Chapter 7.6- Center of Mass in Three-Dimensional Problems

In the previous section, we saw the location of 'C' of some common objects. In this section we will see systems where particles are distributed in space

1. Consider a particle 'P' situated in space
• Let it be situated at a certain distance from the origin 'O' of the frame of reference
    ♦ x-axis is shown in red
    ♦ y-axis is shown in green
    ♦ z-axis is shown in blue
• This is shown in fig.7.42 (a) below:

Fig.7.42

2. The particle is shown as a yellow sphere
• It's mass is 0.5 kg
• We see that:
    ♦ It is not situated on the x-axis
    ♦ It is not situated on the y-axis
    ♦ It is not situated on the xy-plane
    ♦ It is situated in space
3. We know the method to specify the positions of such particles
• The method can be written in 5 steps:
(i) Drop a perpendicular from 'P' onto the xy-plane
• This is indicated by the blue line in fig.7.42(b) above
• Note that, the yellow sphere is a bit transparent. We see the blue line starting from the exact center of the sphere
(ii) Let P' be the foot of the perpendicular
• From P', draw a perpendicular onto the x-axis
• This is indicated by the green line
(iii) From P', draw a perpendicular onto the y-axis
• This is indicated by the red line
(iv) Let the length of the red line be x_P = 2 m
• Let the length of the green line be y_P = 3 m
• Let the length of the blue line be z_P = 2 m
(v) Then the coordinates of P are (x_P, y_P, z_P) = (2,3,2)
4. Consider another particle Q. It's mass is 0.75 kg. This is shown in fig.7.43(a) below:

Fig.7.43

• Q will have it's own red, green and blue lines
• Let the lengths of those lines be 1 m, 2 m, and 1 m respectively
• So the coordinates (x_Q, y_Q, z_Q) will be (1,2,1)
5. So we have a system consisting of two particles P and Q
• We want the location of the 'C' of this system
6. In such cases, the 'C' lies at a point whose coordinates are (X,Y,Z)
• We already know the method to find 'X' and 'Y'
7. We can apply the same method to find 'Z' 
• But to find 'Z', the method should be applied in the z-direction
• As usual, the method involves only 4 steps:
(i) Take the distances (from the xy-plane) for each particle
(ii) Apply the 'due weightage' 
(iii) Find the average of those 'weighted distances'
(iv) This average is the 'Z'
8. So we can write a new equation:
Eq.7.3: $\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}}$ 
9. Now we can form the table:

10. From the table, we get:
$\mathbf\small{\sum{m_i}=1.25}$
$\mathbf\small{\sum{m_ix_i}=1.75}$
$\mathbf\small{\sum{m_iy_i}=3}$
$\mathbf\small{\sum{m_iz_i}=1.75}$
• Thus we get: 
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{3}{1.25}=2.4\, \text{m}}$
$\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
11. Using these coordinates, we can mark 'C'. This is shown as a small white sphere in fig.7.43(b) above
• We see that, 'C' has it's own red, green and blue lines
    ♦ The length of it's red line = X = 1.4 m
    ♦ The length of it's green line = Y = 2.4 m
    ♦ The length of it's blue line = Z = 1.4 m
• Also note that, the 'C' lies on the line joining P and Q

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• Now we know the method to find 'C' of the particles distributed in 3 dimensional space

• In such 3-dimensional problems, another easier method can be used. Let us see the details of that method:

1. In fig.7.44(a) below, a vector is shown in magenta color

Fig.7.44

• It's tail end coincides with 'O'
• It's tip coincides with the center of 'P'
■ So it is the position vector of P
• We can denote it as $\mathbf\small{\vec{r}_P}$. (Details here)
2. Imagine that, a person wants to go from O to P
• He can take either one of the two paths given below:
Path 1:
• This path is along the $\mathbf\small{\vec{r}_P}$. It starts from O and ends at P
Path 2:
This path has 3 segments:
■ Segment 1:
• This segment starts from O
• The person travels a distance x_P
    ♦ x_P is the length of the red line
    ♦ So 'magnitude' of travel is x_P  
• Direction of travel for this segment is: 'along the x-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(x_P)\hat{i}}$
• At the end of this travel, the person is at the foot of the green line
■ Segment 2:
• This segment starts from the foot of the green line
• The person travels a distance y_P
    ♦ y_P is the length of the green line
    ♦ So 'magnitude' of travel is y_P  
• Direction of travel for this segment is: 'parallel to the y-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(y_P)\hat{j}}$
• At the end of this travel, the person is at the foot of the blue line
■ Segment 3:
• This segment starts from the foot of the blue line
• The person travels a distance z_P
    ♦ z_P is the length of the blue line
    ♦ So 'magnitude' of travel is z_P  
• Direction of travel for this segment is: 'parallel to the z-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(z_P)\hat{k}}$
• At the end of this travel, the person reaches the particle P
3. So we see 3 vectors in path 2. If we add those vectors, we will reach P
• That means:
Path 2 = $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
4. The initial and final points in both paths 1 and 2 are the same
• So we can write: Path 1 = Path 2
• That is: $\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
5. The above equality in (4) can be established based on fig.7.44(b) also:
(i) In fig.7.44(b), there are a total of 5 vectors:
Red, green, blue, yellow and magenta
(ii) We want to prove this:
Magenta = red + green + blue
(iii) Consider the following three vectors:
Red, green and yellow
• Applying triangle law of vector addition, we get:
yellow = red + green  
(iv) Consider the following three vectors:
Yellow, blue and magenta
• Applying triangle law of vector addition, we get:
Magenta = yellow + blue
(v) Now we expand 'yellow' using the result in (iii). We get:
Magenta = red + green + blue
• This is the required result that we mentioned in (ii)
6. So it is proved beyond doubt. We can confidently write:
$\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
• In this equation, 
    ♦ $\mathbf\small{\vec{r}_P}$ is the position vector  
    ♦ x_P, y_P and z_P are the coordinates of P
■ So we can write:
The following two items are closely related:
(i) position vector of a particle
(ii) The coordinates of that particle

