Lesson 1: Introduction to Vectors

Video Lesson

Lesson Objective

Dear Learner,

By the end of this lesson, you will be able to:

• Define a vector quantity
• Describe the difference between a vector and a scalar quantities
• Determine the sum or difference of two or more vectors
• Decompose a given vector in to two components
• Determine the magnitude of a vector

Key terms

Brainstorming Question

Dear learner, try to give your answer for the following question. Mansion at least 6 physical quantities and classify them as vectors and scalar quantities.

Your description can contain almost all of the given answers for which they were the right answers:

velocity, force acceleration displacement etc are said to be vector quantities while distance, mass, density, energy, temperature etc are scalar quantities. In general physical quantities having magnitude only are scalars while physical quantities having both magnitude and direction are vectors.

1.1. Revision on Vectors and Scalars

Definition:

A physical quantity having both magnitude and direction is a vector, where as a quantity having only magnitude is scalar.

Vector quantities                      Scalar quantities

• Force                                                           distance
• Displacement                                              work
• Velocity                                                       time
• Acceleration                                                mass
• Torque                                                        temperature

A. Representation of a Vector

A vector is represented by an arrow

• A is the initial point (tail)
• B is the terminal point (head)

$$\begin{gathered} \text{. Length of}v\text{is its magnitude}\left|v\right|=\left|\overrightarrow{\mathrm{AB}}\right| \\ \text{Direction of}\overrightarrow{\mathrm{AB}}\text{is represented by its arrow.} \\ \text{Zero vector is a vector whose initial and terminal points lie at the origin. i.e.,}v=(0,0) \end{gathered}$$

B). Components of vectors

$$\begin{gathered} \text{. If}v\text{is a vector with initial point at the origin}(0,0)\text{and its terminal point at}Q(x,y), \\ \text{then the coordinate form of}v\text{is given by:}v=(x,y)\text{vector or}v=\left[\begin{array}{c}x\\ y\end{array}\right] \\ \text{.Numbers}x\text{and}y\text{are components or coordinates of}v\text{.} \\ \text{. Direction of vector}v\text{is the slope (m) =}{\tan }^{–1}\left(\frac{y}{x}\right) \\ \text{. Let}v\text{be a vector with initial point}P(x_1,y_1)\text{and terminal point}Q(x_2,y_2), \\ \text{then}v=\text{vector}PQ=(x_2–x_1,y_2–y_1)\text{or}v=\left[\begin{array}{c}{x}_2–x_1\\ y_2–y_1\end{array}\right] \\ \text{. the numbers}{x}_2–x_1\text{and}{y}_2–y_1\text{are component vectors.} \\ \text{. direction of vector}v\text{is the slope(m) =}\frac{y_2–y_1}{x_2–x_1} \end{gathered}$$

$\text{. magnitude}\text{of}v=\left|v\right|=\sqrt{(x_2–x_1{)}^2+(y_2–y_1{)}^2}$

Example 1:

$$\begin{gathered} \text{Let}P(x,y)=(1,1)\text{and}Q(x,y)=(5,4)\text{be points where}\overrightarrow{v}\text{starts and terminates respectively, the determine} \\ \text{a)}\text{vector}\overrightarrow{v} \\ \text{b)}\text{magnitude of}\overrightarrow{v} \\ \text{c)}\text{direction of}\overrightarrow{v} \end{gathered}$$

Solution:

$\text{a).}\overrightarrow{v}=\left(\begin{array}{c}{x}_2-x_1\\ y_2-y_1\end{array}\right)=\left(\begin{array}{c}5-1\\ 4-1\end{array}\right)=\left(\begin{array}{c}4\\ 3\end{array}\right)$

$$\begin{gathered} \text{b).}\left|v\right|=\sqrt{4^2+3^2}=\sqrt{25}=5\text{units} \\ \text{c). direction of}v=m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=\frac{4-1}{5-1}=\frac{3}{4} \\ \Rightarrow m={\tan }^{-1}\left(\frac{3}{4}\right)=36.8° \end{gathered}$$

Fig 1. component of a vector

Example 2:

$\text{If the initial point of vector}v\text{is at P(3, 45) and is terminated at Q(}x+5\text{, 3) and its direction is}45°\text{, then find}x\text{.}$

Solution:

Apply the principle of vector represesntation to solve for x i.e.

$$\begin{gathered} \text{Direction of}v={\tan }^{-1}\frac{y_2-y_1}{x_2-x_1}=45° \\ \iff \frac{y_2-y_1}{x_2-x_1}=\tan 45°\iff \frac{3-4}{x+5-3}=1 \\ \Rightarrow -\frac{1}{x+2}=1\Rightarrow x+2=-1\Rightarrow x=-3 \end{gathered}$$

Therefore the value of x is -3.

1.2. operations on vectors

I. Addition and Sbtraction of Vectors

If u and v are two vectors, the sum u + v is the vector determined by translating vector v until its tail coincides with the head of u. Then, the directed line segment from the tail of u to the head of v is the vector u + v.

subtraction of vectors is done with the same principle as addition. i.e. u – v = u + (-v)

fig. 1.2. Graph showing vector addition

Example 3:

$$\begin{gathered} \text{Let}\overrightarrow{u}\text{lies on P(1,3), Q(-3,1) and}\overrightarrow{v}\text{lies on A(2,3) and B(-3,0), then find:} \\ \text{a).}\overrightarrow{u}+\overrightarrow{v} \\ \text{b).}\overrightarrow{u}-\overrightarrow{v} \end{gathered}$$

$$\begin{gathered} \mathbf{u}=(–3–2,1–3)=(–4,–2)\text{and}\mathbf{v}=(–3–2,0–3)=(–5,–3) \\ \text{a).}\mathbf{u}+\mathbf{v}=(–4,–2)+(–5,–3)=(–9,–5) \\ \text{b).}\mathbf{u}–\mathbf{v}=(–4,–2)–(–5,–3)=(1,1) \end{gathered}$$

Properties of Vector Addition

$$\begin{gathered} \text{Let}\mathbf{u},\mathbf{v},\text{and}\mathbf{w}\text{be any three vectors, then} \\ 1.\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\dots \dots \dots \dots .\text{commutative property} \end{gathered}$$

$\text{2).}(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{v}+(\mathbf{u}+\mathbf{w})$

II. Multiplication of vectors by scalars

Definition

$$\begin{gathered} \text{Let}\overrightarrow{v}\text{be a vector and}K\text{be a scalar, then}k\overrightarrow{v}\text{is the scalar product:} \\ \text{• if}k>0\text{,}k\overrightarrow{v}\text{has the same direction with}\overrightarrow{v}\text{.} \\ \text{• if}k<0\text{,}k\overrightarrow{v}\text{is in the opposite direction with}\overrightarrow{v}\text{.} \\ \text{• if}k=0\text{or}\overrightarrow{v}=0\text{, then}k\overrightarrow{v}=0 \end{gathered}$$

Example 4:

$$\begin{gathered} \text{Let}\overrightarrow{u}=\left(5,1\right)\text{and}\overrightarrow{v}=\left(-6,4\right)\text{, then find} \\ \text{a).}\mathrm{2u} \\ \text{b).}3\overrightarrow{u}+5\overrightarrow{v} \end{gathered}$$

Solution:

$$\begin{gathered} \text{a).}2\overrightarrow{u}=2\left(5,1\right)=\left(2\times 5,2\times 1\right)=\left(10,2\right) \\ \text{b).}3\overrightarrow{u}+5\overrightarrow{v}=3\mathrm{(5,1)}+5\left(-6,4\right)=\left(3\times 5,3\times 1\right)+\left(5\times \left(-6\right),5\times 4\right)=\left(-15,23\right) \end{gathered}$$

Properties of Scalar Multiplication

$$\begin{gathered} \text{Let}\overrightarrow{u}\text{and}\overrightarrow{v}\text{be vectors and}{k}_1\text{and}{k}_2\text{be scalars then the following properties holds} \\ \text{•}\left[k_1\overrightarrow{u}\right]=k_1\overrightarrow{u} \\ \text{•}{k}_1\left(k_2\overrightarrow{u}\right)=\left(k_1{k}_2\right)\overrightarrow{u}\text{Associative property} \\ \text{•}\left(k_1+k_2\right)\overrightarrow{u}=k_1\overrightarrow{u}+k_2\overrightarrow{u}\text{and}{k}_1\left(\overrightarrow{u}+\overrightarrow{v}\right)=k_1\overrightarrow{u}+k_1\overrightarrow{v}\text{Distributive property} \\ \text{•}1\overrightarrow{u}=\overrightarrow{u}\text{Identity property} \\ \text{•}0\overrightarrow{u}=0\text{multiplicative property of zero} \end{gathered}$$

Resolution of vectors

$$\begin{gathered} \text{Let}\overrightarrow{u}\text{be a vector and decomposing}\overrightarrow{u}\text{in to its components is said to be resolution.} \\ \text{Components:} \\ \text{• Horizontal component (}\overrightarrow{u_x}\text{)} \\ \text{• Vertical component (}\overrightarrow{u_y}\text{)} \end{gathered}$$

$$\begin{gathered} \cos \theta =\frac{u_x}{u}\Rightarrow u_x=u\cos \theta \text{horizontal component} \\ \sin \theta =\frac{u_y}{u}\Rightarrow u_y=u\sin \theta \text{vertical component} \\ \theta ={\tan }^{-1}\left(\frac{u_y}{u_x}\right)\text{is measured from the positive x – axis direction (position)} \end{gathered}$$

Example 5:

$$\begin{gathered} \text{1. Find}\text{x-}\text{and}\text{y-components}\text{if}\text{a}\text{vector}\text{having}\text{a}\text{magnitude}\text{of}6\text{units}\text{and} \\ \text{makes}\text{an}\text{angle}\text{of}{60}^o\text{with}\text{the}\text{positive}x–\text{axis}\text{?} \end{gathered}$$

Solution

$\text{Let}\mathbf{U}\text{be}\text{the}\text{vector}\text{with}\left|u\right|=6\text{units}\text{and}\theta =60°\text{then:}$

III. Unit Vectors and Standard Unit Vectors

   Definition: 

$$\begin{gathered} \text{• Any vector whose magnitude one is a unit vector.} \\ \text{• Standard unit vectors in the plane are the vectors}\overrightarrow{i}=\left(1,0\right)\text{and}\overrightarrow{j}=\left(0,1\right) \\ \text{•}\left|\overrightarrow{i}\right|=\left|\overrightarrow{j}\right|=1 \\ \text{• Unit vectors}\overrightarrow{i}\text{and}\overrightarrow{j}\text{are always point to the positive x- and y- direction respectively.} \end{gathered}$$

Note:

$$\begin{gathered} \text{A unit vector on the direction of}\overrightarrow{v}\text{is determined by:}\frac{\overrightarrow{v}}{\left|v\right|} \\ \text{Any vector in the plane can be expressed uniquely in the form:} \end{gathered}$$

$$\begin{gathered} \mathbf{v}=xi+yj=x(1,0)+y(0,1) \\ =\left(\begin{array}{c}x\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ y\end{array}\right) \\ =\left(\begin{array}{c}x\\ y\end{array}\right) \end{gathered}$$

Example 6:

Express the following vectors in a coordinate form?

$$\begin{gathered} \text{a).}2i+3j \\ \text{b).}i+5j \end{gathered}$$

Solution:

$$\begin{gathered} \text{a).}2i+3j=2(1,0)+3(0,1)=(2,3) \\ \text{b).}i+5j=(1,0)+5(0,1)=(1,5) \end{gathered}$$

1.3. Magnitude or Norm of a vector

Definition:

$$\begin{gathered} \text{For a vector defined by}\mathbf{v}=(x,y) \\ \text{Norm (magnitude) of}\mathbf{v}=\left|\mathbf{v}\right|=\sqrt{x{}^2+y{}^2} \\ \text{Let}V\text{be a vector with initial point}P(x_1,y_1)\text{and terminal point}Q(x_2,y_2) \\ \text{then}\left|v\right|=\sqrt{(x_2–x_1{)}^2+(y_2–y_1{)}^2} \end{gathered}$$

Example 7:

$$\begin{gathered} \text{a). Find two unit vectors in the same direction and at the opposite of the given vector}\overrightarrow{u}=\left(2,5\right) \\ \text{b). If}\overrightarrow{u}=3\overrightarrow{i}+5\overrightarrow{j}\text{and}\overrightarrow{v}=7\overrightarrow{i}-\overrightarrow{j}\text{then find:} \\ \text{i).}\left|\overrightarrow{u}+\overrightarrow{v}\right| \\ \text{ii).}\left|\overrightarrow{u}-\overrightarrow{v}\right| \end{gathered}$$

Solution

Lets see how these problems are to be solved:

$$\begin{gathered} \text{Lets see how these problems are to be solved:} \\ \text{a).}\overrightarrow{u}=\left(2,5\right)\Rightarrow \left|\overrightarrow{u}\right|=\sqrt{2^2+5^2}=\sqrt{29} \\ \text{• a unit vector in the direction of}\overrightarrow{u}=\frac{\overrightarrow{u}}{\left|\overrightarrow{u}\right|}=\frac{\left(2,5\right)}{\sqrt{29}}=\frac{2}{\sqrt{29}}\overrightarrow{i}+\frac{5}{\sqrt{29}}\overrightarrow{j} \\ \text{• a unit vector on the opposite of}\overrightarrow{u}=-\frac{\overrightarrow{u}}{\left|\overrightarrow{u}\right|}=-\frac{\left(2,5\right)}{\sqrt{29}}=-\frac{2}{\sqrt{29}}\stackrel{}{\mathrm{i</mi\; <mo\; stretchy="true">\to </mo>}}-\frac{5}{\sqrt{29}}\overrightarrow{j} \end{gathered}$$

$$\begin{gathered} \text{b). i.}\left|\overrightarrow{u}+\overrightarrow{v}\right|\text{: First find}\overrightarrow{u}+\overrightarrow{v}=\left(3\overrightarrow{i}+5\overrightarrow{j}\right)+\left(7\overrightarrow{i}-\overrightarrow{j}\right)=\left(3+7\right)\overrightarrow{i}+\left(5-1\right)\overrightarrow{j}=10\overrightarrow{i}+4\overrightarrow{j} \\ \text{then}\left|\overrightarrow{u}+\overrightarrow{v}\right|=\sqrt{{10}^2+4^2}=\sqrt{116}=2\sqrt{29}\text{units} \\ \text{ii.}\left|\stackrel{}{\mathrm{u</mi\; <mo\; stretchy="true">\to </mo>}}-\overrightarrow{v}\right|\text{: Lets determine}\overrightarrow{u}-\overrightarrow{v}=\left(3\overrightarrow{i}+5\overrightarrow{j}\right)-\left(7\overrightarrow{i}-\overrightarrow{j}\right)=\left(3-7\right)\overrightarrow{i}+\left(5+1\right)\overrightarrow{j}=\text{-4i + 6j} \\ \text{then}|u–v|=\sqrt{(-4{)}^2+(6{)}^2)}=\sqrt{52}=2\sqrt{13}\text{units} \end{gathered}$$

Equality of vectors

$$\begin{gathered} \text{Let}\overrightarrow{u}=u_1\overrightarrow{i}+u_2\overrightarrow{j}\text{and}\overrightarrow{v}=v_1\overrightarrow{i}+v_2\overrightarrow{j}\text{are equal vectors if and only if} \\ \text{•}{u}_1=v_1\text{and}{u}_2=v_2 \\ \text{•}\left|\overrightarrow{u}\right|=\left|\overrightarrow{v}\right| \\ \text{•}\overrightarrow{u}\text{and}\overrightarrow{v}\text{have the same direction} \end{gathered}$$

$$\begin{gathered} \text{Let}\overrightarrow{u}\text{and}\overrightarrow{v}\text{are parallel vectors, then:} \\ \text{•}\overrightarrow{u}=k\overrightarrow{v} \\ \text{• one is the scalar multiple of the other} \end{gathered}$$

Example 8:

$$\begin{gathered} \text{a). Show that vectors}\overrightarrow{u}=3\overrightarrow{i}+4\overrightarrow{j}\text{and}\overrightarrow{v}=9\overrightarrow{i}+12\overrightarrow{j}\text{are parallel?} \\ \text{b).}\text{Find}\text{the}\text{values}\text{of}x\text{and}y\text{if}\text{the}\text{vectors}\mathbf{u}=\left(\begin{array}{c}10–2x\\ 8–y\end{array}\right)\text{and}\mathbf{v}=\left(\begin{array}{c}5x–3\\ y+4\end{array}\right)\text{be}\text{parallel?} \end{gathered}$$

Solution :

$$\begin{gathered} \text{a).}\text{For}\overrightarrow{u}=3i+4j\text{and}\overrightarrow{v}=9i+12j \\ \text{determine}k\text{= the constant or scalar in which it is the same in all points} \\ \text{by considering}\overrightarrow{u}=k\overrightarrow{v},\mathrm{implies}3=k\times 9\text{and}4=k\times 12 \\ \mathrm{implies}k=\frac{1}{3},\text{therefore}\overrightarrow{u}\text{and}\overrightarrow{v}\text{are parallel.} \end{gathered}$$

$$\begin{gathered} \text{b).}\mathbf{u}=\mathbf{v}\Rightarrow \left(\begin{array}{c}10–2x\\ 8–y\end{array}\right)=\left(\begin{array}{c}5x–3\\ y+4\end{array}\right) \\ \iff 10–2x=5x–3 \\ \Rightarrow 7x=13 \\ \Rightarrow x=\frac{13}{7} \\ \text{and}8–y=y+4 \\ \Rightarrow y+y=8–4 \\ \Rightarrow 2y=4 \\ \text{then}y=2 \end{gathered}$$