Chapter 2 - Vectors

MATH1050 Vectors

2.1 Vectors

A Vector quantity is something whose specification requires both a magnitude and a direction.

Normally, when writing vectors, a lower case letter, with the tilde (~) symbol below it, for example, $\underline{v}$.

Geometric Representation of a Vector

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A Vector can be represented geometrically in ${\mathbb{R}}^2$ or ${\mathbb{R}}^3$ by an arrow.

The length of the arrow represents the magnitude of the Vector, and the direction of the Vector is indicated by the direction the arrow is pointing.

The actual location of the arrow in the diagram is irrelevant, only its magnitude and direction matter.

If $P$ and $Q$ are points in ${\mathbb{R}}^2$ or ${\mathbb{R}}^3$, then $\overrightarrow{PQ}$ denotes the Vector from $P$ to $Q$. The point $P$ is the tail of the Vector and the point $Q$ is the head of the Vector.

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Matrix Representation of a Vector

For a (geometric) Vector $v\in {\mathbb{R}}^2$ with tail at the point $(x_1,y_1)$ and head at the point $(x_2,y_2)$, the matrix form of the Vector has 2 rows and 1 column and is written as:

x_{2}-x_{1} \\ y_{2}-y_{1} \end{pmatrix}$$ The matrix form of a [[Vector]] is the same for all geometric representations of the [[Vector]]. The usual notation for writing a general [[Vector]] $v$ in matrix form is: $$v=\begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}$$ This is a column [[Vector]]. It can also be written as a row [[Vector]]: $$v=(v_{1},v_{2})$$ Given a [[Vector]] in matrix form, you can find a geometric representation of $v$ by picking any point in the plane as the tail of the [[Vector]], moving $v_{1}$ units in the $x$ direction and the $v_{2}$ units in the $y$ direction to find the point that is the head of the [[Vector]]. For a geometric [[Vector]] $v=\overrightarrow{PQ}$ in $\mathbb{R}^3$, where $P = (x_{P},y_{P},z_{P})$ and $Q=(x_{Q},y_{Q},z_{Q})$, the matrix form is: $$\large v=\begin{pmatrix} x_{Q}-x_{P} \\ y_{Q}-y_{P} \\ z_{Q}-z_{P} \\ \end{pmatrix}=\begin{pmatrix} v_{1} \\ v_{2} \\ v_{3} \end{pmatrix} \text{or} \space v=(v_{1},v_{2},v_{3})

The entries $v_1$ and $v_2$ are called the components of the Vector.

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2.2 Addition of Vectors

Geometric addition of Vectors

Given vectors $v$ and $w$, we define the sum $v+w$ by the triangle rule for Vector addition. Let $P,Q$ and $R$ be points in ${\mathbb{R}}^2$ (or ${\mathbb{R}}^3$) such that $\mathbf{v}=\overrightarrow{PQ}$ and $\mathbf{w}=\overrightarrow{QR}$. Then $\mathbf{v}+\mathbf{w}=\overrightarrow{PR}$.

Note that when adding vectors $v$ and $w$ geometrically you put the tail of $w$ at the head of $v$ and then draw the sum $v+w$ from the tail of $v$ to the head of $w$.

Matrix addition of Vectors

If:

v_{1} \\ v_{2} \end{pmatrix}$$ and: $$w=\begin{pmatrix} w_{1} \\ w_{2} \end{pmatrix}$$ , then $$v+w=\begin{pmatrix} v_{1}+w_{1} \\ v_{2}+w_{2} \end{pmatrix}$$ Addition of vectors in $\mathbb{R}^{3}$ is the same procedure. We define the *zero [[Vector]]* to be the [[Vector]] (of an appropriate size) with each component equal to zero, and denote it by 0. $$0=\begin{pmatrix} 0 \\ 0 \\ \end{pmatrix} or \space 0=\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}$$ The zero [[Vector]] has zero magnitude and unspecified direction, so can be represented geometrically as a point. ### Properties of Vector addition 1. [[Vector]] addition is *commutative* ($v+w=w+v$). 1. To illustrate the commutativity of addition geometrically, consider four points $P,Q,R,S$, arranged in 2-space so that: $$v=\overrightarrow{PQ}=\overrightarrow{SR} \space\text{and}\space w=\overrightarrow{PS}=\overrightarrow{QR}$$ ![[Drawing 2023-02-23 00.15.53.excalidraw|center]] Then: $$\large v+w=\overrightarrow{PQ}+\overrightarrow{QR}=\overrightarrow{PR}=\overrightarrow{PS}+\overrightarrow{SR}=w+v$$ 2. [[Vector]] addition is *associative*, that is: $$u+(v+w)=(u+v)+w$$ 3. $0+v=v+0=v$ ## 2.3 Scalar Multiplication of Vectors ### Geometric Scalar Multiplication of Vectors Given a vector $v$ and a real number $t$, we define the scalar multiple $tv$ to be the vector whose magnitude is $|t|$ times the magnitude of $v$, and whose direction is the same as $v$ if $t>0$ and opposite to $v$ if $t<0$. Note that if $t=0$, then the scalar multiple $tv$ is the zero vector. ### Matrix Scalar Multiplication of Vectors If $t$ is a real number and $v=$: $$v=\begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}$$ then: $$tv=\begin{pmatrix} t \times v_{1} \\ t \times v_{2} \end{pmatrix}$$ Scalar multiplication of vectors in $\mathbb{R}^{3}$ is the same procedure.

t\times \begin{pmatrix} w_{1} \ w_{2} \ w_{3} \end{pmatrix}=\begin{pmatrix} t \times w_{1} \ t \times w_{2} \ t \times w_{3} \end{pmatrix}

Note that we usually write $-1\mathbf{v}$ as $-\mathbf{v}$. We can define *vector subtraction* as a combination of vector addition and scalar multiplication. If $\mathbf{v}$ and $\mathbf{w}$ are two vectors, then:

\mathbf{v}-\mathbf{w}=\mathbf{v}+(\mathbf{-w})

## 2.4 Position Vectors Of all the geometric representations of a vector $\mathbf{v}$, the one with its tail at the origin is special. In $\mathbb{R}^{2}$, let $P$ be the point with coordinates $(x_{P},y_{P},z_{P})$. The vector $\vec{OP}$ with its tail at the origin $O$ and its head at $P$ is called the *position vector* of $P$. The matrix form of $\vec{OP}$ = $(x_{P},y_{P})$. Similarly, in $\mathbb{R}^{3}$, let $P$ be the point with coordinates $(x_{P},y_{P},z_{P})$. The vector $\vec{OP}$ with its tail at the origin $O$ and its head at $P$ is called the *position vector* of P, and the matrix form of $\vec{OP}$ is:

\begin{pmatrix} x_{P}-0 \ y_{P}-0 \ z_{P}-0 \end{pmatrix}

\begin{pmatrix} x_{P} \ y_{P} \ z_{P} \ \end{pmatrix}

which can also be written as $\vec{OP}= (x_{P},y_{P},z_{P})$. The coordinates of the point $P$ are the components of position vector of $P$. ## 2.5 The Norm of a Vector The *norm* (or *length* or *magnitude*) of the vector $\mathbf{v}=\vec{PQ}$ is the (shortest) distance between points $P$ and $Q$. The Norm of the vector $\mathbf{v}$ is denoted $||\mathbf{v||}$. In $\mathbb{R}^{2}$, if $P=(x_{P},y_{P})$ and $Q=(x_{Q},y_{Q})$, then:

\mathbf{v}=\vec{PQ}=\begin{pmatrix} x_{Q}-x_{P} \ y_{Q}-y_{P} \end{pmatrix}

\begin{pmatrix} v_{1} \ v_{2} \ v_{3} \end{pmatrix}

$$and:$$

|\mathbf{v}| = \sqrt{ v_{1}^{2}+v_{2}^{2} }

In $\mathbb{R}^{3}$, if $P=(x_{P},y_{P},z_{P})$ and $Q=(x_{Q},y_{Q},z_{Q})$, then:

\mathbf{v}=\vec{PQ}=\begin{pmatrix} x_{Q}-x_{P} \ y_{Q}-y_{P} \ z_{Q}-z_{P} \end{pmatrix} = \begin{pmatrix} v_{1} \ v_{2} \ v_{3} \end{pmatrix}, \text{and}
|\mathbf{v}|=\sqrt{ v_{1}^{2}+v_{2}^{2}+v_{3}^{2} }$$ Note that for most vectors $\mathbf{v}$ and $\mathbf{w}$, $|\mathbf{v}+\mathbf{w}|\ne |\mathbf{v}|+|\mathbf{w}|$. For any vector $\mathbf{v}$ and any real number $t$, $|t\mathbf{v}|=|t|+|\mathbf{v}|$.

A vector with norm 1 is called a unit vector. The notation $\hat{\mathbf{v}}$ will be used to denote a unit vector having the same direction as the vector $\mathbf{v}$.

For a given vector $\mathbf{v}$, with norm $|\mathbf{v}|$, the vector:

$$\hat{\mathbf{v}}=\frac{1}{|\mathbf{v}|}\mathbf{v}$$

is a unit vector in the direction of $\mathbf{v}$.

2.6 Component Form of a Vector

Component Form in 2-space

In the $(x,y)$ plane, there are two important unit vectors. The unit vector in the direction of the $x$-axis is denoted $\mathbf{i}$, and the unit vector in the direction of the $y$-axis is denoted $\mathbf{j}$, so:

$$\mathbf{i}=\left(\begin{array}{c}1\\ 0\end{array}\right)$$

and:

$$\mathbf{j}=\left(\begin{array}{c}0\\ 1\end{array}\right)$$

Note that $\mathbf{i}$ and $\mathbf{j}$ can also be written as row vectors as $\mathbf{i}=(1,0)$ and $\mathbf{j}=(0,1)$.

Any vector $\mathbf{v}$ in ${\mathbb{R}}^2$can be written as the sum of scalar multiples of $\mathbf{i}$ and $\mathbf{j}$, since:

$$\mathbf{v}=\left(\begin{array}{c}a\\ b\end{array}\right)=a\left(\begin{array}{c}1\\ 0\end{array}\right)+b\left(\begin{array}{c}0\\ 1\end{array}\right)=a\mathbf{i}+b\mathbf{j}$$

The component form of the vector $\mathbf{v}$ is $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$.

Component Form in 3-space

In 3 space, there are three important unit vectors. In the $(x,y,z)$ plane, there are there important unit vectors. The unit vector in the direction of the $x$-axis is denoted $\mathbf{i}$, and the unit vector in the direction of the $y$-axis is denoted $\mathbf{j}$, and the unit vector in the direction of the $z$-axis is denoted $\mathbf{k}$, so:

$$\mathbf{i}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\mathbf{j}=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\mathbf{k}=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$$

Note that $\mathbf{i},\mathbf{j},\mathbf{k}$ can also be written as row vectors.

Converting Vectors from Geometric to Component Form

A vector $\mathbf{v}$ in ${\mathbb{R}}^2$, with magnitude $|\mathbf{v}|$ and direction $\theta$ measured anti-clockwise from the positive $x$-axis, has component form.

$$\mathbf{v}=|\mathbf{v}|\cos \theta \underline{\mathbf{i}}+|\mathbf{v}|\sin \theta \underline{\mathbf{j}}$$

Converting Vectors from Component to Geometric Form

If $\mathbf{v}=v_1\underline{\mathbf{i}}+0\underline{\mathbf{j}}$ or $\mathbf{v}=0\underline{\mathbf{i}}+v_2\underline{\mathbf{j}}$, then $\theta$ will be one of 0, $\frac{\pi }{2}$,$\pi ,\text{or}\frac{3\pi }{2}$, and you can determine which it is from the sketch.

If neither of $v_1$ not $v_2$ is zero, then calculate $\varphi =\arctan (|\frac{v_2}{v_1}|)$. The value of $\varphi$ will be between 0 and $\frac{\pi }{2}$. The value of $\theta$ will be one of $\varphi ,\pi -\varphi ,\pi +\varphi ,\text{or}2\pi -\varphi$. You can identify which it is from your sketch.

2.7 The Scalar Product

Let $\mathbf{v}=\overrightarrow{OP}\ne 0$ and $\mathbf{w}-\overrightarrow{OQ}\ne 0$ be two vectors. Then the angle between $\mathbf{v}$ and $\mathbf{w}$ is the angle $\theta$ between $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ at the origin, with $0\le \theta \le \pi -\pi$.

The scalar product of two vectors $\mathbf{v}$ and $\mathbf{w}$ is:

$$\mathbf{v}\cdot \mathbf{w}=\{\begin{array}{ll}0& \text{if}\mathbf{v}=0\text{or}\mathbf{w}=0\\ |\mathbf{v}||\mathbf{w}|\cos \theta & \text{otherwise.}\end{array}$$

Where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$.

The scalar product is also called the dot product or inner product.

If $\mathbf{v}$ and $\mathbf{w}$ are in matrix form, then the scalar product is easy to calculate. If $\mathbf{v}=(v_1,v_2)$ and $\mathbf{w}=(w_1,w_2)$, then:

$$\mathbf{v}\cdot \mathbf{w}=v_1{w}_1+v_2{w}_2$$

If $\mathbf{v}=(v_1,v_2,v_3)$ and $\mathbf{w}=(w_1,w_2,w_3)$, then:

$$\mathbf{v}\cdot \mathbf{w}=v_1{w}_1+v_2{w}_2+v_3{w}_3$$

If we are given two vectors and $\mathbf{v}$ and $\mathbf{w}$ in matrix or component form, then we can use the scalar product to calculate the angle between $\mathbf{v}$ and $\mathbf{w}$.

Example 2.7.2

Calculate the angle (in both degrees and radians) between vectors $\mathbf{u}=(4,3,1)$ and $\mathbf{v}=(2,-3,-2)$.

\Large \begin{gather*} \begin{aligned} \underline{\mathbf{u}} \cdot \underline{\mathbf{v}} &= |\mathbf{u}|\times|\mathbf{v}|\times \cos \theta \ \ \begin{pmatrix} 4 \ 3 \ 1 \
\end{pmatrix} \cdot \begin{pmatrix} 2 \ -3 \ -2 \ \end{pmatrix} &= \sqrt{ 4^{2}+3^{2}+1^{2} }\times \sqrt{ 2^{2}+(-3)^{2}+(-2)^{2} }\times \cos \theta \ \ 8+(-9)+(-2) &=\sqrt{26 }\times \sqrt{ 17 }\times \cos \theta \ -\frac{3}{\sqrt{ 26 }\sqrt{ 17 }}&=\cos \theta \ \theta &= \arccos\left( -\frac{3}{\sqrt{ 26 }\sqrt{ 17} } \right) \ \ &\approx 98.2^\circ \ &\approx 1.71 \text{rad} \end{aligned} \end{gather*}