2−(a2−b2)2−(a3−b3)2=a12+a22+a32+b12+b22+b32−(a12−2a1b1+b12)−(a22−2a2b2+b22)−(a32−2a3b3+b32)=2a1b1+2a2b2+2a3b3cos⁡(θ)=2a1b1+2a2b2+2a3b32∣A∣∣B∣cos⁡(θ)=a1b1+a2b2+a3b3∣A∣∣B∣cos⁡(θ)=A⋅B∣A∣∣B∣\begin{align*} |A - B|^2 &= |A|^2+|B|^2-2|A||B|\cos(\theta)\\ 2|A||B|\cos(\theta) &= |A|^2+|B|^2-|A - B|^2\\\\ &= a_1^2 + a_2^2 + a_3^2+b_1^2 + b_2^2 + b_3^2 - \\&\quad(a_1-b_1)^2-(a_2-b_2)^2-(a_3-b_3)^2\\\\ &= a_1^2 + a_2^2 + a_3^2+b_1^2 + b_2^2 + b_3^2 - \\&\quad (a_1^2-2a_1b_1+b_1^2)-(a_2^2-2a_2b_2+b_2^2)-(a_3^2-2a_3b_3+b_3^2)\\\\ &= 2a_1b_1+2a_2b_2+2a_3b_3 \\ \cos(\theta) &= \frac{2a_1b_1+2a_2b_2+2a_3b_3}{2|A||B|}\\ \cos(\theta) &= \frac{a_1b_1+a_2b_2+a_3b_3}{|A||B|}\\ \cos(\theta) &= \frac{A\cdot B}{|A||B|}\\ \end{align*}

The Law of Cosines is very much like Pythagorean Theorem. My confusion is around this 22 that goes missing. When you expand the vectors out, amazingly all the elements end up with a factor of 2. So it divides out. I hope I was able to make that painfully clear.

There’s also this great Dot product - Wikipedia article. Basically, we named the numerator the dot product.

Examples on p. 307

The projection of one vector onto another is to determine how much of a certain vector is pointing in the same direction as another. If VV is the vector that represents the projection of AA onto BB, the scalar projection of A onto B is:

∣V∣=∣A∣cos⁡(θ)=∣A∣A⋅B∣A∣∣B∣=A⋅B∣B∣\begin{align*} |V| &= |A|\cos(\theta)\\ &= |A| \frac {A \cdot B} {|A||B|}\\ &= \frac {A \cdot B} {|B|}\\ \end{align*}

But to get the vector of length VV, we need to give the magnitude a direction. That direction is BB. Get the unit vector in the BB direction with b⃗=B/∣B∣\vec{b} = B/|B|, and therefore:

∣V∣=∣A∣cos⁡(θ)=∣A∣B∣B∣A⋅B∣A∣∣B∣=BA⋅B∣B∣2\begin{align*} |V| &= |A|\cos(\theta)\\ &= |A| \frac{B}{|B|} \frac {A \cdot B} {|A||B|}\\ &= B\frac {A \cdot B} {|B|^2}\\ \end{align*}

It is assumed 0≤θ≤π/20 \leq \theta \leq \pi / 2. If π/2≤θ≤π\pi / 2 \leq \theta \leq \pi then the vector VV, projected from AA onto BB is anti-parallel to BB, meaning it’s magnitude points in the opposite direction. In this case the dot product A⋅BA \cdot B is negative.

Don’t think of it as projection. Think of it like, vector BB merely provides the direction and we want to know how much of AA points in that direction.

Going to list the properties now… but they are the same already listed from the course book.

pp. 311-312 has many problem to go over for the brave.

12.4 The Cross Product

p. 312

The cross product is a non-zero vector perpendicular to vectors AA and BB. Because there are infinite vector lengths perpendicular, we define the cross product vector, with a specific magnitude… in just a minute.

Apparently, it is useful to know what the determinant is. It’s a linear algebra concept …

If AA is a 2×22\times 2 matrix like (abcd)\big(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\big), the determinant of AA is:

det⁡(A)=det⁡[abcd]=∣abcd∣=ad−bc\det(A) = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

That turns a matrix into a single value. Well, it goes into several other concepts that I will save for the linear algebra section, but includes a determinant of a 3×33 \times 3 matrix and eventually becomes:

∣ijka1a2a3b1b2b3∣=(a2b3−a3−b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k=(a2b3−a3−b2)i+(a3b1−a1b3)j+(a1b2−a2b1)k=⟨a2b3−a3−b2,  a3b1−a1b3,  a1b2−a2b1⟩=A×B\begin{align*} \begin{vmatrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} &= (a_2b_3 - a_3-b_2)i - (a_1b_3-a_3b_1)j + (a_1b_2-a_2b_1)k \\ &= (a_2b_3 - a_3-b_2)i + (a_3b_1-a_1b_3)j + (a_1b_2-a_2b_1)k \\ &= \langle a_2b_3 - a_3-b_2,\;a_3b_1-a_1b_3,\;a_1b_2-a_2b_1 \rangle \\ &= A \times B \end{align*}

So the cross product is the determinant of a 3×33 \times 3 matrix. The cross product has nice properties like:

∣A×B∣2=(a2b3−a3b1)2+(a3b1−a1b3)2+(a1b2−a2b1)2=+  A lot of factoring=(a12+a22+a32)(b12+b22+b32)−(a1b1+a2b2+a3b3)2=∣A∣2∣B∣2−(A⋅B)2=∣A∣2∣B∣2−∣A∣2∣B∣2cos⁡2(θ)=∣A∣2∣B∣2(1−cos⁡2(θ))=∣A∣2∣B∣2sin⁡2(θ)∴∣A×B∣=∣A∣∣B∣sin⁡(θ)\begin{align*} |A \times B|^2 &= (a_2b_3-a_3b_1)^2 + (a_3b_1-a_1b_3)^2 + (a_1b_2-a_2b_1)^2 \\ &= +\;\text{A lot of factoring}\\ &= (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) - (a_1b_1 + a_2b_2 + a_3b_3)^2 \\ &= |A|^2|B|^2-(A \cdot B)^2 \\ &= |A|^2|B|^2-|A|^2|B|^2 \cos^2(\theta)\\ &= |A|^2|B|^2(1-\cos^2(\theta))\\ &= |A|^2|B|^2\sin^2(\theta)\\ &\therefore \\ |A \times B| &= |A||B|\sin(\theta) \end{align*}

Because we have a factor of sine, if the angle between the vectors is 0, then the magnitude of the cross product is 0.

Additionally, if we take two vectors and put them “tail to tail” so that they cast parallel vectors to form a parallelogram but only for visual purposes, the area of the parallelogram is height∗width=∣A∣sin⁡(θ)∗∣B∣=∣A×B∣\text{height} * \text{width} = |A|\sin(\theta) * |B|=|A \times B|, the magnitude of the cross product.

The direction of the cross product is determined by a simple rule called the right hand rule. The right hand rule basically means if you rotate the projected vector AA counter-clockwise to project onto BB, then curling the fingers into the palm of your right hand, you should note your thumb points up, which is the direction of the vector.

This means, because of direction, A×B≠B×AA \times B \neq B \times A.

The course book didn’t capture all of the properties.

Theorem 12.4.2: if u⃗\vec{u}, v⃗\vec{v}, and w⃗\vec{w} are all vectors and cc is just a real number, then the following:

u⃗×(v⃗+w⃗)=u⃗×v⃗+u⃗×w⃗(v⃗+w⃗)×u⃗=v⃗×u⃗+w⃗×u⃗(cu⃗)×v⃗=c(u⃗×v⃗)=u⃗×(cv⃗)(u⃗⋅(v⃗×w⃗))=(u⃗×v⃗)⋅w⃗u⃗×(v⃗×w⃗)=(u⃗⋅w⃗)v⃗−(u⃗⋅v⃗)w⃗\begin{align*} \vec{u} \times (\vec{v}+\vec{w}) &= \vec{u} \times \vec{v}+\vec{u} \times \vec{w}\\ (\vec{v}+\vec{w}) \times \vec{u} &= \vec{v} \times \vec{u}+\vec{w} \times \vec{u}\\ (c\vec{u}) \times \vec{v} &= c(\vec{u} \times \vec{v}) = \vec{u} \times (c\vec{v})\\ (\vec{u} \cdot (\vec{v} \times \vec{w})) &= (\vec{u} \times \vec{v}) \cdot \vec{w}\\ \vec{u} \times (\vec{v} \times \vec{w}) &= (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \end{align*}

That is it for now. Lots of practice problems on p. 316.

Knowledge Check

What vector is equal to the directed line segment from point (0,0)(0,0) to (−2,−3)(-2,-3)?

The directed line segment from (0,0)(0,0) to (2,3)(2,3) would have equal magnitude but an opposite direction. This implies they are not equal.

The directed line segment from (−2,−3)(-2,-3) to (0,0)(0,0) would also have equal magnitude but an opposite direction. This implies they are not equal.

The directed line segment from (0,0)(0,0) to (−1,−3/2)(-1, -3/2) would also have half the magnitude but points in the same direction. This implies they are not equal.

The directed line segment from (10,10)(10, 10) to (8,7)(8, 7), although at a different position, would have equal magnitude and equal direction. This implies they are equal.

Consider the basis {v⃗1,v⃗2}\{\vec{v}_1, \vec{v}_2\} of R2\mathbb{R}^2.

The following are false:

The following is true:

What is the unit vector for u⃗=(2,4,4)\vec{u} = (2,4,4)?

Remember that unit vector u⃗\vec{u} is:

u⃗=v⃗∣v⃗∣=(2,4,4)22+42+42=(2,4,4)4+16+16=16∗(2,4,4)\begin{align*} \vec{u} &= \frac{\vec{v}} {|\vec{v}|} = \frac{(2,4,4)} {\sqrt{2^2+4^2+4^2}} = \frac{(2,4,4)} {\sqrt{4+16+16}}\\ &= \frac{1}{6} * (2,4,4) \end{align*}

Find the angle between vectors u⃗=(0,5)\vec{u} = (0,5) and v⃗=(3,4)\vec{v} = (3,4).

We know that the angle between vectors is:

cos⁡(θ)=a⃗⋅b⃗∣a⃗∣∣b⃗∣=0∗3+5∗402+52∗32+42=205∗5=45θ=arccos⁡(45)≡cos⁡−1(45)=0.6435rad\begin{align*} \cos(\theta) &= \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{0*3+5*4}{\sqrt{0^2+5^2}*\sqrt{3^2+4^2}} = \frac{20}{5*5} = \frac{4}{5}\\ \theta &= \arccos \left( \frac{4}{5} \right) \equiv \cos^{-1} \left( \frac{4}{5} \right) = 0.6435 rad \end{align*}

arccosine is the inverse of cosine and hence is often also denoted as cos⁡−1\cos^{-1}.

The following statements are considered true: