# migrations

dir: embeddings_queue
filename: 00001-embeddings.sqlite.sql
hash: d3755dfd232be8e8301f4d7fcfb3a486
sql:
CREATE TABLE embeddings_queue (
    seq_id INTEGER PRIMARY KEY,
    created_at TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
    operation INTEGER NOT NULL,
    topic TEXT NOT NULL,
    id TEXT NOT NULL,
    vector BLOB,
    encoding TEXT,
    metadata TEXT
);


dir: sysdb
filename: 00001-collections.sqlite.sql
hash: 38352d725ad1c16074fac420b22b4633
sql:
CREATE TABLE collections (
    id TEXT PRIMARY KEY,
    name TEXT NOT NULL,
    topic TEXT NOT NULL,
    UNIQUE (name)
);

CREATE TABLE collection_metadata (
    collection_id TEXT REFERENCES collections(id) ON DELETE CASCADE,
    key TEXT NOT NULL,
    str_value TEXT,
    int_value INTEGER,
    float_value REAL,
    PRIMARY KEY (collection_id, key)
);


dir: sysdb
filename: 00002-segments.sqlite.sql
hash: 2913cb6a503055a95f625448037e8912
sql:
CREATE TABLE segments (
    id TEXT PRIMARY KEY,
    type TEXT NOT NULL,
    scope TEXT NOT NULL,
    topic TEXT,
    collection TEXT REFERENCES collection(id)
);

CREATE TABLE segment_metadata (
    segment_id TEXT  REFERENCES segments(id) ON DELETE CASCADE,
    key TEXT NOT NULL,
    str_value TEXT,
    int_value INTEGER,
    float_value REAL,
    PRIMARY KEY (segment_id, key)
);


dir: sysdb
filename: 00004-tenants-databases.sqlite.sql
hash: 048867ce8fcdefe4023c7110e4433591
sql:
CREATE TABLE IF NOT EXISTS tenants (
    id TEXT PRIMARY KEY,
    UNIQUE (id)
);

CREATE TABLE IF NOT EXISTS databases (
    id TEXT PRIMARY KEY, -- unique globally
    name TEXT NOT NULL, -- unique per tenant
    tenant_id TEXT NOT NULL REFERENCES tenants(id) ON DELETE CASCADE,
    UNIQUE (tenant_id, name) -- Ensure that a tenant has only one database with a given name
);

CREATE TABLE IF NOT EXISTS collections_tmp (
    id TEXT PRIMARY KEY, -- unique globally
    name TEXT NOT NULL, -- unique per database
    topic TEXT NOT NULL,
    dimension INTEGER,
    database_id TEXT NOT NULL REFERENCES databases(id) ON DELETE CASCADE,
    UNIQUE (name, database_id)
);

-- Create default tenant and database
INSERT OR REPLACE INTO tenants (id) VALUES ('default_tenant'); -- The default tenant id is 'default_tenant' others are UUIDs
INSERT OR REPLACE INTO databases (id, name, tenant_id) VALUES ('00000000-0000-0000-0000-000000000000', 'default_database', 'default_tenant');

INSERT OR REPLACE INTO collections_tmp (id, name, topic, dimension, database_id)
    SELECT id, name, topic, dimension, '00000000-0000-0000-0000-000000000000' FROM collections;
DROP TABLE collections;
ALTER TABLE collections_tmp RENAME TO collections;


dir: metadb
filename: 00001-embedding-metadata.sqlite.sql
hash: 2b4cf52c4bb2676e21d6860a4409f856
sql:
CREATE TABLE embeddings (
    id INTEGER PRIMARY KEY,
    segment_id TEXT NOT NULL,
    embedding_id TEXT NOT NULL,
    seq_id BLOB NOT NULL,
    created_at TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
    UNIQUE (segment_id, embedding_id)
);

CREATE TABLE embedding_metadata (
    id INTEGER REFERENCES embeddings(id),
    key TEXT NOT NULL,
    string_value TEXT,
    int_value INTEGER,
    float_value REAL,
    PRIMARY KEY (id, key)
);

CREATE TABLE max_seq_id (
    segment_id TEXT PRIMARY KEY,
    seq_id BLOB NOT NULL
);

CREATE VIRTUAL TABLE embedding_fulltext USING fts5(id, string_value);


dir: metadb
filename: 00002-embedding-metadata.sqlite.sql
hash: 12a570f7121b3a8ce750a2a7c36da20f
sql:
-- SQLite does not support adding check with alter table, as a result, adding a check
-- involve creating a new table and copying the data over. It is over kill with adding
-- a boolean type column. The application write to the table needs to ensure the data
-- integrity.
ALTER TABLE embedding_metadata ADD COLUMN bool_value INTEGER


dir: metadb
filename: 00003-full-text-tokenize.sqlite.sql
hash: f97ad6334aeaa8f419f01110b648b97a
sql:
CREATE VIRTUAL TABLE embedding_fulltext_search USING fts5(string_value, tokenize='trigram');
INSERT INTO embedding_fulltext_search (rowid, string_value) SELECT rowid, string_value FROM embedding_metadata;
DROP TABLE embedding_fulltext;


# embeddings_queue

created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 7aef73d5-bb96-4595-af57-317ea4a38240
encoding: FLOAT32
metadata:
{"chroma:document": "The objects of study in advanced calculus are differentiable functions of several variables. To set the stage for the study, the Euclidean space as a vector space endowed with the dot product is defined in Section 1.1. To aid visualizing points in the Euclidean space, the notion of a vector is introduced in Section 1.2. In Section 1.3 Euclidean motions, mappings preserving the Euclidean distance, are briefly discussed. The last Section 1.4 contains a discussion on the cross product which is only defined"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: c1aef786-d8d9-4a72-b3e5-621b0306df36
encoding: FLOAT32
metadata:
{"chroma:document": "a discussion on the cross product which is only defined for vectors in the three dimensional Euclidean space, that is, our physical space."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 369082f2-1298-45a0-bb56-bb5fa1881c50
encoding: FLOAT32
metadata:
{"chroma:document": "### 1.1 The Dot Product\n\nAn $n$-tuple is given by\n\n$${\\bf x}=(x_{1},x_{2},\\cdots,x_{n})\\,\\quad x_{j}\\in\\mathbb{R},\\ \\ j=1,\\cdots,n\\.$$\n\nIt is called an ordered pair when $n=2$. Denote by $\\mathbb{R}^{n}$ the collection of all $n$-tuples. The zero $n$-tuple, $(0,0,\\cdots,0)$, will be written as ${\\bf 0}$ from time to time. There are two algebraic operations defined on $\\mathbb{R}^{n}$, namely, the addition\n\n$${\\bf x}+{\\bf y}\\ =(x_{1}+y_{1},x_{2}+y_{2},\\cdots,x_{n}+y_{n})\\,$$\n\nand the scalar multiplication"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: c9c866fe-16e0-42f6-9a04-807de48fc592
encoding: FLOAT32
metadata:
{"chroma:document": "and the scalar multiplication\n\n$$\\alpha{\\bf x}\\ =(\\alpha x_{1},\\alpha x_{2},\\cdots,\\alpha x_{n})\\,\\quad\\alpha \\in\\mathbb{R}\\,$$\n\nwhere\n\n$${\\bf x}=(x_{1},x_{2},\\cdots,x_{n}),\\quad{\\bf y}=(y_{1},y_{2},\\cdots,y_{n})\\.$$\n\nRecall that the ordinary multiplication assigns a number as the product of two numbers, so it can be regarded as a map from $\\mathbb{R}\\times\\mathbb{R}$ to $\\mathbb{R}$. One may expect a multiplication on\n\n[MISSING_PAGE_EMPTY:2]"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: bd079079-bc58-4f49-be2b-04010e7fb122
encoding: FLOAT32
metadata:
{"chroma:document": "[MISSING_PAGE_EMPTY:2]\n\nOne has no difficulty in verifying the dot product satisfies these three axioms. Alternatively one may use $\\langle\\textbf{x},\\textbf{y}\\rangle$ to denote $\\textbf{x}\\cdot\\textbf{y}$. We will do this to avoid confusion occasionally. Note that $\\textbf{x}\\cdot\\textbf{y}=\\textbf{y}\\cdot\\ \\textbf{x}$, so the dot product between **x** and **y** is the same as the dot product between **x** and **y**. Things would be very different when we study the cross product later."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b1d6e750-c992-487c-b693-3be1deed4091
encoding: FLOAT32
metadata:
{"chroma:document": "At this point let us make a digression to establish a fundamental inequality. Some of you may already learn this inequality, but its interesting proof is worth to go through once more.\n\n**Theorem 1.1** (**Cauchy-Schwarz Inequality**).: _For $\\textbf{x},\\textbf{y}\\in\\mathbb{R}^{n}$,_\n\n$$\\left|\\sum_{j=1}^{n}x_{j}y_{j}\\right|\\leq\\sqrt{\\sum_{j=1}^{n}x_{j}^{2}}\\sqrt{ \\sum_{j=1}^{n}y_{j}^{2}}\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 1f88f963-4881-4972-8374-ec2271e24630
encoding: FLOAT32
metadata:
{"chroma:document": "_Furthermore, equality sign holds if and only if either one of $\\textbf{x},\\textbf{y}$ is zero $n$-tuple or there is some $\\alpha\\neq 0$ such that $\\textbf{y}=\\alpha\\textbf{x}.$_\n\nThe condition $\\textbf{x}=\\alpha\\textbf{y}$ means **x** and **y** are proportional to each other. Using the language of linear algebra, the equality condition is simply **x** and **y** are linearly dependent.\n\nProof.: First assume not all $x_{j}$'s are zero in **x**. Consider the expression"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: cbae2be9-5e2c-4a68-937a-f34f059c11ce
encoding: FLOAT32
metadata:
{"chroma:document": "$$\\sum_{j=1}^{n}(x_{j}t-y_{j})^{2}\\,$$\n\nwhich is a sum of squares and so must be non-negative for all $t\\in\\mathbb{R}$. We can express it as a quadratic polynomial in $t$ as\n\n$$p(t)\\equiv at^{2}-2bt+c\\,$$\n\nwhere\n\n$$a=\\sum_{j=1}^{n}x_{j}^{2}\\,\\quad b=\\sum_{j=1}^{n}x_{j}y_{j},\\quad c=\\sum_{j=1} ^{n}y_{j}^{2}\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b7dd15be-7544-412b-9d5c-8c874dbcb240
encoding: FLOAT32
metadata:
{"chroma:document": "Since $a>0$, $p(t)$ tends to $\\infty$ as $t\\rightarrow\\pm\\infty$. Therefore, it is non-negative if and only if its discriminant is non-positive, that is, $4b^{2}-4ac\\leq 0$, which yields $|b|\\leq\\sqrt{ac}$ after taking square root. Our inequality follows. Moreover, the equality sign holds if and only if $4b^{2}-4ac=0$. In this case the quadratic equation $at^{2}-2bt+c=0$ has a (double) root, say, $t_{1}$. Going back to the original expression, we have"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e2e40853-c7fd-45ed-a0cf-8248063e96c9
encoding: FLOAT32
metadata:
{"chroma:document": "$$\\sum_{j=1}^{n}(x_{j}t_{1}-y_{j})^{2}=0\\,$$which forces $t_{1}x_{j}=y_{j}$ for all $j=1,\\cdots,n$. So we can take $\\alpha=t_{1}$ in case $c=\\sum_{j}y_{j}^{2}>0$.\n\nWhen all $x_{j}$'s vanish but not all $y_{j}$'s, we exchange $\\mathbf{x}$ and $\\mathbf{y}$ to get the same conclusion.\n\nFinally, when all $x_{j}$'s and $y_{j}$'s vanish, the inequality clearly holds."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 4072fb52-b9ac-4561-b11d-4ad11bfae643
encoding: FLOAT32
metadata:
{"chroma:document": "Accompanying with the notion of the inner product are those of the norm and the distance. Indeed, the **Euclidean norm** of an $n$-tuple is defined to be\n\n$$|\\mathbf{x}| = (\\mathbf{x}\\cdot\\mathbf{x})^{1/2}$$ $$= \\left(\\sum_{j=1}^{n}x_{j}^{2}\\right)^{1/2}$$ $$= \\sqrt{x_{1}^{2}+x_{2}^{2}+\\cdots x_{n}^{2}}\\.$$\n\nThe **Euclidean distance** between $\\mathbf{x}$ and $\\mathbf{y}$ is defined by\n\n$$|\\mathbf{x}-\\mathbf{y}|=\\sqrt{(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+\\cdots+(x_{ n}-y_{n})^{2}}\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 800c886f-6d23-4b8f-9c20-f17a821b077d
encoding: FLOAT32
metadata:
{"chroma:document": "In terms of these notions, Cauchy-Schwarz Inequality can be rewritten in a compact form\n\n$$|\\mathbf{x}\\cdot\\mathbf{y}|\\leq|\\mathbf{x}||\\mathbf{y}|\\.$$\n\nIn mathematics, a distance is a rule to assign a non-negative number to any pair of elements in a set under consideration. The rule consists of three \"axioms\": For $a,b,c$ in this set,\n\n* $d(a,b)\\geq 0\\,$ and equal to 0 iff $a=b$,\n* $d(a,b)=d(b,a)$, and\n* $d(a,b)\\leq d(a,c)+d(c,b)$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: a53d64e1-73a1-4fd8-ba20-9e043f578fa5
encoding: FLOAT32
metadata:
{"chroma:document": "Now, taking $d(\\mathbf{x},\\mathbf{y})=|\\mathbf{x}-\\mathbf{y}|$, we see that it satisfies all these three axioms: $\\mathbf{x},\\mathbf{y},\\mathbf{z}\\in\\mathbb{R}^{n}$,\n\n* $|\\mathbf{x}-\\mathbf{y}|\\geq 0$ and equal to 0 if and only if $\\mathbf{x}=\\mathbf{y}$,\n* $|\\mathbf{x}-\\mathbf{y}|=|\\mathbf{y}-\\mathbf{x}|$,\n* $|\\mathbf{x}-\\mathbf{y}|\\leq|\\mathbf{x}-\\mathbf{z}|+|\\mathbf{z}-\\mathbf{y}|$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 9af7fbf4-6060-4fbe-9a3d-945a8e2b7efe
encoding: FLOAT32
metadata:
{"chroma:document": "### 1.1 The Dot Product\n\nIndeed, (a) and (b) are obvious. To prove (c), write ${\\bf u}={\\bf x}-{\\bf z},{\\bf v}={\\bf z}-{\\bf y}$ to get\n\n$$|{\\bf u}+{\\bf v}|\\leq|{\\bf u}|+|{\\bf v}|\\,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e0b4b64c-8541-4ad6-8189-79f7a97af41e
encoding: FLOAT32
metadata:
{"chroma:document": "$$|{\\bf u}+{\\bf v}|\\leq|{\\bf u}|+|{\\bf v}|\\,$$\n\nwhich holds after taking square of both sides and then applying the Cauchy-Schwarz Inequality. I let you work out the details in the exercise. Note that $|{\\bf x}|=|{\\bf x}-{\\bf 0}|$, so the norm of ${\\bf x}$ is its distance to the zero $n$-tuple. In these notes, norm and distance are referred to Euclidean norm and Euclidean distance without further specification."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 2f119758-97be-4160-9f6d-8e4acce20168
encoding: FLOAT32
metadata:
{"chroma:document": "Note that the same notation $|x|$ stands for the absolute value of $x$ when $x$ is a real number. The norm of ${\\bf x}$ is the same as its absolute value when $n=1$. When $n\\geq 2$, the notation $|{\\bf x}|$ stands for the norm only as there is no such a definition of absolute value for an $n$-tuple. The notation $\\|{\\bf x}\\|$ is also used to denote the norm of ${\\bf x}$, but it will not be used here."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e3ff3b6b-3483-45da-914a-096f7b119d23
encoding: FLOAT32
metadata:
{"chroma:document": "Recall that the cosine function $\\cos t$ is strictly decreasing from $1$ to $-1$ as $t$ goes from $0$ to $\\pi$. Keeping this in mind, we are going to define the angle between two non-zero $n$-tuples. By Cauchy-Schwarz Inequality, the absolute value of the expression ${\\bf x}\\cdot{\\bf y}/|{\\bf x}||{\\bf y}|$ lies in the interval $[-1,1]$. Therefore, by what we just said, there exists a unique $\\theta\\in[0,\\pi]$ satisfying $\\cos\\theta={\\bf x}\\cdot{\\bf y}/|{\\bf x}||{\\bf y}|$, that is,"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 0de5351d-3955-4f8a-b921-8a6d9452e092
encoding: FLOAT32
metadata:
{"chroma:document": "We define the **angle** between two non-zero n-tuples ${\\bf x}$ and ${\\bf y}$ to be $\\theta$ where ${\\bf x},{\\bf y}$ are arbitrary $n$-tuples. The angle between two $n$-tuples makes no sense when of them is zero. By definition this angle must belong to $[0,\\pi]$. Moreover, it is symmetric, that is, the angle between ${\\bf x}$ and ${\\bf y}$ is the same as the angle between ${\\bf y}$ and ${\\bf x}$. At this stage, the notion of an angle is defined purely in an analytical manner and does not bear any geometric"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 312ab3ab-fb64-4865-96a6-5353ecdfa4c4
encoding: FLOAT32
metadata:
{"chroma:document": "purely in an analytical manner and does not bear any geometric meaning. We will link it to geometry in the next section."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 83121e87-263d-4feb-89e1-4e0d8d9f6e52
encoding: FLOAT32
metadata:
{"chroma:document": "Two $n$-tuples ${\\bf x}$ and ${\\bf y}$ are **perpendicular** or **orthogonal** to each other if ${\\bf x}\\cdot{\\bf y}=0$. In terms of the angle, they are perpendicular if and only if their angle is $\\pi/2$. The zero $n$-tuple is perpendicular to all $n$-tuples. By Cauchy-Schwarz Inequality, we also know that two non-zero ${\\bf x}$ and ${\\bf y}$ satisfy ${\\bf x}=c{\\bf y}$ for some $c>0$ when their angle $\\theta=0$ and satisfy ${\\bf x}=c{\\bf y},c<0$ when $\\theta=\\pi$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 2ed0492a-1601-4478-b7e5-0a9fccd2a2d3
encoding: FLOAT32
metadata:
{"chroma:document": "**Example 1.1**.: Find all $n$-tuples ${\\bf x}$ that are perpendicular to $(1,-1,2)$ and $(-1,0,3)$. These points satisfy\n\n$$(1,-1,2)\\cdot{\\bf x}=0\\,\\quad(-1,0,3)\\cdot{\\bf x}=0\\,$$\n\nthat is, the linear system\n\n$$\\begin{cases}x-y+2z&=0\\,\\\\ -x&+3z&=0\\.\\end{cases}$$We solve this system (see Comments at the end of this chapter) to get ${\\bf x}=(x,y,z)=a(3,5,1),a\\in\\mathbb{R}$. By varying $a$, we obtain infinitely many solutions."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b2504e2f-da41-4f16-b121-6e0d371747f1
encoding: FLOAT32
metadata:
{"chroma:document": "Summing up, we have defined the Euclidean space $(\\mathbb{R}^{n},+,\\cdot,\\langle\\cdot,\\cdot\\rangle)$ which is the set of all $n$-tuples with two algebraic operations-- the addition and the scalar multiplication-- as well as the dot product. From now on it will be abbreviated in a single symbol $\\mathbb{R}^{n}$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: ea674a5a-245d-41e9-ba19-0b82186e2cff
encoding: FLOAT32
metadata:
{"chroma:document": "### 1.2 Vector Representation\n\nVisualizing $n$-tuples for $n=2,3$ as vectors has been used widely in physics and engineering. In this section we discuss how to do it. However, despite its convenience and usefulness, it should be understood that the notion of a vector is only an auxiliary tool. Analysis on the Euclidean space can be carried out without referring to this notion."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 4b17bfd0-fee8-4da2-a0ea-6cb953a32d2e
encoding: FLOAT32
metadata:
{"chroma:document": "To start off, imagine that the coordinates axes have been introduced in the plane. The $x$-axis consists of all ordered pairs of the form $(x,0),\\ x\\in\\mathbb{R}$, and the $y$-axis all ordered pairs of the form $(0,y),\\ y\\in\\mathbb{R}$. So every ordered pair $(x,y)$ can be written as $x(1,0)+y(0,1)$ and $(x,y)$ is a point in the coordinate plane. Points in the plane are in one-to-one correspondence with ordered pairs. The same idea applies to all other dimensions. However, since the two dimensional case is"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 7b1c4e6b-c7a7-4821-8b0b-a05f82b46342
encoding: FLOAT32
metadata:
{"chroma:document": "other dimensions. However, since the two dimensional case is easy to see and the three dimensional case can be seen with some imagination, in the following we will focus on these two spaces. It will be apparent that most of our discussions can be extended to all dimensions."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 906e7fff-23b1-4c49-b6f0-a2e860147680
encoding: FLOAT32
metadata:
{"chroma:document": "For a point $(x,y)$ in the plane, it associates to a **vector** which is an arrow pointing from the base $(0,0)$ to the tip $(x,y)$. The vector degenerates into a point when it is the zero ordered pair $(0,0)$. We call it the **zero vector** and denoted it by $(0,0)$ or simply ${\\bf 0}$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 0a708537-5972-4b32-9f8c-2908766e2713
encoding: FLOAT32
metadata:
{"chroma:document": "After the introduction of the vector for an ordered pair, we interpret the algebraic operations of $\\mathbb{R}^{2}$ as follows. Indeed, by drawing pictures, it is not hard to convince oneself that the addition of two ordered pairs is accomplished by the parallelogram law. Specifically, first form the parallelogram using the two vectors as line segments corresponding to the two points. Then the diagonal of this resulting parallelogram, regarded as a vector pointing from the origin to the other end, is the"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: a0c65054-472d-4996-8c69-f2bdb69e4d9a
encoding: FLOAT32
metadata:
{"chroma:document": "as a vector pointing from the origin to the other end, is the sum of these two vectors. The same situation holds in $n\\geq 3$, as one can always restrict to the plane containing these two vectors provided they are linearly independent. When they are linearly dependent, the geometric interpretation is apparent."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e1db4df7-949e-4a4f-8746-2d0fc4982ace
encoding: FLOAT32
metadata:
{"chroma:document": "The scalar multiplication $(x,y)\\mapsto\\alpha(x,y),\\alpha>0$, means changing the vector by a scale of $\\alpha$ along the same direction. It is a prolongation if $\\alpha>1$ and a shortening if $\\alpha\\in(0,1)$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: baa19cbf-555e-4142-bbcc-847715dc2c0b
encoding: FLOAT32
metadata:
{"chroma:document": "### 1.2 Vector Representation\n\nOn the other hand, when $\\alpha<0$, the resulting vector points in the opposite direction of the original vector with size changes equal to $|\\alpha|$.\n\nHow about substraction of two vectors? Let ${\\bf w}={\\bf v}-{\\bf u}$. Then ${\\bf w}$ can be obtained by first drawing the triangle with vertices at $(0,0),{\\bf u}$ and ${\\bf v}$ and then translate the side from ${\\bf u}$ to ${\\bf v}$ so that its base is located at the origin. The translated side is ${\\bf w}$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 498ba165-41ba-4c71-a3e8-7df24fb5365c
encoding: FLOAT32
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{"chroma:document": "We may also find the midpoint of two ordered pairs. For ${\\bf u},{\\bf v}$, its midpoint is given by $({\\bf u}+{\\bf v})/2$. Regarding as a vector, this midpoint can be described in the following way. First draw the parallelogram formed by ${\\bf u}$ and ${\\bf v}$. Then the intersection point of the two diagonal lines of this parallelogram is the tip of the midpoint (vector)."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 6585faef-36dc-426e-a25c-b71f170240f6
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{"chroma:document": "When we regard an $n$-tuple ${\\bf x}$ as a vector, it is more convenient to call its norm the **magnitude** of the vector. It is a **unit vector** if its magnitude is equal to 1. Likewise, the distance between two points may be called the **length** of the line segment connecting ${\\bf x}$ and ${\\bf y}$. It is consistent with the classical Pythagoras theorem. In fact, the definition of the Euclidean norm and distance were inspired by this classical theorem."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
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{"chroma:document": "Next we show that the angle defined in the last section, which purely depends on analytical terms, is the same as the \"geometric angle\". To see it, let ${\\bf x}=(a,b)$ and ${\\bf y}=(c,d)$ be two non-zero vectors in the plane. By the Law of Cosines in trigonometry (see Comments at the end of this chapter),\n\n$$(c-a)^{2}+(d-b)^{2}=(a^{2}+b^{2})+(c^{2}+d^{2})-2\\sqrt{c^{2}+d^{2}}\\sqrt{a^{2} +b^{2}}\\cos\\phi\\,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b8da422a-8574-4ce2-b8dc-72960e65ff4d
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{"chroma:document": "where $\\phi\\in[0,\\pi]$ is the \"geometric angle\" between ${\\bf x}$ and ${\\bf y}$. Simplifying, we have\n\n$$-2(ac+db)=-2\\sqrt{c^{2}+d^{2}}\\sqrt{a^{2}+b^{2}}\\cos\\phi\\,$$\n\nwhich is equal to\n\n$${\\bf x}\\cdot{\\bf y}=|{\\bf x}||{\\bf y}|\\cos\\phi\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 91b5b229-649a-4fdb-828e-e1cbb55feb63
encoding: FLOAT32
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{"chroma:document": "$${\\bf x}\\cdot{\\bf y}=|{\\bf x}||{\\bf y}|\\cos\\phi\\.$$\n\nComparing with the definition of $\\theta$, we have $\\cos\\phi=\\cos\\theta$ so that $\\phi=\\theta$. In other words, the geometric angle coincides with the analytical angle. The same argument works in higher dimensions as we can restrict to the plane containing any two given vectors."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 74a4c458-9bb9-42bf-9e87-b8ca703eaa7c
encoding: FLOAT32
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{"chroma:document": "A vector is uniquely determined by its magnitude and direction. To be more precise we fix them in definition. Any vector with unit length is called a **direction.** Each non-zero vector ${\\bf x}$ can be written as\n\n$${\\bf x}=|{\\bf x}|\\ \\mathbf{\\xi},$$\n\nwhere $|{\\bf x}|$ is its magnitude and\n\n$$\\mathbf{\\xi}=\\frac{{\\bf x}}{|{\\bf x}|}$$its direction. Every direction $\\boldsymbol{\\xi}=(\\xi_{1},\\xi_{2},\\cdots,\\xi_{n})$ can further be expressed as"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 18c1e797-85c6-48ce-9e5f-b0b5cd9035b1
encoding: FLOAT32
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{"chroma:document": "$$\\boldsymbol{\\xi}=(\\cos\\alpha_{1},\\cos\\alpha_{2},\\cdots,\\cos\\alpha_{n})\\,$$\n\nwhere $\\alpha_{k}\\in[0,\\pi]$ are called the **direction angles** of $\\boldsymbol{\\xi}$. From $\\boldsymbol{\\xi}\\cdot\\mathbf{e}_{k}=\\cos\\alpha_{k}$ we see that $\\alpha_{k}$ is the angle between $\\boldsymbol{\\xi}$ and the $e_{k}$-axis. These $\\cos\\alpha_{k}$'s are called the **direction cosines** of $\\mathbf{x}$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e5f40a4b-0783-4e31-b9e4-6b0677e6e774
encoding: FLOAT32
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{"chroma:document": "**Example 1.2**.: Find the magnitude and direction of $(1,2,-7)$ and determine the vector $(2,a,6)$ that is perpendicular to $(1,2,-7)$. The magnitude of $(1,2,-7)$ is\n\n$$|(1,2,-7)|=\\sqrt{1^{2}+2^{2}+(-7)^{2}}=\\sqrt{54}\\,$$\n\nand its direction is $(1,2,-7)/\\sqrt{54}$. By orthogonality,\n\n$$0=(1,2,-7)\\cdot(2,a,6)=2+2a-42=0\\,$$\n\nwhich implies $a=20$. The vector $(2,20,6)$ is perpendicular to $(1,2,-7)$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: cd791c83-3f91-434e-a67b-bfa8fd9ac7f3
encoding: FLOAT32
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{"chroma:document": "One may also consider the **vector from the initial point $\\mathbf{x}$ to the terminal point $\\mathbf{y}$,** or **the vector based at some point**. Unlike a vector, a vector from $\\mathbf{x}$ to $\\mathbf{y}$ is an arrow whose base and tip are $\\mathbf{x}$ and $\\mathbf{y}$ respectively. Obviously such a \"vector\" is parallel to the position vector of $\\mathbf{y}-x$ whose base is now at the origin. The length and direction of a vector from $\\mathbf{x}$ to $\\mathbf{y}$ are defined as the respective length and"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: af8ddd9c-7d69-49cf-b27f-7498d766d213
encoding: FLOAT32
metadata:
{"chroma:document": "to $\\mathbf{y}$ are defined as the respective length and direction of $\\mathbf{y}-\\mathbf{x}$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e4762a2c-2e77-4758-9fc2-589440a1fd13
encoding: FLOAT32
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{"chroma:document": "**Example 1.3**.: Consider the triangle with vertices at $(1,2),(3,4),(0,-1)$. Find the direction of the vector pointing at the midpoint of the side connecting $(1,2)$ and $(3,4)$ from $(0,-1)$. Well, first we translate $(0,-1)$ to the origin so that the triangle is congruent to the one whose vertices are $((1,2)-(0,-1),(3,4)-(0,-1),(0,-1)-(0,-1)$, that is, $(1,3),(3,5),(0,0)$. The midpoint of the side from $(1,3)$ and $(3,5)$ is given by\n\n$$\\frac{1}{2}\\left((1,3)+(3,5)\\right)=(2,4)\\,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 23e75400-9674-4fc0-acaa-4047f9a48c4f
encoding: FLOAT32
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{"chroma:document": "$$\\frac{1}{2}\\left((1,3)+(3,5)\\right)=(2,4)\\,$$\n\nand its direction is given by\n\n$$\\frac{(2,4)}{\\sqrt{2^{2}+4^{2}}}=\\frac{(2,4)}{\\sqrt{20}}=\\frac{(1,2)}{\\sqrt{5 }}\\.$$\n\n(No need to simplify further.)\n\n**Example 1.4**.: (a) Find the magnitude and direction of the vector from $(1,-1)$ to $(-2,5)$. (b) Find all directions that are perpendicular to the vector in (a)."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e9d5dcd0-81ec-45e3-9b4e-8828f55e3d48
encoding: FLOAT32
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{"chroma:document": "The magnitude and direction of the vector from $(1,-1)$ to $(-2,5)$ are the same as those of the position vector $(-2,5)-(1,-1)=(-3,6)$. Its magnitude is given by\n\n$$|(-3,6)|=\\sqrt{(-3)^{2}+6^{2}}=3\\sqrt{5}\\,$$\n\nand the direction is given by\n\n$$\\frac{(-3,6)}{3\\sqrt{5}}=\\frac{(-1,2)}{\\sqrt{5}}\\.$$\n\n(No need to simplify further.)\n\nA vector $(a,b)$ perpendicular to $(-3,6)$ satisfies\n\n$$(-3,6)\\cdot(a,b)=-3a+6b=0\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 89bf039e-a41f-4320-a5df-fcae725efe90
encoding: FLOAT32
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{"chroma:document": "$$(-3,6)\\cdot(a,b)=-3a+6b=0\\.$$\n\nBy varying $a$ and $b$ according to this relation, there are infinitely many vectors $(a,b)$ satisfying this requirement. For instance, we may take $a=2,b=1$ so $(2,1)$ is one choice. However, to be a direction there is another condition, namely, the length of the vector has to equal to one. There are two such vectors:\n\n$$\\frac{(2,1)}{\\sqrt{5}}\\,\\quad-\\frac{(2,1)}{\\sqrt{5}}\\.$$\n\n(Again no need to simplify.)"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: d5594194-d7ab-41d1-a35d-22f67fba8082
encoding: FLOAT32
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{"chroma:document": "### 1.3 Euclidean Motions\n\nA **Euclidean motion** is a map from $\\mathbb{R}^{n}$ to itself of the form\n\n$$T\\mathbf{x}=A\\mathbf{x}+\\mathbf{b}\\,$$\n\nwhere $\\mathbf{b}\\in\\mathbb{R}^{n}$ and $A$ is an $n\\times n$-matrix, that preserves the distance between two points, that is, for $\\mathbf{x},\\mathbf{y}\\in\\mathbb{R}^{n}$,\n\n$$|T\\mathbf{x}-T\\mathbf{y}|=|\\mathbf{x}-\\mathbf{y}|\\.$$\n\nHere in $A\\mathbf{x}$ the vector $\\mathbf{x}$ should be understood as a column vector."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 497039cc-5ed2-4948-ad40-72490a390cf0
encoding: FLOAT32
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{"chroma:document": "Recall that a square matrix $R$ is called an orthogonal matrix if $R^{\\prime}R=RR^{\\prime}=I$, where $R^{\\prime}$ is the transpose of $R$ and $I$ is the identity matrix.\n\n**Proposition 1.2**.: _A map $T\\mathbf{x}=A\\mathbf{x}+\\mathbf{b}$ is a Euclidean motion if and only if $A$ is an orthogonal matrix._"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: c270b6bb-1f3b-4919-94ab-c21c88a0e6df
encoding: FLOAT32
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{"chroma:document": "Proof.: In the following we use $\\langle\\mathbf{x},\\mathbf{y}\\rangle$ instead $\\mathbf{x}\\cdot\\mathbf{y}$ to denote the dot product. First of all, let $T$ be a Euclidean motion. Then it follows from the definition that\n\n$$|\\mathbf{x}-\\mathbf{y}|=|T\\mathbf{x}-T\\mathbf{y}|=|A\\mathbf{x}-A\\mathbf{y}|= |A(\\mathbf{x}-\\mathbf{y})|,$$which yields immediately that"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b74a4c5d-5c0f-4142-a42f-b07c49ba5335
encoding: FLOAT32
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{"chroma:document": "$$|A({\\bf x}-{\\bf y})|^{2} = \\langle A({\\bf x}-{\\bf y}),A({\\bf x}-{\\bf y})\\rangle$$ $$= |{\\bf x}-{\\bf y}|^{2}$$ $$= |{\\bf x}|^{2}-2\\langle{\\bf x},{\\bf y}\\rangle+|{\\bf y}|^{2}\\.$$\n\nOn the other hand, a direct calculation shows that\n\n$$\\langle A({\\bf x}-{\\bf y}),A({\\bf x}-{\\bf y})\\rangle = \\langle A{\\bf x}-A{\\bf y},A{\\bf x}-A{\\bf y}\\rangle$$ $$= |A{\\bf x}|^{2}-2\\langle A{\\bf x},A{\\bf y}\\rangle+|A{\\bf y}|^{2}.$$\n\nBy comparing, we see that for all ${\\bf x},{\\bf y}$,"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 9c01178c-db32-44c0-9b9d-bc862da18533
encoding: FLOAT32
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{"chroma:document": "By comparing, we see that for all ${\\bf x},{\\bf y}$,\n\n$$\\langle A^{\\prime}A{\\bf x},{\\bf y}\\rangle=\\langle A{\\bf x},A{\\bf y}\\rangle= \\langle{\\bf x},{\\bf y}\\rangle\\,$$\n\nwhich implies that $A^{\\prime}A=I$. Thus $A$ is a orthogonal matrix. Finally, this relation also shows that $T$ is a Euclidean motion whenever $A$ is orthogonal. $\\Box$\n\nHere we have used the following derivation in linear algebra: For a matrix $(B{\\bf x})_{j}=\\sum_{k}b_{kj}x_{k}$,"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: f466c2c1-c02f-4eaa-b72a-3bb37a0aa11b
encoding: FLOAT32
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{"chroma:document": "$$\\langle{\\bf x},B{\\bf y}\\rangle=\\langle B^{\\prime}{\\bf x},{\\bf y}\\rangle\\.$$\n\nIndeed,\n\n$$\\langle{\\bf x},B{\\bf y}\\rangle = \\sum_{j}x_{j}\\sum_{k}b_{kj}y_{k}$$ $$= \\sum_{k}\\sum_{j}b_{kj}x_{j}y_{k}$$ $$= \\sum_{k}y_{k}\\sum_{j}b^{\\prime}_{jk}x_{j}=\\langle B^{\\prime}{ \\bf x},{\\bf y}\\rangle\\.$$\n\nHere are some examples of Euclidean motions."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 467e21f0-d157-416d-9b0d-55eca89839cc
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{"chroma:document": "1. Take $A$ to be the identity and ${\\bf b}$ a nonzero vector. Then $T{\\bf x}={\\bf x}+{\\bf b}$ is a translation. The origin is moved to ${\\bf b}$ after the motion."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 10e4f084-bf35-4a97-90af-30729500dc0e
encoding: FLOAT32
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{"chroma:document": "2. When $n=2$, the Euclidean motion $$T{\\bf x}=\\begin{bmatrix}1&0\\\\ 0&-1\\end{bmatrix}\\begin{bmatrix}x_{1}\\\\ x_{2}\\end{bmatrix}$$ is the reflection with respect to the $x$-axis and $$T{\\bf x}=\\begin{bmatrix}-1&0\\\\ 0&1\\end{bmatrix}\\begin{bmatrix}x_{1}\\\\ x_{2}\\end{bmatrix}$$is the reflection with respect to the $y$-axis. (In matrix form the vector $\\mathbf{x}$ is understood as a column vector.) On the other hand, given any plane in $\\mathbb{R}^{3}$, one may consider the reflection with respect to this plane."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: d0d722e5-d954-4992-be5c-d7dab28ac0e8
encoding: FLOAT32
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{"chroma:document": "one may consider the reflection with respect to this plane. For instance, $$T\\mathbf{x}=\\begin{bmatrix}1&0&0\\\\ 0&1&0\\\\ 0&0&-1\\end{bmatrix}\\begin{bmatrix}x_{1}\\\\ x_{2}\\\\ x_{3}\\end{bmatrix}$$ is the reflection with respect to the $xy$-plane in $\\mathbb{R}^{3}$. The reflection with respect to any straight line in $\\mathbb{R}^{2}$ or with respect to any plane in $\\mathbb{R}^{3}$ can be defined similarly."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 6a38b826-8a21-462c-8fe1-c8ce863be4bd
encoding: FLOAT32
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{"chroma:document": "3. The (counterclockwise) rotation of angle $\\theta$ in $\\mathbb{R}^{2}$ is given by the Euclidean motion $$T\\mathbf{x}=\\begin{bmatrix}\\cos\\theta&-\\sin\\theta\\\\ \\sin\\theta&\\cos\\theta\\end{bmatrix}\\begin{bmatrix}x_{1}\\\\ x_{2}\\end{bmatrix}\\,\\quad\\theta\\in(0,2\\pi)\\.$$ In $\\mathbb{R}^{3}$, one can perform a rotation around a fixed axis. For instance, the rotation $$T\\mathbf{x}=\\begin{bmatrix}\\cos\\theta&-\\sin\\theta&0\\\\ \\sin\\theta&\\cos\\theta&0\\\\ 0&0&1\\end{bmatrix}\\begin{bmatrix}x_{1}\\\\ x_{2}\\\\ x_{3}\\end{bmatrix}$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b116c17a-d5bc-429d-a0f5-7a214d0f43ba
encoding: FLOAT32
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{"chroma:document": "Let us verify that Euclidean motions are closed under compositions. Let $T\\mathbf{x}=A\\mathbf{x}+\\mathbf{b}$ and $S\\mathbf{x}=B\\mathbf{x}+\\mathbf{c}$ be two Euclidean motions. Its composition is given by\n\n$$ST\\mathbf{x}=B(A\\mathbf{x}+\\mathbf{b})+\\mathbf{c}=C\\mathbf{x}+\\mathbf{d}\\, \\quad C\\equiv BA,\\quad\\mathbf{d}=B\\mathbf{b}+\\mathbf{c}\\.$$\n\nAs\n\n$$C^{\\prime}C = (BA)^{\\prime}BA$$ $$= A^{\\prime}B^{\\prime}BA$$ $$= A^{\\prime}IA$$ $$= I,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 0c89518d-a5ff-429b-97b9-e2c09b6ad866
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{"chroma:document": "we conclude that $ST$ is again a Euclidean motion. Furthermore, we claim that each Euclidean motion admits an inverse. Indeed, letting $U\\mathbf{x}=A^{-1}\\mathbf{x}-A^{-1}\\mathbf{b}$ which is obviously an Euclidean motion, we have\n\n$$UT\\mathbf{x}=A^{-1}(A\\mathbf{x}+\\mathbf{b})-A^{-1}\\mathbf{b}=\\mathbf{x}\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: fd5649aa-c7f8-4925-aea8-0d4cf2cf1c81
encoding: FLOAT32
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{"chroma:document": "Summing up, the collection of all Euclidean motions forms a group called the Euclidean group of $\\mathbb{R}^{n}$. (It is alright if you have not learned what a group is. You will learn it in MATH2070.)\n\nIn the following we study the structure of Euclidean motions for $n=2,3$. Apparently it suffices to look at the orthogonal matrix $A$.\n\n**Theorem 1.3**.: _In $\\mathbb{R}^{2}$, every orthogonal matrix can be written as_"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 89da1207-fd22-40cf-938c-37e774c0eae2
encoding: FLOAT32
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{"chroma:document": "1. $$\\begin{bmatrix}\\cos\\theta&-\\sin\\theta\\\\ \\sin\\theta&\\cos\\theta\\end{bmatrix}\\,$$\n2. $$\\begin{bmatrix}\\cos\\theta&-\\sin\\theta\\\\ \\sin\\theta&\\cos\\theta\\end{bmatrix}\\begin{bmatrix}1&0\\\\ 0&-1\\end{bmatrix}\\,\\quad\\theta\\in[0,2\\pi)\\.$$\n\n_Case (a) is a genuine rotation for $\\theta\\in(0,2\\pi)$ and reduces to the identity at $\\theta=0$. Case (b) is the reflection with respect to the $x$-axis and then followed by a rotation._\n\nProof.: Let\n\n$$A=\\begin{bmatrix}a&b\\\\ c&d\\end{bmatrix}\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: c16fb7ec-a68f-47f7-9e9c-f0382be0672e
encoding: FLOAT32
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{"chroma:document": "Proof.: Let\n\n$$A=\\begin{bmatrix}a&b\\\\ c&d\\end{bmatrix}\\.$$\n\nBy orthogonality $A^{\\prime}A=I$ we have\n\n$$a^{2}+c^{2}=1\\,\\quad b^{2}+d^{2}=1\\,\\quad ab+cd=0\\.$$\n\nSince $a$ is a number between $-1$ and $1$, we can find a unique $\\theta\\in[0,2\\pi)$ such that $a=\\cos\\theta,\\ c=\\sin\\theta$. Then either $b=-\\sin\\theta,\\ d=\\cos\\theta$ or $b=\\sin\\theta,d=-\\cos\\theta$, so (a) or (b) must hold."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 9fb1f8a9-c164-4681-9177-f214afb0a45a
encoding: FLOAT32
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{"chroma:document": "In the following we consider the three dimensional case. Denote by $R_{z}(\\theta)$ the rotation around the $z$-axis by an angle $\\theta$:\n\n$$R_{z}(\\theta)=\\begin{bmatrix}\\cos\\theta&-\\sin\\theta&0\\\\ \\sin\\theta&\\cos\\theta&0\\\\ 0&0&1\\end{bmatrix}\\.$$\n\nSimilarly we define\n\n$$R_{x}(\\theta)=\\begin{bmatrix}1&0&0\\\\ 0&\\cos\\theta&-\\sin\\theta\\\\ 0&\\sin\\theta&\\cos\\theta\\end{bmatrix}\\,$$\n\nand\n\n$$R_{y}(\\theta)=\\begin{bmatrix}\\cos\\theta&0&-\\sin\\theta\\\\ 0&1&0\\\\ \\sin\\theta&0&\\cos\\theta\\end{bmatrix}\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 6cfd9bf7-04e6-42b7-bb85-0ad4987894be
encoding: FLOAT32
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{"chroma:document": "Also denote by $L_{z}$ the reflection with respect to the $xy$-plane:\n\n$$L_{z}=\\begin{bmatrix}1&0&0\\\\ 0&1&0\\\\ 0&0&-1\\end{bmatrix}\\.$$Similarly\n\n$$L_{x}=\\begin{bmatrix}-1&0&0\\\\ 0&1&0\\\\ 0&0&1\\end{bmatrix}\\,$$\n\nand\n\n$$L_{y}=\\begin{bmatrix}1&0&0\\\\ 0&-1&0\\\\ 0&0&1\\end{bmatrix}\\.$$\n\n**Theorem 1.4**.: _* In $\\mathbb{R}^{3}$, every orthogonal matrix can be written as_\n\n1. $R_{z}(\\alpha)R_{x}(\\beta)R_{z}(\\gamma)$__, _or_\n2. $R_{z}(\\alpha)R_{x}(\\beta)R_{z}(\\gamma)L_{z}$__,_\n\n_for some $\\alpha,\\beta,$ and $\\gamma$._"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 95fe3832-16e2-49bb-b269-fd51df863cda
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{"chroma:document": "_for some $\\alpha,\\beta,$ and $\\gamma$._\n\nProof.: * Let $A=\\left(a_{ij}\\right),i,j=1,2,3,$ be orthogonal. We have\n\n$$\\begin{bmatrix}\\cos\\theta&-\\sin\\theta&0\\\\ \\sin\\theta&\\cos\\theta&0\\\\ 0&0&1\\end{bmatrix}\\begin{bmatrix}a_{11}&a_{12}&a_{13}\\\\ a_{21}&a_{22}&a_{23}\\\\ a_{31}&a_{32}&a_{33}\\end{bmatrix}$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 2de36edf-f15b-4540-9ed3-c885217a316c
encoding: FLOAT32
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{"chroma:document": "$$=\\begin{bmatrix}\\cos\\theta a_{11}-\\sin\\theta a_{21}&\\cos\\theta a_{12}-\\sin \\theta a_{22}&\\cos\\theta a_{13}-\\sin\\theta a_{23}\\\\ \\sin\\theta a_{11}+\\cos\\theta a_{21}&\\sin\\theta a_{12}+\\cos\\theta a_{22}&\\sin \\theta a_{13}+\\cos\\theta a_{23}\\\\ a_{31}&a_{32}&a_{33}\\end{bmatrix}\\.$$\n\nChoose $\\theta$ so that $\\cos\\theta a_{13}-\\sin\\theta a_{23}=0$ and write the resulting matrix as\n\n$$\\begin{bmatrix}b_{11}&b_{12}&0\\\\ b_{21}&b_{22}&b_{23}\\\\ b_{31}&b_{32}&b_{33}\\end{bmatrix}\\.$$\n\nWe further have"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 0517f7a6-88a2-4d9a-a204-20615e4b5c2e
encoding: FLOAT32
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{"chroma:document": "We further have\n\n$$=\\begin{bmatrix}b_{11}&b_{12}&0\\\\ \\cos\\varphi b_{21}-\\sin\\varphi b_{31}&\\cos\\varphi b_{22}-\\sin\\varphi b_{32}& \\cos\\varphi b_{23}-\\sin\\varphi b_{33}\\\\ \\sin\\varphi b_{21}+\\cos\\varphi b_{31}&\\sin\\varphi b_{22}+\\cos\\varphi b_{32}& \\sin\\varphi b_{23}+\\cos\\varphi b_{33}\\end{bmatrix}\\.$$\n\nChoose $\\varphi$ so that $\\cos\\varphi b_{23}-\\sin\\varphi b_{33}=0$ and write the resulting matrix as"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 99776b81-84ff-4dc2-8483-6559d175458f
encoding: FLOAT32
metadata:
{"chroma:document": "$$\\begin{bmatrix}c_{11}&c_{12}&0\\\\ c_{21}&c_{22}&0\\\\ c_{31}&c_{32}&c_{33}\\end{bmatrix}\\.$$This matrix is the product of three orthogonal matrices, again it is orthogonal. Therefore, $c_{33}=\\pm 1$. Moreover, from\n\n$$c_{11}\\times 0+c_{21}\\times 0+c_{31}\\times c_{33}=0\\,$$\n\nwe deduce $c_{31}=0$. Similarly, $c_{32}=0$. The matrix is in fact of the form\n\n$$\\begin{bmatrix}c_{11}&c_{12}&0\\\\ c_{21}&c_{22}&0\\\\ 0&0&\\pm 1\\end{bmatrix}\\,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: d5ef6f37-8f7c-4736-a5f6-efa0bfa19e57
encoding: FLOAT32
metadata:
{"chroma:document": "where the $2\\times 2$-matrix is orthogonal. It can be written as $R_{z}(\\gamma)$ or $R_{z}(\\gamma)L_{z}$ for some $\\gamma$ according to Proposition 1.3. We conclude that\n\n$$R_{x}(\\varphi)R_{z}(\\theta)A=R_{z}(\\gamma)\\,$$\n\nor\n\n$$R_{x}(\\varphi)R_{z}(\\theta)A=R_{z}(\\gamma)L_{z}\\,$$\n\nthat is,\n\n$$A=R_{z}(-\\theta)R_{x}(-\\varphi)R_{z}(\\gamma)\\,$$\n\nor\n\n$$A=R_{z}(-\\theta)R_{x}(-\\varphi)R_{z}(\\gamma)L_{z}\\.$$\n\nThe desired result follows by taking $\\alpha=-\\theta$ and $\\beta=-\\varphi$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 09c16bcf-6240-4485-a71b-88707ae3a67e
encoding: FLOAT32
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{"chroma:document": "### 1.4 The Cross Product in $\\mathbb{R}^{3}$\n\nThe cross product assigns a 3-vector to two given 3-vectors. There is no such product in the general dimension. Somehow it shows how special our physical space is. The cross product is important due to its relevance in physics and engineering.\n\nNotations like $\\mathbf{x},\\mathbf{y},\\mathbf{u}$ and $\\mathbf{v}$ are common for vectors. We have used $\\mathbf{x},\\mathbf{y}$ in the previous sections. Now we use $\\mathbf{u},\\mathbf{v}$ in this one."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 1d3aa631-40d2-47ea-b4cf-f3100f827dcd
encoding: FLOAT32
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{"chroma:document": "First the definition. Let $\\mathbf{u},\\mathbf{v}\\in\\mathbb{R}^{3}$, define the cross product of $\\mathbf{u}=(u_{1},u_{2},u_{3})$ and $\\mathbf{v}=(v_{1},v_{2},v_{3})$ to be\n\n$$\\mathbf{u}\\times\\mathbf{v}=\\left(u_{2}v_{3}-u_{3}v_{2},-(u_{1}v_{3}-u_{3}v_{1 }),u_{1}v_{2}-u_{2}v_{1}\\right)\\.$$\n\nIn particular, we have"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: e2f12323-3503-4f42-9127-a8c766a2d9ca
encoding: FLOAT32
metadata:
{"chroma:document": "In particular, we have\n\n$$\\mathbf{e}_{1}\\times\\mathbf{e}_{2}=\\mathbf{e}_{3},\\quad\\mathbf{e}_{2}\\times \\mathbf{e}_{3}=\\mathbf{e}_{1},\\quad\\mathbf{e}_{3}\\times\\mathbf{e}_{1}= \\mathbf{e}_{2}\\.$$To aid memorizing, we can formally express it as the determinant\n\n$$\\left|\\begin{array}{ccc}\\mathbf{e}_{1}&\\mathbf{e}_{2}&\\mathbf{e}_{3}\\\\ u_{1}&u_{2}&u_{3}\\\\ v_{1}&v_{2}&v_{3}\\end{array}\\right|\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 1f93b7b7-8c73-46da-8ac7-9cc580d87d76
encoding: FLOAT32
metadata:
{"chroma:document": "Expand the determinant by the first row yields the formula above. Here it is formal because it does not make sense to put the unit vectors $\\mathbf{e}_{1},\\mathbf{e}_{2},\\mathbf{e}_{3}$ as entries in a matrix.\n\nThe cross product is in many aspects in sharp contrast with an ordinary product. Some of its properties are list below:\n\n**Theorem 1.5**.: _For $\\boldsymbol{u},\\boldsymbol{v},\\boldsymbol{w}\\in\\mathbb{R}^{3}$,_"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 595d5636-72f4-4abd-8acc-4a421fdc5cfd
encoding: FLOAT32
metadata:
{"chroma:document": "1. $(\\alpha\\boldsymbol{u}+\\beta\\boldsymbol{v})\\times\\boldsymbol{w}=\\alpha \\boldsymbol{u}\\times\\boldsymbol{w}+\\beta\\boldsymbol{v}\\times\\boldsymbol{w}\\,\\quad\\forall \\alpha,\\beta\\in\\mathbb{R}$_._\n2. $\\boldsymbol{u}\\times\\boldsymbol{v}=-\\boldsymbol{v}\\times\\boldsymbol{u}$ _. In particular,_ $\\boldsymbol{u}\\times\\boldsymbol{u}=0$ _._\n3. $(\\boldsymbol{u}\\times\\boldsymbol{v})\\times\\boldsymbol{w}=\\boldsymbol{u}\\times (\\boldsymbol{v}\\times\\boldsymbol{w})$ _is not always true._"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 1de4ac98-0033-40a0-80de-4ec7cf6141a8
encoding: FLOAT32
metadata:
{"chroma:document": "The proofs of Theorem (a) and (b) are straightforward from definition. As for (c), which asserts that the associative law does not hold, some examples suffice:\n\n$$(\\mathbf{e}_{1}\\times\\mathbf{e}_{2})\\times\\mathbf{e}_{2}=-\\mathbf{e}_{1}\\, \\quad\\mathbf{e}_{1}\\times(\\mathbf{e}_{2}\\times\\mathbf{e}_{2})=\\boldsymbol{0}\\,$$\n\nand\n\n$$(1,1,1)\\times\\big{(}(1,0,-1)\\times(2,1,1)\\big{)}=(3,-1,-5)\\,\\quad\\big{(}(1,1,1) \\times(1,0,-1)\\big{)}\\times(2,1,1)=(4,0,-4).$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b6b3fb4b-004b-4ae6-bcd0-54bacf1cb1d6
encoding: FLOAT32
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{"chroma:document": "As a vector is completely determined by its magnitude and direction, let us consider the magnitude and direction of the cross product. First of all, using the definition of the cross product, one can verify directly that\n\n$$\\mathbf{u}\\cdot(\\mathbf{u}\\times\\mathbf{v})=0\\,\\quad\\mathbf{v}\\cdot(\\mathbf{u} \\times\\mathbf{v})=0\\,$$\n\nso\n\n$$(\\alpha\\mathbf{u}+\\beta\\mathbf{v})\\cdot\\mathbf{u}\\times\\mathbf{v}=0\\,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 84b31419-13a8-4103-b757-b1239eb4c01c
encoding: FLOAT32
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{"chroma:document": "that is, it is perpendicular to the two dimensional subspace spanned by the vectors $\\mathbf{u}$ and $\\mathbf{v}$. After the definition of a plane is introduced in the next chapter, we can say that the cross product of two linearly independent vectors points in the normal direction of the plane spanned by $\\mathbf{u}$ and $\\mathbf{v}$. When they are linearly dependent, their cross product is the zero vector which does not form a normal direction. There are two normal directions, pointing upward or"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 7c434c03-0485-4aaf-8b35-26422ab7d219
encoding: FLOAT32
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{"chroma:document": "direction. There are two normal directions, pointing upward or downward, so to speak. The choice of the direction of the cross product is determined by the right hand rule. That is, with the thumb making a right angle with the other four fingers of your right hand, first point the four fingers along the direction of $\\mathbf{u}$ and then move them to $\\mathbf{v}$ in an angle less than $\\pi$. The direction of is where your thumb points to. To see this, one identifies the direction of $\\mathbf{u}$ with"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b7975c75-ccc7-4739-8c9e-5e6488ceb620
encoding: FLOAT32
metadata:
{"chroma:document": "To see this, one identifies the direction of $\\mathbf{u}$ with $\\mathbf{e}_{1}$. If $\\mathbf{v}$ lies in the first or second quadrants, $\\mathbf{v}=\\alpha\\mathbf{e}_{1}+\\beta\\mathbf{e}_{2}\\,\\beta>0$, and $\\mathbf{u}\\times\\mathbf{v}$ points to $\\mathbf{e}_{3}$. If $\\mathbf{v}$ lies in the third or fourth quadrants, $\\mathbf{v}=\\alpha\\mathbf{e}_{1}+\\beta\\mathbf{e}_{2}\\,\\beta<0$, and $\\mathbf{u}\\times\\mathbf{v}$ points to $-\\mathbf{e}_{3}$. This is the right hand rule."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 322c59e3-b856-4727-b9f4-002c78f475e7
encoding: FLOAT32
metadata:
{"chroma:document": "We have described the direction of the cross product. How about its magnitude? We have\n\n**Theorem 1.6**.: _For $\\textbf{u},\\textbf{v}\\in\\mathbb{R}^{3}$,_\n\n$$|\\textbf{u}\\times\\textbf{v}|=|\\textbf{u}||\\textbf{v}|\\sin\\theta,\\quad\\theta \\in[0,\\pi]\\,$$\n\n_where $\\theta$ is the angle between **u** and **v**._\n\nProof.: The proof depends on the identity\n\n$$|\\textbf{u}\\times\\textbf{v}|^{2}=|\\textbf{u}|^{2}|\\textbf{v}|^{2}-(\\textbf{u} \\cdot\\textbf{v})^{2}\\.$$\n\nIndeed, by brute force"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 2aaa384a-6a7b-4ea2-b6c2-f77a5dc0c9f4
encoding: FLOAT32
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{"chroma:document": "Indeed, by brute force\n\n$$|\\textbf{u}\\times\\textbf{v}|^{2} = (u_{2}v_{3}-u_{3}v_{2})^{2}+(u_{1}v_{3}-u_{3}v_{1})^{2}+(u_{1}v_{ 2}-u_{2}v_{1})^{2}$$ $$= u_{2}^{2}v_{3}^{2}+u_{3}^{2}v_{2}^{2}+u_{1}^{2}v_{3}^{2}+u_{3}^{ 2}v_{1}^{2}+u_{1}^{2}v_{2}^{2}+u_{2}^{2}v_{1}^{2}-2u_{2}v_{3}u_{3}v_{2}-2u_{1}v _{3}u_{3}v_{1}-2u_{1}v_{2}u_{2}v_{1}\\.$$\n\nOn the other hand,"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 0036f95f-aae1-4aec-82d9-d24c5eed5419
encoding: FLOAT32
metadata:
{"chroma:document": "On the other hand,\n\n$$|\\textbf{u}|^{2}|\\textbf{v}|^{2}-(\\textbf{u}\\cdot\\textbf{v})^{2}$$ $$= (u_{1}^{2}+u_{2}^{2}+u_{3}^{2})(v_{1}^{2}+v_{2}^{2}+v_{3}^{2})-(u_ {1}v_{1}+u_{2}v_{2}+u_{3}v_{3})^{2}$$ $$= u_{2}^{2}v_{3}^{2}+u_{3}^{2}v_{2}^{2}+u_{1}^{2}v_{3}^{2}+u_{3}^{ 2}v_{1}^{2}+u_{1}^{2}v_{2}^{2}+u_{2}^{2}v_{1}^{2}-2u_{2}v_{3}u_{3}v_{2}-2u_{1} v_{3}u_{3}v_{1}-2u_{1}v_{2}u_{2}v_{1}\\,$$\n\nwhence the identity holds. Now, by the Cosine Law,"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 3db7dce1-794f-4d96-9aeb-239165f5a833
encoding: FLOAT32
metadata:
{"chroma:document": "whence the identity holds. Now, by the Cosine Law,\n\n$$|\\textbf{u}\\times\\textbf{v}| = \\sqrt{|\\textbf{u}|^{2}|\\textbf{v}|^{2}-|\\textbf{u}|^{2}|\\textbf{v }|^{2}\\cos^{2}\\theta}$$ $$= |\\textbf{u}||\\textbf{v}||\\sin\\theta|$$ $$= |\\textbf{u}||\\textbf{v}||\\sin\\theta\\,$$\n\nas $\\sin\\theta\\geq 0$ on $[0,\\pi]$. \n\nIn conclusion the magnitude and direction of the decomposition of the cross product is given by\n\n$$\\textbf{u}\\times\\textbf{v}=|\\textbf{u}||\\textbf{v}|\\sin\\theta\\ \\textbf{n}\\,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 0d3013d1-bf8e-44f4-b795-329eba26d7ac
encoding: FLOAT32
metadata:
{"chroma:document": "where **n** is the unit vector determined by the right hand rule (when **u** and **v** are linearly independent, that is, when $\\sin\\theta\\neq 0$)."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 9d59c00c-9e18-4720-b95f-992a23eb6c4e
encoding: FLOAT32
metadata:
{"chroma:document": "**Corollary 1.7**.:\n1. _The area of the parallelogram spanned by_ **u** _and_ **v** _is equal to_ $|\\textbf{u}\\times\\textbf{v}|$_._\n2. _The area of the triangle with two sides given by_ $u$ _and_ $v$ _is equal to_ $1/2|\\textbf{u}\\times\\textbf{v}|$_._\n3. _The volume of the parallelepiped spanned by_ $u$,_ $v$ _and_ $w$ _is equal to_ $$V=|\\textbf{w}\\cdot(\\textbf{u}\\times\\textbf{v})|\\ \\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 20d81635-0ae9-45ff-8a15-9034ee7b9fa9
encoding: FLOAT32
metadata:
{"chroma:document": "Proof.: (a) follows immediately from Theorem 1.6 and (b) from (a). To prove (c), we may assume **u** and **v** lie on the $xy$-plane after a rotation. The volume of the parallelepiped is given by the product of the area of the parallelogram spanned by **u** and **v** with its height. Now $|\\textbf{u}\\times\\textbf{v}|$ is equal to the area of this parallelogram. On the other hand, its height is given by $|\\textbf{w}\\cdot\\textbf{e}_{3}|$. Therefore, letting $\\alpha$ be the angle between **w** and the"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: f9e82580-4de2-4c28-8dc0-55b651bd2d02
encoding: FLOAT32
metadata:
{"chroma:document": "$$|\\textbf{w}\\cdot(\\textbf{u}\\times\\textbf{v})| = |\\textbf{w}||\\textbf{u}\\times\\textbf{v}||\\cos\\alpha|$$ $$= |\\textbf{u}\\times\\textbf{v}|\\ |\\textbf{w}\\cdot\\textbf{e}_{3}|$$ $$= V\\.$$\n\n**Example 1.5**.: Determine if the four points\n\n$$(1,0,1),\\ (2,4,-6),\\ (3,-1,5),\\ (1,-9,19)\\,$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: b4e802ef-6b86-41e0-8b37-34b4505bfc9b
encoding: FLOAT32
metadata:
{"chroma:document": "$$(1,0,1),\\ (2,4,-6),\\ (3,-1,5),\\ (1,-9,19)\\,$$\n\nlie on the same plane in $\\mathbb{R}^{3}$. Well, they lie on the same plane if and only if the parallelepiped formed by these vectors has zero volume. We compute the volume using this corollary after subtracting the first vector from the last three vectors (to make sure that the vectors are based at the origin):\n\n$$(1,4,-7)\\cdot\\big{(}(2,-1,4)\\times(0,-9,18)\\big{)} = (1,4,-7)\\cdot(18,-36,-18)$$ $$= 0\\,$$\n\nso they lie on the same plane."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 5dc1f2c1-dcb5-4b6e-ba05-a79f762bfa77
encoding: FLOAT32
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{"chroma:document": "so they lie on the same plane.\n\n**Comments on Chapter 1.**\n\n**1.1.** A helping hand from linear algebra. In several occasions we need to solve homogeneous systems of linear equations. Let us review it by looking at some examples. First, consider the single equation\n\n$$x-2y+6z=0,\\quad(x,y,z)\\in\\mathbb{R}^{3}\\.$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: d37cf377-3362-461f-90ba-3e56b5e3ac42
encoding: FLOAT32
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{"chroma:document": "To solve this equation means to find all possible $(x,y,z)$ satisfying this relation. Obviously $(0,0,0)$ is a solution, but there are many others, for instance, $(-6,0,1)$ and $(2,1,0)$ are also solutions. To find all solutions, we set $y=a$ and $z=b$. Then $x=2a-6b$, so $(2a-6b,a,b)=a(2,1,0)+b(-6,0,1)$ gives all solutions. A solution is obtained whenever values to $a$ and $b$ are assigned. We may say the solution is described by two parameters. Next, consider the system"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 2c6b96f0-ffbf-4a64-be13-5c326dac2281
encoding: FLOAT32
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{"chroma:document": "$$\\begin{cases}x-y+5z&=0\\,\\\\ x+y-3z&=0\\.\\end{cases}$$\n\nSetting $z=a$, the system becomes\n\n$$\\begin{cases}x-y&=-5a\\,\\\\ x+y&=3a\\,\\end{cases}$$\n\nwhich is readily solved to yield $x=-a$ and $y=4a$, so the general solution is given by $(x,y,z)=a(-1,4,1)$ where $a$ is the only parameter. From these two examples, we see there are three principles governing homogeneous linear systems."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 2ed33a5c-5c24-41e4-87c2-bbae32e11039
encoding: FLOAT32
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{"chroma:document": "* $\\mathbf{0}$ is always a solution (the trivial solution).\n* The general solution consists of several free parameters. In most cases, the number of parameters is equal to $n-m$ where $n$ is the number of unknowns and $m$ is the number of equations.\n* In some exceptional cases, the number of parameters is less than $n-m$.\n\nExceptional cases come up when the linear system is kind of cheating us. For instance, look at\n\n$$\\begin{cases}x-y+5z&=0\\,\\\\ 2x-2y+10z&=0\\.\\end{cases}$$"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 9717cf65-aa08-4465-b335-cbfc5e3f77cf
encoding: FLOAT32
metadata:
{"chroma:document": "$$\\begin{cases}x-y+5z&=0\\,\\\\ 2x-2y+10z&=0\\.\\end{cases}$$\n\nThe second equation in this system comes from multiplying the first equation by 2, so essentially there is only one equation in this system. Its general solution contains two instead of one parameters. This situation occurs in the study of standard forms in Section 2.4.\n\n**1.2.** The Law of Cosines states that, let $\\Delta ABC$ be a triangle and $a=\\overline{CB},\\ b=\\overline{AC}$ and $c=\\overline{BA}$, and $\\phi=\\angle ACB$. Then"}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 705fd0ab-1661-407d-84d5-d31d9195997f
encoding: FLOAT32
metadata:
{"chroma:document": "$$c^{2}=a^{2}+b^{2}-2ab\\cos\\phi\\.$$\n\nTo prove it let $h=\\overline{AH}$ be the height of the triangle from $A$ and $s$ be $\\overline{CH}$. By Pythagoras Theorem,\n\n$$c^{2}=h^{2}+(a-s)^{2}=b^{2}-s^{2}+(a-s)^{2}=b^{2}-2as+a^{2}\\,$$\n\nimplies $s=(c^{2}-a^{2}-b^{2})/2a$ and the Cosine Law follows after noting $s=b\\cos\\phi$."}


created_at: 2024-04-20 10:01:57
topic: persistent://default/default/7d7d9584-af30-44ee-b496-95b182965ea2
id: 78b277f5-31ec-413f-91e0-1510481af813
encoding: FLOAT32
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{"chroma:document": "**1.3.** You may look up Wikipedia under \"Euclidean motion\" and \"orthogonal matrix\" to find further information on these interesting topics. How the cross product is used in physics can be also be found in Wikipedia under \"cross product\". From the same source you will see how the cross product arises from the Lie algebra of the orthogonal group.\n\n**Supplementary Reading**\n\n1.1 and 1.2 in chapter 1, [Au]."}


# embedding_metadata

key: chroma:document
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The objects of study in advanced calculus are differentiable functions of several variables. To set the stage for the study, the Euclidean space as a vector space endowed with the dot product is defined in Section 1.1. To aid visualizing points in the Euclidean space, the notion of a vector is introduced in Section 1.2. In Section 1.3 Euclidean motions, mappings preserving the Euclidean distance, are briefly discussed. The last Section 1.4 contains a discussion on the cross product which is only defined


key: chroma:document
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a discussion on the cross product which is only defined for vectors in the three dimensional Euclidean space, that is, our physical space.


key: chroma:document
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### 1.1 The Dot Product

An $n$-tuple is given by

$${\bf x}=(x_{1},x_{2},\cdots,x_{n})\,\quad x_{j}\in\mathbb{R},\ \ j=1,\cdots,n\.$$

It is called an ordered pair when $n=2$. Denote by $\mathbb{R}^{n}$ the collection of all $n$-tuples. The zero $n$-tuple, $(0,0,\cdots,0)$, will be written as ${\bf 0}$ from time to time. There are two algebraic operations defined on $\mathbb{R}^{n}$, namely, the addition

$${\bf x}+{\bf y}\ =(x_{1}+y_{1},x_{2}+y_{2},\cdots,x_{n}+y_{n})\,$$

and the scalar multiplication


key: chroma:document
string_value:
and the scalar multiplication

$$\alpha{\bf x}\ =(\alpha x_{1},\alpha x_{2},\cdots,\alpha x_{n})\,\quad\alpha \in\mathbb{R}\,$$

where

$${\bf x}=(x_{1},x_{2},\cdots,x_{n}),\quad{\bf y}=(y_{1},y_{2},\cdots,y_{n})\.$$

Recall that the ordinary multiplication assigns a number as the product of two numbers, so it can be regarded as a map from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$. One may expect a multiplication on

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One has no difficulty in verifying the dot product satisfies these three axioms. Alternatively one may use $\langle\textbf{x},\textbf{y}\rangle$ to denote $\textbf{x}\cdot\textbf{y}$. We will do this to avoid confusion occasionally. Note that $\textbf{x}\cdot\textbf{y}=\textbf{y}\cdot\ \textbf{x}$, so the dot product between **x** and **y** is the same as the dot product between **x** and **y**. Things would be very different when we study the cross product later.


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At this point let us make a digression to establish a fundamental inequality. Some of you may already learn this inequality, but its interesting proof is worth to go through once more.

**Theorem 1.1** (**Cauchy-Schwarz Inequality**).: _For $\textbf{x},\textbf{y}\in\mathbb{R}^{n}$,_

$$\left|\sum_{j=1}^{n}x_{j}y_{j}\right|\leq\sqrt{\sum_{j=1}^{n}x_{j}^{2}}\sqrt{ \sum_{j=1}^{n}y_{j}^{2}}\.$$


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_Furthermore, equality sign holds if and only if either one of $\textbf{x},\textbf{y}$ is zero $n$-tuple or there is some $\alpha\neq 0$ such that $\textbf{y}=\alpha\textbf{x}.$_

The condition $\textbf{x}=\alpha\textbf{y}$ means **x** and **y** are proportional to each other. Using the language of linear algebra, the equality condition is simply **x** and **y** are linearly dependent.

Proof.: First assume not all $x_{j}$'s are zero in **x**. Consider the expression


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$$\sum_{j=1}^{n}(x_{j}t-y_{j})^{2}\,$$

which is a sum of squares and so must be non-negative for all $t\in\mathbb{R}$. We can express it as a quadratic polynomial in $t$ as

$$p(t)\equiv at^{2}-2bt+c\,$$

where

$$a=\sum_{j=1}^{n}x_{j}^{2}\,\quad b=\sum_{j=1}^{n}x_{j}y_{j},\quad c=\sum_{j=1} ^{n}y_{j}^{2}\.$$


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Since $a>0$, $p(t)$ tends to $\infty$ as $t\rightarrow\pm\infty$. Therefore, it is non-negative if and only if its discriminant is non-positive, that is, $4b^{2}-4ac\leq 0$, which yields $|b|\leq\sqrt{ac}$ after taking square root. Our inequality follows. Moreover, the equality sign holds if and only if $4b^{2}-4ac=0$. In this case the quadratic equation $at^{2}-2bt+c=0$ has a (double) root, say, $t_{1}$. Going back to the original expression, we have


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$$\sum_{j=1}^{n}(x_{j}t_{1}-y_{j})^{2}=0\,$$which forces $t_{1}x_{j}=y_{j}$ for all $j=1,\cdots,n$. So we can take $\alpha=t_{1}$ in case $c=\sum_{j}y_{j}^{2}>0$.

When all $x_{j}$'s vanish but not all $y_{j}$'s, we exchange $\mathbf{x}$ and $\mathbf{y}$ to get the same conclusion.

Finally, when all $x_{j}$'s and $y_{j}$'s vanish, the inequality clearly holds.


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Accompanying with the notion of the inner product are those of the norm and the distance. Indeed, the **Euclidean norm** of an $n$-tuple is defined to be

$$|\mathbf{x}| = (\mathbf{x}\cdot\mathbf{x})^{1/2}$$ $$= \left(\sum_{j=1}^{n}x_{j}^{2}\right)^{1/2}$$ $$= \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots x_{n}^{2}}\.$$

The **Euclidean distance** between $\mathbf{x}$ and $\mathbf{y}$ is defined by

$$|\mathbf{x}-\mathbf{y}|=\sqrt{(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+\cdots+(x_{ n}-y_{n})^{2}}\.$$


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In terms of these notions, Cauchy-Schwarz Inequality can be rewritten in a compact form

$$|\mathbf{x}\cdot\mathbf{y}|\leq|\mathbf{x}||\mathbf{y}|\.$$

In mathematics, a distance is a rule to assign a non-negative number to any pair of elements in a set under consideration. The rule consists of three "axioms": For $a,b,c$ in this set,

* $d(a,b)\geq 0\,$ and equal to 0 iff $a=b$,
* $d(a,b)=d(b,a)$, and
* $d(a,b)\leq d(a,c)+d(c,b)$.


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Now, taking $d(\mathbf{x},\mathbf{y})=|\mathbf{x}-\mathbf{y}|$, we see that it satisfies all these three axioms: $\mathbf{x},\mathbf{y},\mathbf{z}\in\mathbb{R}^{n}$,

* $|\mathbf{x}-\mathbf{y}|\geq 0$ and equal to 0 if and only if $\mathbf{x}=\mathbf{y}$,
* $|\mathbf{x}-\mathbf{y}|=|\mathbf{y}-\mathbf{x}|$,
* $|\mathbf{x}-\mathbf{y}|\leq|\mathbf{x}-\mathbf{z}|+|\mathbf{z}-\mathbf{y}|$.


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### 1.1 The Dot Product

Indeed, (a) and (b) are obvious. To prove (c), write ${\bf u}={\bf x}-{\bf z},{\bf v}={\bf z}-{\bf y}$ to get

$$|{\bf u}+{\bf v}|\leq|{\bf u}|+|{\bf v}|\,$$


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$$|{\bf u}+{\bf v}|\leq|{\bf u}|+|{\bf v}|\,$$

which holds after taking square of both sides and then applying the Cauchy-Schwarz Inequality. I let you work out the details in the exercise. Note that $|{\bf x}|=|{\bf x}-{\bf 0}|$, so the norm of ${\bf x}$ is its distance to the zero $n$-tuple. In these notes, norm and distance are referred to Euclidean norm and Euclidean distance without further specification.


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Note that the same notation $|x|$ stands for the absolute value of $x$ when $x$ is a real number. The norm of ${\bf x}$ is the same as its absolute value when $n=1$. When $n\geq 2$, the notation $|{\bf x}|$ stands for the norm only as there is no such a definition of absolute value for an $n$-tuple. The notation $\|{\bf x}\|$ is also used to denote the norm of ${\bf x}$, but it will not be used here.


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Recall that the cosine function $\cos t$ is strictly decreasing from $1$ to $-1$ as $t$ goes from $0$ to $\pi$. Keeping this in mind, we are going to define the angle between two non-zero $n$-tuples. By Cauchy-Schwarz Inequality, the absolute value of the expression ${\bf x}\cdot{\bf y}/|{\bf x}||{\bf y}|$ lies in the interval $[-1,1]$. Therefore, by what we just said, there exists a unique $\theta\in[0,\pi]$ satisfying $\cos\theta={\bf x}\cdot{\bf y}/|{\bf x}||{\bf y}|$, that is,


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We define the **angle** between two non-zero n-tuples ${\bf x}$ and ${\bf y}$ to be $\theta$ where ${\bf x},{\bf y}$ are arbitrary $n$-tuples. The angle between two $n$-tuples makes no sense when of them is zero. By definition this angle must belong to $[0,\pi]$. Moreover, it is symmetric, that is, the angle between ${\bf x}$ and ${\bf y}$ is the same as the angle between ${\bf y}$ and ${\bf x}$. At this stage, the notion of an angle is defined purely in an analytical manner and does not bear any geometric


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purely in an analytical manner and does not bear any geometric meaning. We will link it to geometry in the next section.


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Two $n$-tuples ${\bf x}$ and ${\bf y}$ are **perpendicular** or **orthogonal** to each other if ${\bf x}\cdot{\bf y}=0$. In terms of the angle, they are perpendicular if and only if their angle is $\pi/2$. The zero $n$-tuple is perpendicular to all $n$-tuples. By Cauchy-Schwarz Inequality, we also know that two non-zero ${\bf x}$ and ${\bf y}$ satisfy ${\bf x}=c{\bf y}$ for some $c>0$ when their angle $\theta=0$ and satisfy ${\bf x}=c{\bf y},c<0$ when $\theta=\pi$.


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**Example 1.1**.: Find all $n$-tuples ${\bf x}$ that are perpendicular to $(1,-1,2)$ and $(-1,0,3)$. These points satisfy

$$(1,-1,2)\cdot{\bf x}=0\,\quad(-1,0,3)\cdot{\bf x}=0\,$$

that is, the linear system

$$\begin{cases}x-y+2z&=0\,\\ -x&+3z&=0\.\end{cases}$$We solve this system (see Comments at the end of this chapter) to get ${\bf x}=(x,y,z)=a(3,5,1),a\in\mathbb{R}$. By varying $a$, we obtain infinitely many solutions.


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Summing up, we have defined the Euclidean space $(\mathbb{R}^{n},+,\cdot,\langle\cdot,\cdot\rangle)$ which is the set of all $n$-tuples with two algebraic operations-- the addition and the scalar multiplication-- as well as the dot product. From now on it will be abbreviated in a single symbol $\mathbb{R}^{n}$.


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### 1.2 Vector Representation

Visualizing $n$-tuples for $n=2,3$ as vectors has been used widely in physics and engineering. In this section we discuss how to do it. However, despite its convenience and usefulness, it should be understood that the notion of a vector is only an auxiliary tool. Analysis on the Euclidean space can be carried out without referring to this notion.


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To start off, imagine that the coordinates axes have been introduced in the plane. The $x$-axis consists of all ordered pairs of the form $(x,0),\ x\in\mathbb{R}$, and the $y$-axis all ordered pairs of the form $(0,y),\ y\in\mathbb{R}$. So every ordered pair $(x,y)$ can be written as $x(1,0)+y(0,1)$ and $(x,y)$ is a point in the coordinate plane. Points in the plane are in one-to-one correspondence with ordered pairs. The same idea applies to all other dimensions. However, since the two dimensional case is


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other dimensions. However, since the two dimensional case is easy to see and the three dimensional case can be seen with some imagination, in the following we will focus on these two spaces. It will be apparent that most of our discussions can be extended to all dimensions.


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For a point $(x,y)$ in the plane, it associates to a **vector** which is an arrow pointing from the base $(0,0)$ to the tip $(x,y)$. The vector degenerates into a point when it is the zero ordered pair $(0,0)$. We call it the **zero vector** and denoted it by $(0,0)$ or simply ${\bf 0}$.


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After the introduction of the vector for an ordered pair, we interpret the algebraic operations of $\mathbb{R}^{2}$ as follows. Indeed, by drawing pictures, it is not hard to convince oneself that the addition of two ordered pairs is accomplished by the parallelogram law. Specifically, first form the parallelogram using the two vectors as line segments corresponding to the two points. Then the diagonal of this resulting parallelogram, regarded as a vector pointing from the origin to the other end, is the


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as a vector pointing from the origin to the other end, is the sum of these two vectors. The same situation holds in $n\geq 3$, as one can always restrict to the plane containing these two vectors provided they are linearly independent. When they are linearly dependent, the geometric interpretation is apparent.


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The scalar multiplication $(x,y)\mapsto\alpha(x,y),\alpha>0$, means changing the vector by a scale of $\alpha$ along the same direction. It is a prolongation if $\alpha>1$ and a shortening if $\alpha\in(0,1)$


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### 1.2 Vector Representation

On the other hand, when $\alpha<0$, the resulting vector points in the opposite direction of the original vector with size changes equal to $|\alpha|$.

How about substraction of two vectors? Let ${\bf w}={\bf v}-{\bf u}$. Then ${\bf w}$ can be obtained by first drawing the triangle with vertices at $(0,0),{\bf u}$ and ${\bf v}$ and then translate the side from ${\bf u}$ to ${\bf v}$ so that its base is located at the origin. The translated side is ${\bf w}$.


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We may also find the midpoint of two ordered pairs. For ${\bf u},{\bf v}$, its midpoint is given by $({\bf u}+{\bf v})/2$. Regarding as a vector, this midpoint can be described in the following way. First draw the parallelogram formed by ${\bf u}$ and ${\bf v}$. Then the intersection point of the two diagonal lines of this parallelogram is the tip of the midpoint (vector).


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When we regard an $n$-tuple ${\bf x}$ as a vector, it is more convenient to call its norm the **magnitude** of the vector. It is a **unit vector** if its magnitude is equal to 1. Likewise, the distance between two points may be called the **length** of the line segment connecting ${\bf x}$ and ${\bf y}$. It is consistent with the classical Pythagoras theorem. In fact, the definition of the Euclidean norm and distance were inspired by this classical theorem.


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Next we show that the angle defined in the last section, which purely depends on analytical terms, is the same as the "geometric angle". To see it, let ${\bf x}=(a,b)$ and ${\bf y}=(c,d)$ be two non-zero vectors in the plane. By the Law of Cosines in trigonometry (see Comments at the end of this chapter),

$$(c-a)^{2}+(d-b)^{2}=(a^{2}+b^{2})+(c^{2}+d^{2})-2\sqrt{c^{2}+d^{2}}\sqrt{a^{2} +b^{2}}\cos\phi\,$$


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where $\phi\in[0,\pi]$ is the "geometric angle" between ${\bf x}$ and ${\bf y}$. Simplifying, we have

$$-2(ac+db)=-2\sqrt{c^{2}+d^{2}}\sqrt{a^{2}+b^{2}}\cos\phi\,$$

which is equal to

$${\bf x}\cdot{\bf y}=|{\bf x}||{\bf y}|\cos\phi\.$$


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$${\bf x}\cdot{\bf y}=|{\bf x}||{\bf y}|\cos\phi\.$$

Comparing with the definition of $\theta$, we have $\cos\phi=\cos\theta$ so that $\phi=\theta$. In other words, the geometric angle coincides with the analytical angle. The same argument works in higher dimensions as we can restrict to the plane containing any two given vectors.


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A vector is uniquely determined by its magnitude and direction. To be more precise we fix them in definition. Any vector with unit length is called a **direction.** Each non-zero vector ${\bf x}$ can be written as

$${\bf x}=|{\bf x}|\ \mathbf{\xi},$$

where $|{\bf x}|$ is its magnitude and

$$\mathbf{\xi}=\frac{{\bf x}}{|{\bf x}|}$$its direction. Every direction $\boldsymbol{\xi}=(\xi_{1},\xi_{2},\cdots,\xi_{n})$ can further be expressed as


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$$\boldsymbol{\xi}=(\cos\alpha_{1},\cos\alpha_{2},\cdots,\cos\alpha_{n})\,$$

where $\alpha_{k}\in[0,\pi]$ are called the **direction angles** of $\boldsymbol{\xi}$. From $\boldsymbol{\xi}\cdot\mathbf{e}_{k}=\cos\alpha_{k}$ we see that $\alpha_{k}$ is the angle between $\boldsymbol{\xi}$ and the $e_{k}$-axis. These $\cos\alpha_{k}$'s are called the **direction cosines** of $\mathbf{x}$.


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**Example 1.2**.: Find the magnitude and direction of $(1,2,-7)$ and determine the vector $(2,a,6)$ that is perpendicular to $(1,2,-7)$. The magnitude of $(1,2,-7)$ is

$$|(1,2,-7)|=\sqrt{1^{2}+2^{2}+(-7)^{2}}=\sqrt{54}\,$$

and its direction is $(1,2,-7)/\sqrt{54}$. By orthogonality,

$$0=(1,2,-7)\cdot(2,a,6)=2+2a-42=0\,$$

which implies $a=20$. The vector $(2,20,6)$ is perpendicular to $(1,2,-7)$.


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One may also consider the **vector from the initial point $\mathbf{x}$ to the terminal point $\mathbf{y}$,** or **the vector based at some point**. Unlike a vector, a vector from $\mathbf{x}$ to $\mathbf{y}$ is an arrow whose base and tip are $\mathbf{x}$ and $\mathbf{y}$ respectively. Obviously such a "vector" is parallel to the position vector of $\mathbf{y}-x$ whose base is now at the origin. The length and direction of a vector from $\mathbf{x}$ to $\mathbf{y}$ are defined as the respective length and


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**Example 1.3**.: Consider the triangle with vertices at $(1,2),(3,4),(0,-1)$. Find the direction of the vector pointing at the midpoint of the side connecting $(1,2)$ and $(3,4)$ from $(0,-1)$. Well, first we translate $(0,-1)$ to the origin so that the triangle is congruent to the one whose vertices are $((1,2)-(0,-1),(3,4)-(0,-1),(0,-1)-(0,-1)$, that is, $(1,3),(3,5),(0,0)$. The midpoint of the side from $(1,3)$ and $(3,5)$ is given by

$$\frac{1}{2}\left((1,3)+(3,5)\right)=(2,4)\,$$


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$$\frac{1}{2}\left((1,3)+(3,5)\right)=(2,4)\,$$

and its direction is given by

$$\frac{(2,4)}{\sqrt{2^{2}+4^{2}}}=\frac{(2,4)}{\sqrt{20}}=\frac{(1,2)}{\sqrt{5 }}\.$$

(No need to simplify further.)

**Example 1.4**.: (a) Find the magnitude and direction of the vector from $(1,-1)$ to $(-2,5)$. (b) Find all directions that are perpendicular to the vector in (a).


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The magnitude and direction of the vector from $(1,-1)$ to $(-2,5)$ are the same as those of the position vector $(-2,5)-(1,-1)=(-3,6)$. Its magnitude is given by

$$|(-3,6)|=\sqrt{(-3)^{2}+6^{2}}=3\sqrt{5}\,$$

and the direction is given by

$$\frac{(-3,6)}{3\sqrt{5}}=\frac{(-1,2)}{\sqrt{5}}\.$$

(No need to simplify further.)

A vector $(a,b)$ perpendicular to $(-3,6)$ satisfies

$$(-3,6)\cdot(a,b)=-3a+6b=0\.$$


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$$(-3,6)\cdot(a,b)=-3a+6b=0\.$$

By varying $a$ and $b$ according to this relation, there are infinitely many vectors $(a,b)$ satisfying this requirement. For instance, we may take $a=2,b=1$ so $(2,1)$ is one choice. However, to be a direction there is another condition, namely, the length of the vector has to equal to one. There are two such vectors:

$$\frac{(2,1)}{\sqrt{5}}\,\quad-\frac{(2,1)}{\sqrt{5}}\.$$

(Again no need to simplify.)


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### 1.3 Euclidean Motions

A **Euclidean motion** is a map from $\mathbb{R}^{n}$ to itself of the form

$$T\mathbf{x}=A\mathbf{x}+\mathbf{b}\,$$

where $\mathbf{b}\in\mathbb{R}^{n}$ and $A$ is an $n\times n$-matrix, that preserves the distance between two points, that is, for $\mathbf{x},\mathbf{y}\in\mathbb{R}^{n}$,

$$|T\mathbf{x}-T\mathbf{y}|=|\mathbf{x}-\mathbf{y}|\.$$

Here in $A\mathbf{x}$ the vector $\mathbf{x}$ should be understood as a column vector.


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Recall that a square matrix $R$ is called an orthogonal matrix if $R^{\prime}R=RR^{\prime}=I$, where $R^{\prime}$ is the transpose of $R$ and $I$ is the identity matrix.

**Proposition 1.2**.: _A map $T\mathbf{x}=A\mathbf{x}+\mathbf{b}$ is a Euclidean motion if and only if $A$ is an orthogonal matrix._


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Proof.: In the following we use $\langle\mathbf{x},\mathbf{y}\rangle$ instead $\mathbf{x}\cdot\mathbf{y}$ to denote the dot product. First of all, let $T$ be a Euclidean motion. Then it follows from the definition that

$$|\mathbf{x}-\mathbf{y}|=|T\mathbf{x}-T\mathbf{y}|=|A\mathbf{x}-A\mathbf{y}|= |A(\mathbf{x}-\mathbf{y})|,$$which yields immediately that


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$$|A({\bf x}-{\bf y})|^{2} = \langle A({\bf x}-{\bf y}),A({\bf x}-{\bf y})\rangle$$ $$= |{\bf x}-{\bf y}|^{2}$$ $$= |{\bf x}|^{2}-2\langle{\bf x},{\bf y}\rangle+|{\bf y}|^{2}\.$$

On the other hand, a direct calculation shows that

$$\langle A({\bf x}-{\bf y}),A({\bf x}-{\bf y})\rangle = \langle A{\bf x}-A{\bf y},A{\bf x}-A{\bf y}\rangle$$ $$= |A{\bf x}|^{2}-2\langle A{\bf x},A{\bf y}\rangle+|A{\bf y}|^{2}.$$

By comparing, we see that for all ${\bf x},{\bf y}$,


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By comparing, we see that for all ${\bf x},{\bf y}$,

$$\langle A^{\prime}A{\bf x},{\bf y}\rangle=\langle A{\bf x},A{\bf y}\rangle= \langle{\bf x},{\bf y}\rangle\,$$

which implies that $A^{\prime}A=I$. Thus $A$ is a orthogonal matrix. Finally, this relation also shows that $T$ is a Euclidean motion whenever $A$ is orthogonal. $\Box$

Here we have used the following derivation in linear algebra: For a matrix $(B{\bf x})_{j}=\sum_{k}b_{kj}x_{k}$,


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$$\langle{\bf x},B{\bf y}\rangle=\langle B^{\prime}{\bf x},{\bf y}\rangle\.$$

Indeed,

$$\langle{\bf x},B{\bf y}\rangle = \sum_{j}x_{j}\sum_{k}b_{kj}y_{k}$$ $$= \sum_{k}\sum_{j}b_{kj}x_{j}y_{k}$$ $$= \sum_{k}y_{k}\sum_{j}b^{\prime}_{jk}x_{j}=\langle B^{\prime}{ \bf x},{\bf y}\rangle\.$$

Here are some examples of Euclidean motions.


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1. Take $A$ to be the identity and ${\bf b}$ a nonzero vector. Then $T{\bf x}={\bf x}+{\bf b}$ is a translation. The origin is moved to ${\bf b}$ after the motion.


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2. When $n=2$, the Euclidean motion $$T{\bf x}=\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}$$ is the reflection with respect to the $x$-axis and $$T{\bf x}=\begin{bmatrix}-1&0\\ 0&1\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}$$is the reflection with respect to the $y$-axis. (In matrix form the vector $\mathbf{x}$ is understood as a column vector.) On the other hand, given any plane in $\mathbb{R}^{3}$, one may consider the reflection with respect to this plane.


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one may consider the reflection with respect to this plane. For instance, $$T\mathbf{x}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&-1\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}$$ is the reflection with respect to the $xy$-plane in $\mathbb{R}^{3}$. The reflection with respect to any straight line in $\mathbb{R}^{2}$ or with respect to any plane in $\mathbb{R}^{3}$ can be defined similarly.


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3. The (counterclockwise) rotation of angle $\theta$ in $\mathbb{R}^{2}$ is given by the Euclidean motion $$T\mathbf{x}=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}\,\quad\theta\in(0,2\pi)\.$$ In $\mathbb{R}^{3}$, one can perform a rotation around a fixed axis. For instance, the rotation $$T\mathbf{x}=\begin{bmatrix}\cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}$$


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Let us verify that Euclidean motions are closed under compositions. Let $T\mathbf{x}=A\mathbf{x}+\mathbf{b}$ and $S\mathbf{x}=B\mathbf{x}+\mathbf{c}$ be two Euclidean motions. Its composition is given by

$$ST\mathbf{x}=B(A\mathbf{x}+\mathbf{b})+\mathbf{c}=C\mathbf{x}+\mathbf{d}\, \quad C\equiv BA,\quad\mathbf{d}=B\mathbf{b}+\mathbf{c}\.$$

As

$$C^{\prime}C = (BA)^{\prime}BA$$ $$= A^{\prime}B^{\prime}BA$$ $$= A^{\prime}IA$$ $$= I,$$


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we conclude that $ST$ is again a Euclidean motion. Furthermore, we claim that each Euclidean motion admits an inverse. Indeed, letting $U\mathbf{x}=A^{-1}\mathbf{x}-A^{-1}\mathbf{b}$ which is obviously an Euclidean motion, we have

$$UT\mathbf{x}=A^{-1}(A\mathbf{x}+\mathbf{b})-A^{-1}\mathbf{b}=\mathbf{x}\.$$


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Summing up, the collection of all Euclidean motions forms a group called the Euclidean group of $\mathbb{R}^{n}$. (It is alright if you have not learned what a group is. You will learn it in MATH2070.)

In the following we study the structure of Euclidean motions for $n=2,3$. Apparently it suffices to look at the orthogonal matrix $A$.

**Theorem 1.3**.: _In $\mathbb{R}^{2}$, every orthogonal matrix can be written as_


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1. $$\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}\,$$
2. $$\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\,\quad\theta\in[0,2\pi)\.$$

_Case (a) is a genuine rotation for $\theta\in(0,2\pi)$ and reduces to the identity at $\theta=0$. Case (b) is the reflection with respect to the $x$-axis and then followed by a rotation._

Proof.: Let

$$A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\.$$


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Proof.: Let

$$A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\.$$

By orthogonality $A^{\prime}A=I$ we have

$$a^{2}+c^{2}=1\,\quad b^{2}+d^{2}=1\,\quad ab+cd=0\.$$

Since $a$ is a number between $-1$ and $1$, we can find a unique $\theta\in[0,2\pi)$ such that $a=\cos\theta,\ c=\sin\theta$. Then either $b=-\sin\theta,\ d=\cos\theta$ or $b=\sin\theta,d=-\cos\theta$, so (a) or (b) must hold.


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In the following we consider the three dimensional case. Denote by $R_{z}(\theta)$ the rotation around the $z$-axis by an angle $\theta$:

$$R_{z}(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1\end{bmatrix}\.$$

Similarly we define

$$R_{x}(\theta)=\begin{bmatrix}1&0&0\\ 0&\cos\theta&-\sin\theta\\ 0&\sin\theta&\cos\theta\end{bmatrix}\,$$

and

$$R_{y}(\theta)=\begin{bmatrix}\cos\theta&0&-\sin\theta\\ 0&1&0\\ \sin\theta&0&\cos\theta\end{bmatrix}\.$$


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Also denote by $L_{z}$ the reflection with respect to the $xy$-plane:

$$L_{z}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&-1\end{bmatrix}\.$$Similarly

$$L_{x}=\begin{bmatrix}-1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}\,$$

and

$$L_{y}=\begin{bmatrix}1&0&0\\ 0&-1&0\\ 0&0&1\end{bmatrix}\.$$

**Theorem 1.4**.: _* In $\mathbb{R}^{3}$, every orthogonal matrix can be written as_

1. $R_{z}(\alpha)R_{x}(\beta)R_{z}(\gamma)$__, _or_
2. $R_{z}(\alpha)R_{x}(\beta)R_{z}(\gamma)L_{z}$__,_

_for some $\alpha,\beta,$ and $\gamma$._


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_for some $\alpha,\beta,$ and $\gamma$._

Proof.: * Let $A=\left(a_{ij}\right),i,j=1,2,3,$ be orthogonal. We have

$$\begin{bmatrix}\cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{bmatrix}$$


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$$=\begin{bmatrix}\cos\theta a_{11}-\sin\theta a_{21}&\cos\theta a_{12}-\sin \theta a_{22}&\cos\theta a_{13}-\sin\theta a_{23}\\ \sin\theta a_{11}+\cos\theta a_{21}&\sin\theta a_{12}+\cos\theta a_{22}&\sin \theta a_{13}+\cos\theta a_{23}\\ a_{31}&a_{32}&a_{33}\end{bmatrix}\.$$

Choose $\theta$ so that $\cos\theta a_{13}-\sin\theta a_{23}=0$ and write the resulting matrix as

$$\begin{bmatrix}b_{11}&b_{12}&0\\ b_{21}&b_{22}&b_{23}\\ b_{31}&b_{32}&b_{33}\end{bmatrix}\.$$

We further have


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We further have

$$=\begin{bmatrix}b_{11}&b_{12}&0\\ \cos\varphi b_{21}-\sin\varphi b_{31}&\cos\varphi b_{22}-\sin\varphi b_{32}& \cos\varphi b_{23}-\sin\varphi b_{33}\\ \sin\varphi b_{21}+\cos\varphi b_{31}&\sin\varphi b_{22}+\cos\varphi b_{32}& \sin\varphi b_{23}+\cos\varphi b_{33}\end{bmatrix}\.$$

Choose $\varphi$ so that $\cos\varphi b_{23}-\sin\varphi b_{33}=0$ and write the resulting matrix as


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$$\begin{bmatrix}c_{11}&c_{12}&0\\ c_{21}&c_{22}&0\\ c_{31}&c_{32}&c_{33}\end{bmatrix}\.$$This matrix is the product of three orthogonal matrices, again it is orthogonal. Therefore, $c_{33}=\pm 1$. Moreover, from

$$c_{11}\times 0+c_{21}\times 0+c_{31}\times c_{33}=0\,$$

we deduce $c_{31}=0$. Similarly, $c_{32}=0$. The matrix is in fact of the form

$$\begin{bmatrix}c_{11}&c_{12}&0\\ c_{21}&c_{22}&0\\ 0&0&\pm 1\end{bmatrix}\,$$


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where the $2\times 2$-matrix is orthogonal. It can be written as $R_{z}(\gamma)$ or $R_{z}(\gamma)L_{z}$ for some $\gamma$ according to Proposition 1.3. We conclude that

$$R_{x}(\varphi)R_{z}(\theta)A=R_{z}(\gamma)\,$$

or

$$R_{x}(\varphi)R_{z}(\theta)A=R_{z}(\gamma)L_{z}\,$$

that is,

$$A=R_{z}(-\theta)R_{x}(-\varphi)R_{z}(\gamma)\,$$

or

$$A=R_{z}(-\theta)R_{x}(-\varphi)R_{z}(\gamma)L_{z}\.$$

The desired result follows by taking $\alpha=-\theta$ and $\beta=-\varphi$.


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### 1.4 The Cross Product in $\mathbb{R}^{3}$

The cross product assigns a 3-vector to two given 3-vectors. There is no such product in the general dimension. Somehow it shows how special our physical space is. The cross product is important due to its relevance in physics and engineering.

Notations like $\mathbf{x},\mathbf{y},\mathbf{u}$ and $\mathbf{v}$ are common for vectors. We have used $\mathbf{x},\mathbf{y}$ in the previous sections. Now we use $\mathbf{u},\mathbf{v}$ in this one.


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First the definition. Let $\mathbf{u},\mathbf{v}\in\mathbb{R}^{3}$, define the cross product of $\mathbf{u}=(u_{1},u_{2},u_{3})$ and $\mathbf{v}=(v_{1},v_{2},v_{3})$ to be

$$\mathbf{u}\times\mathbf{v}=\left(u_{2}v_{3}-u_{3}v_{2},-(u_{1}v_{3}-u_{3}v_{1 }),u_{1}v_{2}-u_{2}v_{1}\right)\.$$

In particular, we have


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In particular, we have

$$\mathbf{e}_{1}\times\mathbf{e}_{2}=\mathbf{e}_{3},\quad\mathbf{e}_{2}\times \mathbf{e}_{3}=\mathbf{e}_{1},\quad\mathbf{e}_{3}\times\mathbf{e}_{1}= \mathbf{e}_{2}\.$$To aid memorizing, we can formally express it as the determinant

$$\left|\begin{array}{ccc}\mathbf{e}_{1}&\mathbf{e}_{2}&\mathbf{e}_{3}\\ u_{1}&u_{2}&u_{3}\\ v_{1}&v_{2}&v_{3}\end{array}\right|\.$$


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Expand the determinant by the first row yields the formula above. Here it is formal because it does not make sense to put the unit vectors $\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}$ as entries in a matrix.

The cross product is in many aspects in sharp contrast with an ordinary product. Some of its properties are list below:

**Theorem 1.5**.: _For $\boldsymbol{u},\boldsymbol{v},\boldsymbol{w}\in\mathbb{R}^{3}$,_


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1. $(\alpha\boldsymbol{u}+\beta\boldsymbol{v})\times\boldsymbol{w}=\alpha \boldsymbol{u}\times\boldsymbol{w}+\beta\boldsymbol{v}\times\boldsymbol{w}\,\quad\forall \alpha,\beta\in\mathbb{R}$_._
2. $\boldsymbol{u}\times\boldsymbol{v}=-\boldsymbol{v}\times\boldsymbol{u}$ _. In particular,_ $\boldsymbol{u}\times\boldsymbol{u}=0$ _._
3. $(\boldsymbol{u}\times\boldsymbol{v})\times\boldsymbol{w}=\boldsymbol{u}\times (\boldsymbol{v}\times\boldsymbol{w})$ _is not always true._


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The proofs of Theorem (a) and (b) are straightforward from definition. As for (c), which asserts that the associative law does not hold, some examples suffice:

$$(\mathbf{e}_{1}\times\mathbf{e}_{2})\times\mathbf{e}_{2}=-\mathbf{e}_{1}\, \quad\mathbf{e}_{1}\times(\mathbf{e}_{2}\times\mathbf{e}_{2})=\boldsymbol{0}\,$$

and

$$(1,1,1)\times\big{(}(1,0,-1)\times(2,1,1)\big{)}=(3,-1,-5)\,\quad\big{(}(1,1,1) \times(1,0,-1)\big{)}\times(2,1,1)=(4,0,-4).$$


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As a vector is completely determined by its magnitude and direction, let us consider the magnitude and direction of the cross product. First of all, using the definition of the cross product, one can verify directly that

$$\mathbf{u}\cdot(\mathbf{u}\times\mathbf{v})=0\,\quad\mathbf{v}\cdot(\mathbf{u} \times\mathbf{v})=0\,$$

so

$$(\alpha\mathbf{u}+\beta\mathbf{v})\cdot\mathbf{u}\times\mathbf{v}=0\,$$


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that is, it is perpendicular to the two dimensional subspace spanned by the vectors $\mathbf{u}$ and $\mathbf{v}$. After the definition of a plane is introduced in the next chapter, we can say that the cross product of two linearly independent vectors points in the normal direction of the plane spanned by $\mathbf{u}$ and $\mathbf{v}$. When they are linearly dependent, their cross product is the zero vector which does not form a normal direction. There are two normal directions, pointing upward or


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direction. There are two normal directions, pointing upward or downward, so to speak. The choice of the direction of the cross product is determined by the right hand rule. That is, with the thumb making a right angle with the other four fingers of your right hand, first point the four fingers along the direction of $\mathbf{u}$ and then move them to $\mathbf{v}$ in an angle less than $\pi$. The direction of is where your thumb points to. To see this, one identifies the direction of $\mathbf{u}$ with


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To see this, one identifies the direction of $\mathbf{u}$ with $\mathbf{e}_{1}$. If $\mathbf{v}$ lies in the first or second quadrants, $\mathbf{v}=\alpha\mathbf{e}_{1}+\beta\mathbf{e}_{2}\,\beta>0$, and $\mathbf{u}\times\mathbf{v}$ points to $\mathbf{e}_{3}$. If $\mathbf{v}$ lies in the third or fourth quadrants, $\mathbf{v}=\alpha\mathbf{e}_{1}+\beta\mathbf{e}_{2}\,\beta<0$, and $\mathbf{u}\times\mathbf{v}$ points to $-\mathbf{e}_{3}$. This is the right hand rule.


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We have described the direction of the cross product. How about its magnitude? We have

**Theorem 1.6**.: _For $\textbf{u},\textbf{v}\in\mathbb{R}^{3}$,_

$$|\textbf{u}\times\textbf{v}|=|\textbf{u}||\textbf{v}|\sin\theta,\quad\theta \in[0,\pi]\,$$

_where $\theta$ is the angle between **u** and **v**._

Proof.: The proof depends on the identity

$$|\textbf{u}\times\textbf{v}|^{2}=|\textbf{u}|^{2}|\textbf{v}|^{2}-(\textbf{u} \cdot\textbf{v})^{2}\.$$

Indeed, by brute force


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Indeed, by brute force

$$|\textbf{u}\times\textbf{v}|^{2} = (u_{2}v_{3}-u_{3}v_{2})^{2}+(u_{1}v_{3}-u_{3}v_{1})^{2}+(u_{1}v_{ 2}-u_{2}v_{1})^{2}$$ $$= u_{2}^{2}v_{3}^{2}+u_{3}^{2}v_{2}^{2}+u_{1}^{2}v_{3}^{2}+u_{3}^{ 2}v_{1}^{2}+u_{1}^{2}v_{2}^{2}+u_{2}^{2}v_{1}^{2}-2u_{2}v_{3}u_{3}v_{2}-2u_{1}v _{3}u_{3}v_{1}-2u_{1}v_{2}u_{2}v_{1}\.$$

On the other hand,


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On the other hand,

$$|\textbf{u}|^{2}|\textbf{v}|^{2}-(\textbf{u}\cdot\textbf{v})^{2}$$ $$= (u_{1}^{2}+u_{2}^{2}+u_{3}^{2})(v_{1}^{2}+v_{2}^{2}+v_{3}^{2})-(u_ {1}v_{1}+u_{2}v_{2}+u_{3}v_{3})^{2}$$ $$= u_{2}^{2}v_{3}^{2}+u_{3}^{2}v_{2}^{2}+u_{1}^{2}v_{3}^{2}+u_{3}^{ 2}v_{1}^{2}+u_{1}^{2}v_{2}^{2}+u_{2}^{2}v_{1}^{2}-2u_{2}v_{3}u_{3}v_{2}-2u_{1} v_{3}u_{3}v_{1}-2u_{1}v_{2}u_{2}v_{1}\,$$

whence the identity holds. Now, by the Cosine Law,


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whence the identity holds. Now, by the Cosine Law,

$$|\textbf{u}\times\textbf{v}| = \sqrt{|\textbf{u}|^{2}|\textbf{v}|^{2}-|\textbf{u}|^{2}|\textbf{v }|^{2}\cos^{2}\theta}$$ $$= |\textbf{u}||\textbf{v}||\sin\theta|$$ $$= |\textbf{u}||\textbf{v}||\sin\theta\,$$

as $\sin\theta\geq 0$ on $[0,\pi]$. 

In conclusion the magnitude and direction of the decomposition of the cross product is given by

$$\textbf{u}\times\textbf{v}=|\textbf{u}||\textbf{v}|\sin\theta\ \textbf{n}\,$$


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where **n** is the unit vector determined by the right hand rule (when **u** and **v** are linearly independent, that is, when $\sin\theta\neq 0$).


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**Corollary 1.7**.:
1. _The area of the parallelogram spanned by_ **u** _and_ **v** _is equal to_ $|\textbf{u}\times\textbf{v}|$_._
2. _The area of the triangle with two sides given by_ $u$ _and_ $v$ _is equal to_ $1/2|\textbf{u}\times\textbf{v}|$_._
3. _The volume of the parallelepiped spanned by_ $u$,_ $v$ _and_ $w$ _is equal to_ $$V=|\textbf{w}\cdot(\textbf{u}\times\textbf{v})|\ \.$$


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Proof.: (a) follows immediately from Theorem 1.6 and (b) from (a). To prove (c), we may assume **u** and **v** lie on the $xy$-plane after a rotation. The volume of the parallelepiped is given by the product of the area of the parallelogram spanned by **u** and **v** with its height. Now $|\textbf{u}\times\textbf{v}|$ is equal to the area of this parallelogram. On the other hand, its height is given by $|\textbf{w}\cdot\textbf{e}_{3}|$. Therefore, letting $\alpha$ be the angle between **w** and the


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$$|\textbf{w}\cdot(\textbf{u}\times\textbf{v})| = |\textbf{w}||\textbf{u}\times\textbf{v}||\cos\alpha|$$ $$= |\textbf{u}\times\textbf{v}|\ |\textbf{w}\cdot\textbf{e}_{3}|$$ $$= V\.$$

**Example 1.5**.: Determine if the four points

$$(1,0,1),\ (2,4,-6),\ (3,-1,5),\ (1,-9,19)\,$$


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$$(1,0,1),\ (2,4,-6),\ (3,-1,5),\ (1,-9,19)\,$$

lie on the same plane in $\mathbb{R}^{3}$. Well, they lie on the same plane if and only if the parallelepiped formed by these vectors has zero volume. We compute the volume using this corollary after subtracting the first vector from the last three vectors (to make sure that the vectors are based at the origin):

$$(1,4,-7)\cdot\big{(}(2,-1,4)\times(0,-9,18)\big{)} = (1,4,-7)\cdot(18,-36,-18)$$ $$= 0\,$$

so they lie on the same plane.


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so they lie on the same plane.

**Comments on Chapter 1.**

**1.1.** A helping hand from linear algebra. In several occasions we need to solve homogeneous systems of linear equations. Let us review it by looking at some examples. First, consider the single equation

$$x-2y+6z=0,\quad(x,y,z)\in\mathbb{R}^{3}\.$$


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To solve this equation means to find all possible $(x,y,z)$ satisfying this relation. Obviously $(0,0,0)$ is a solution, but there are many others, for instance, $(-6,0,1)$ and $(2,1,0)$ are also solutions. To find all solutions, we set $y=a$ and $z=b$. Then $x=2a-6b$, so $(2a-6b,a,b)=a(2,1,0)+b(-6,0,1)$ gives all solutions. A solution is obtained whenever values to $a$ and $b$ are assigned. We may say the solution is described by two parameters. Next, consider the system


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$$\begin{cases}x-y+5z&=0\,\\ x+y-3z&=0\.\end{cases}$$

Setting $z=a$, the system becomes

$$\begin{cases}x-y&=-5a\,\\ x+y&=3a\,\end{cases}$$

which is readily solved to yield $x=-a$ and $y=4a$, so the general solution is given by $(x,y,z)=a(-1,4,1)$ where $a$ is the only parameter. From these two examples, we see there are three principles governing homogeneous linear systems.


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* $\mathbf{0}$ is always a solution (the trivial solution).
* The general solution consists of several free parameters. In most cases, the number of parameters is equal to $n-m$ where $n$ is the number of unknowns and $m$ is the number of equations.
* In some exceptional cases, the number of parameters is less than $n-m$.

Exceptional cases come up when the linear system is kind of cheating us. For instance, look at

$$\begin{cases}x-y+5z&=0\,\\ 2x-2y+10z&=0\.\end{cases}$$


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$$\begin{cases}x-y+5z&=0\,\\ 2x-2y+10z&=0\.\end{cases}$$

The second equation in this system comes from multiplying the first equation by 2, so essentially there is only one equation in this system. Its general solution contains two instead of one parameters. This situation occurs in the study of standard forms in Section 2.4.

**1.2.** The Law of Cosines states that, let $\Delta ABC$ be a triangle and $a=\overline{CB},\ b=\overline{AC}$ and $c=\overline{BA}$, and $\phi=\angle ACB$. Then


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$$c^{2}=a^{2}+b^{2}-2ab\cos\phi\.$$

To prove it let $h=\overline{AH}$ be the height of the triangle from $A$ and $s$ be $\overline{CH}$. By Pythagoras Theorem,

$$c^{2}=h^{2}+(a-s)^{2}=b^{2}-s^{2}+(a-s)^{2}=b^{2}-2as+a^{2}\,$$

implies $s=(c^{2}-a^{2}-b^{2})/2a$ and the Cosine Law follows after noting $s=b\cos\phi$.


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**1.3.** You may look up Wikipedia under "Euclidean motion" and "orthogonal matrix" to find further information on these interesting topics. How the cross product is used in physics can be also be found in Wikipedia under "cross product". From the same source you will see how the cross product arises from the Lie algebra of the orthogonal group.

**Supplementary Reading**

1.1 and 1.2 in chapter 1, [Au].


