Video Lesson

Lesson objective

Dear learners,

At the end of the lesson you will be able to

• Apply algebraic method to find the resultant of two collinear vectors, and perpendicular vectors.
• Apply trigonometry to determine the direction of a resultant vector in terms of the angle it forms with a reference direction.
• Compare the resultant of two given vectors as determined by way of geometric construction and by analytic method.
• Resolve a vector into its components.
• Find the resultant of two or more vectors applying component method.
• Define unit vector and determine a unit vector in the direction of given vector.
• Apply the unit representation of vectors to determine the resultant of two or more vectors.

Brainstorming question

Do you recall the different ways in which vectors that represent similar quantities can be added to/ subtracted from one another? Mention two ways of vector addition based on your physics lesson in grade 10.

Key Terms and Concepts

Adding two collinear vectors

When force vectors $\overrightarrow{F_{1}}and \overrightarrow{F_{1}}$ are acting in same directions, The resultant of these vectors $\overrightarrow{R}$ has a magnitude equal to the sum of the magnitude of each of the vectors and it is directed along the direction of any one of the vectors

When force vectors$\overrightarrow{F_{1}}and \overrightarrow{F_{1}}$ are acting in opposite directions, their resultant $\overrightarrow{R}$ will have magnitude equal to the absolute value of the difference in magnitudes of the vectors and it is directed along the direction of the larger vector

• We write the resultant vector $\overrightarrow{R}$ symbolically as
• $\overrightarrow{R}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

Examples 3.1

A man rows his boat along the direction of flow of a river (downstream). If the boat can sail in still water at 0.50 m/s, and the river flows at 0.30 m/s, what is the resultant velocity of the boat?

Solution

Let the magnitude of velocity of the river be A and that of the boat be B.
A = 0.30 m/s
B = 0.50m/s
The resultant velocity is $\overrightarrow{R}=\overrightarrow{A_{1}}+\overrightarrowB_{2}}$

Since both vectors are along the same direction, the magnitude of the resultant velocity is given by:
$\overrightarrow{R}=\overrightarrow{A_{1}}+\overrightarrowB_{2}}$ = 0.30 m/s + 0.50m/s.
$\overrightarrow{R}$ = 0.80 m/s.
The direction of the resultant is along the direction of the river flow. Therefore, the resultant velocity is:
$\overrightarrow{R}$=0.8m/s, downstream

Examples 3.2

A block is pulled by two forces of 15 N and 25 N to the left, and by three forces of 10 N, 20 N, 30 N to the right. Find the magnitude and direction of the resultant force

Solution

If you sum the forces pulling to the left, you get 40 N to the left, and if you sum the forces pulling to the right, you get 60 N to the right.

Thus, the resultant force is 20 N to the right

Examples 3.3

Find the resultant for the following force

Solution

The resultant force is in the same direction as the two forces, and has the magnitude equal to the sum of the two magnitudes: N7

Examples 3.4

Two men are pushing a block along a horizontal surface by exerting oppositely directed forces of magnitudes 200 N and 100 N respectively, as shown in Figure 2.13. What is the resultant force applied on the block by the two men?

Solution

The two men applied two different forces on the same block.
1Let the magnitude of the force exerted by the man on the right be $\overrightarrow{F_{1}}$ = 200 N and that of the man on the left be $\overrightarrow{F_{1}}$= 100 N.

$\overrightarrow{R}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

The magnitude of the resultant force is $\overrightarrow{R}=\overrightarrow{F_{1}}-\overrightarrow{F_{2}}$ = 200N – 100N = 100 N
The direction of the resultant is along the direction of the 200 N (larger) force.
$\overrightarrow{R}$=100N,toward the right.

Figure 2.10 Two men pushing the block in opposite directions

Adding two perpendicular vectors

• when they are in two dimensions become more complicated than the rules for addition of vector in one dimension.
• We have to use graphical method of vector addition to construct resultant vectors, followed by trigonometry to find magnitudes and directions.
• We have to use graphical method of vector addition to construct resultant vectors, followed by trigonometry to find magnitudes and directions

According to Pythagoras theorem applied on a right angled triangle, the square of the hypotenuse, c, is equal to the sum of squares of the other two sides, a and b.

$Sin\Theta=\frac{opposite}{hypotenuse}$

$Cos\Theta=\frac{adjacent}{hypotenuse}$

$Tan\Theta=\frac{opposite}{adjacent}$

Example 3.5

A car moves from A to B for 4 km and then from B to C at a right angle for 3 Km. It then returns from point C to A for 5 km. What will be the distance and displacement?

Solution

Given that

• Distance from A to B: 4 km
• Distance from B to C: 3 km
• Distance from C to A: 5 km

We know that distance is the total path covered by an object. So,

Distance Covered = AB + BC + AC = 4 + 3 + 5 = 12km

Displacement is the difference between the initial and final position. However, here the final position and initial position are the same i.e. A.

Thus, the displacement here will be 0.

Components of a vector

• This section introduces a simple but general method for adding vectors known as Component Method.
• The projection of the vector on the coordinate axes is known as component of the vector.
• The x-component is the part of the vector that points along the x-axis, and
• The y-component is the part of the vector that points along the y-axis.
• The projection of vector $\overrightarrow{A}$ on the x axis,$ \overrightarrow{A_{x}}$ , is known as the x component
• of vector $\overrightarrow{A}$ and that on the y axis, $ \overrightarrow{A_{y}}$ , is known as the y component of vector $\overrightarrow{A}$
• Components A and A are numbers but they are not vectors. $ \overrightarrow{A_{x}}$ and $ \overrightarrow{A_{y}}$ are the vector components of $\overrightarrow{A}$ and in symbols
• $\overrightarrow{A}= \overrightarrow{A_{x}}+ \overrightarrow{A_{y}}$

Figure 2.11 Vector A in terms of its vector components

• The x and y components of vector A (Figure 2.16a) have magnitudes of $\overrightarrow{A_{x}}=Acos\Theta$ and $\overrightarrow{A_{y}}=Asin\Theta$, respectively.
• components may be positive or negative umbers depending on how the vector is oriented, i.e., the angle it makes with the x axis.
• The process of breaking up a vector into its components is known as resolution of a vector.
• Note that: In finding the x- and y- components of a vector, we associate cosine with the x-component and sine with the y-component, as in Figure 2.16a. This association is due to the fact that we chose to measure the angle $\alpha$ with respect to the positive x-axis.
• If the angle were measured with respect to the y-axis, as in Figure 2.16b, the components would be given by
• $A_{x}=Asin\alpha$ and $ A_{y}=Acos\alpha$

Example 3.6

Find the x and y components of a vector having a magnitude of 12 and making an angle of 45 degrees with the positive x-axis.

Solution:

The given vector is V= 12, and it makes an angle θ = 45º.

The x component of the vector = 𝑉𝑥 = VCosθ = 12.Cos45º = 12.(1/√2) = 6√2.

The y component of the vector = 𝑉𝑦 = VSinθ = 12.Sin45º = 12.(1/√2) = 6√2.

Therefore, the x component and the y components of the vector are both equal to 6√2.​​​​​​

Calculating a Resultant Vector

If the perpendicular components A_x and 𝐴_𝑦 of a vector 𝐴 are known, then A can also be found analytically. To find the magnitude 𝐴 and direction θ of a vector from its perpendicular components 𝐴_𝑥 and 𝐴_𝑦, we use the following relationships:

Magnitude

$A=\sqrt{A_{x}^{2}+A_{y}^{2}}$

Direction

$\Theta=tan^{-1}\left( \frac{A_{y}}{A_{x}} \right)$

Attention:

• If both Ax and Ay are positive numbers, angle θ is above the +x axis
• If Ax is negative number and Ay is positive number, angle θ is above the -x axis
• If both Ax and Ay are negative numbers, angle θ is below the -x axis
• If Ax is positive number and Ay is negative number, angle θ is below the +x axis

Figure 2.12 Resultant vector

Example 3.7

Adding Vectors Using Analytical Methods

Add the vector A𝐴 to the vector B shown in Figure, using perpendicular components along the x– and y-axes. The x– and y-axes are along the east–west and north–south directions, respectively. Vector A represents the first leg of a walk in which a person walks 53.0m in a direction 20.0º north of east. Vector B represents the second leg, a displacement of 34.0m in a direction 63.0º north of east.

• Figure 2.13 Vector 𝐴 has magnitude 53.0m and direction 20.0º north of the x-axis. Vector B has magnitude 34.0m and direction 63.0º north of the x-axis. You can use analytical methods to determine the magnitude and direction of R

Strategy

• The components of A and B along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution

• Following the method outlined above, we first find the components of A and B along the x- and y-axes. Note that 𝐴=53.0𝑚,θ_𝐴=20.0º,𝐵=34.0𝑚, and θ_B=63.0º
• We find the x-components by using Ax=Acosθ which gives
• A_x=AcosθA=(53.0m)(cos20.0º)(53.0m)(0.940)=49.8m and
• B_x=Bcosθ = B=(34.0m)(cos63.0º)(34.0m)(0.454)=15.4m.
• Similarly, the y-components are found using Ay=Asinθ
• A_y=Asinθ =A_y=(53.0m)(sin20.0º)(53.0m)(0.342)=18.1m and
• B_y=Bsinθ = B_y=(34.0m)(sin63.0º)(34.0m)(0.891)=30.3m.
• The x- and y-components of the resultant are thus
• R_x=A_x+ B _x =49.8m+15.4m=65.2m and
  • 𝑅𝑦=𝐴𝑦+𝐵𝑦=18.1𝑚+30.3𝑚=48.4𝑚.
• Now we can find the magnitude of the resultant by using the Pythagorean theorem:
• 𝑅= √𝑅𝑥²+𝑅𝑦²= (65.2)²+(48.4)²𝑚 so that
  • R=81.2m.
• Finally, we find the direction of the resultant:
• θ=tan−1(Ry/Rx)=+tan−1(48.4/65.2).thus
• θ=tan−1(0.742)=36.6^o

Unit Vectors

• A unit vector is a vector that has the magnitude equal to 1.
• Unit vectors are used to represent the direction of vectors in space.
• The symbol of unit vectors is denoted by a ‘hat’ (^) to distinguish it from ordinary vectors.
• The first one is to write the coordinates in brackets: $\overrightarrow{V}=\left( X,Y,Z \right)$
• Furthermore the other way is to use three unit vectors, in which each point along one of the axes:$\overrightarrow{V}=x\hat{i}+y\hat{j}+z\hat{k}$
• Moreover, the magnitude of vector is:
• $\left| V \right|=\sqrt{X^{2}+Y^{2}+Z^{2}}$
• Unit vector=$\frac{vector}{Magnituid of the vector}$ or
• $\hat{V}=\frac{\overrightarrow{V}}{\left| \overrightarrow{V} \right|}$
• If we write it in bracket format then:
• $\hat{V}=\frac{\overrightarrow{V}}{\left| \overrightarrow{V} \right|}=\hat{V}=\frac{\overrightarrow{V}}{\left| \overrightarrow{V} \right|}=\frac{\left( x\hat{i,}y\hat{j,}z\hat{k} \right)}{\sqrt{X^{2}+Y^{2}+Z^{2}}}$=$\left( \frac{X}{\sqrt{X^{2}+Y^{2}+Z^{2}}},\frac{Y}{\sqrt{X^{2}+Y^{2}+Z^{2}}},\frac{Z}{\sqrt{X^{2}+Y^{2}+Z^{2}}} \right)$
• If we write it in unit vector component format:
• $\hat{V}=\frac{\overrightarrow{V}}{\left| \overrightarrow{V} \right|}=\frac{\left( X\hat{i,Y\hat{j,Z\hat{k}}} \right)}{\sqrt{X^{2}+Y^{2}+Z^{2}}}=\frac{X}{\sqrt{X^{2}+Y^{2}+Z^{2}}}\hat{i}+\frac{Y}{\sqrt{X^{2}+Y^{2}+Z^{2}}}\hat{j}+\frac{Z}{\sqrt{X^{2}+Y^{2}+Z^{2}}}\hat{k}$

Example 3.8

There is a vector$\overrightarrow{r}=12\hat{i}-3\hat{j}-4\hat{k}$. Calculate the unit vector $\overrightarrow{r}$. Also, express it in both unit vector component format and bracket format.

Solution:

• $\left| \overrightarrow{r} \right|=\sqrt{X^{2}+Y^{2}+Z^{2}}$
• $\left| \overrightarrow{r} \right|=\sqrt{12^{2}+\left( -3 \right)^{2}+\left( -4 \right)^{2}}$
• $\left| \overrightarrow{r} \right|=\sqrt{144+9+16}$
• $\left| \overrightarrow{r} \right|=\sqrt{169}$
• $\left| \overrightarrow{r} \right|$ =13
• So, now we can use the magnitude to find the unit vector $\hat{b}$
• $\hat{r}=\frac{\overrightarrow{r}}{\left| \overrightarrow{r} \right|} =\frac{\left( x\hat{i,}y\hat{j,}z\hat{k} \right)}{\sqrt{X^{2}+Y^{2}+Z^{2}}}$=
• $\hat{r}=\frac{12\hat{i}-3\hat{j}+4\hat{k}}{13}$
• $\left| \overrightarrow{r} \right|=\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}-\frac{4}{13}\hat{k}$
• Besides, in bracket format the unit vector is

$\hat{r}=\left( \frac{12}{13},-\frac{3}{13},-\frac{4}{13}\right)$

Product of Vectors

• There are two different kinds of products of vectors: Scalar (Dot) product and Vector (Cross) product.
• The scalar product of two vectors yields a result that is a scalar quantity while the vector product of two vectors yields another vector.

Multiplying or dividing a vector by a number or scalar

• A vector quantity such as a force can be multiplied by a scalar quantity (an ordinary number).
• Multiplying a vector by a number gives a vector.
• When we multiply a vector A by a scalar cA, the result c has a magnitude equal to |c| A
• If c is positive, A and cA are in the same direction (they are parallel) but if c is negative, A and cA are opposite directions (they are antiparallel).
• Multiplication of Vectors
• Multiplication of vectors can be of two types:

(i) Scalar Multiplication

(ii) Vector Multiplication

Multiplication of vectors with scalar:

When a vector is multiplied by a scalar quantity, then the magnitude of the vector changes in accordance with the magnitude of the scalar but the direction of the vector remains unchanged.

Suppose we have a vector $\overrightarrow{a}$

then if this vector is multiplied by a scalar quantity k then we get a new vector with magnitude as |$k\overrightarrow{a}$

if  k is positive and if  k is negative then the direction of k becomes just opposite of the direction of vector $\overrightarrow{a}$

Example 3.9

let U=$\left\langle -1,3 \right\rangle$ Find 7U

7U=7$\left\langle -1,3 \right\rangle$

$\left\langle -7,21 \right\rangle$

Dot product

• The scalar product of two vectors is denoted by AB: . Because of this notion, the vv
• scalar product is also called the dot product. Although A and B are vectors, the vv
• quantity AB. is a scala
• The scalar product of two vectors a and b of magnitude |a| and |b| is given as |a||b| cos θ, where θ represents the angle between the vectors a and b taken in the direction of the vectors.
• We can express the scalar product as:

a.b=|a||b| cosθ

where |a| and |b| represent the magnitude of the vectors a and b while cos θ denotes the cosine of the angle between both the vectors and a.b indicate the dot product of the two vectors.

Example 3.10

 Let there be two vectors [6, 2, -1] and [5, -8, 2]. Find the dot product of the vectors.

Solution:

Given vectors: [6, 2, -1] and [5, -8, 2] be a and b respectively.

a.b = (6)(5) + (2)(-8) + (-1)(2)

a.b = 30 – 16 – 2

a.b = 12