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7. Now, if 'Q' is another particle with coordinates (x_Q,y_Q,z_Q), we can write:
$\mathbf\small{\vec{r}_Q=(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}}$
8. Consider the system consisting of the two particles P and Q
• Let 'C' be the center of mass of the system
• The coordinates of 'C' are (X,Y,Z)
9. If we know those coordinates, we can easily write the position vector of 'C'
• Based on (6) above, we get: $\mathbf\small{\vec{r}_C=(X)\hat{i}+(Y)\hat{j}+(Z)\hat{k}}$
10. Now we expand X, Y and Z. We have:
• $\mathbf\small{X=\frac{m_p\,x_P+m_Q\,x_Q}{m_P+m_Q}=\frac{m_p\,x_P+m_Q\,x_Q}{M}}$
• $\mathbf\small{Y=\frac{m_p\,y_P+m_Q\,y_Q}{M}}$
• $\mathbf\small{Z=\frac{m_p\,z_P+m_Q\,z_Q}{M}}$
11. So the equation in (9) becomes:
$\mathbf\small{\vec{r}_C=\left[\frac{m_p\,x_P+m_Q\,x_Q}{M} \right]\hat{i}+\left[\frac{m_p\,y_P+m_Q\,y_Q}{M} \right]\hat{j}+\left[\frac{m_p\,z_P+m_Q\,z_Q}{M} \right]\hat{k}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{(m_P\,x_P)\hat{i}+(m_Q\,x_Q)\hat{i}+(m_P\,y_P)\hat{j}+(m_Q\,y_Q)\hat{j}+(m_P\,z_P)\hat{k}+(m_Q\,z_Q)\hat{k}}{M}}$
12. Now we rearrange the above equation by the two steps:
(i) Bringing the terms with m_P together
(ii) Bringing the terms with m_Q together
• We get:
$\mathbf\small{\vec{r}_C=\frac{[(m_P\,x_P)\hat{i}+(m_P\,y_P)\hat{j}+(m_P\,z_P)\hat{k}]+[(m_Q\,x_Q)\hat{i}+(m_Q\,y_Q)\hat{j}+(m_Q\,z_Q)\hat{k}]}{M}}$
• Taking m_P and m_Q outside, we get:
$\mathbf\small{\vec{r}_C=\frac{m_P[(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}]+m_Q[(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}]}{M}}$
13. But:
    ♦ $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}=\vec{r}_P}$ 
    ♦ $\mathbf\small{(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}=\vec{r}_Q}$
• Thus (12) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_P[\vec{r}_P]+m_Q[\vec{r}_Q]}{M}}$
14. This gives us an easy method to write the 'position vector of C'
• Let us check and see if we will get the same answer as before:
(i) The coordinates of P are (2,3,2)
• So $\mathbf\small{\vec{r}_P=2\hat{i}+3\hat{j}+2\hat{k}}$
• So $\mathbf\small{m_P\,\vec{r}_P=0.5(2\hat{i}+3\hat{j}+2\hat{k})=\hat{i}+1.5\hat{j}+\hat{k}}$
(ii) The coordinates of Q are (1,2,1)
• So $\mathbf\small{\vec{r}_Q=\hat{i}+2\hat{j}+\hat{k}}$
• So $\mathbf\small{m_Q\,\vec{r}_P=0.75(\hat{i}+2\hat{j}+\hat{k})=0.75\hat{i}+1.5\hat{j}+0.75\hat{k}}$
(iii) Thus the numerator in (13) becomes:
$\mathbf\small{(\hat{i}+1.5\hat{j}+\hat{k})+(0.75\hat{i}+1.5\hat{j}+0.75\hat{k})}$
$\mathbf\small{=(1.75\hat{i}+3.0\hat{j}+1.75\hat{k})}$
• The denominator M is the total mass = (m_P + m_Q) = (0.5+0.75) = 1.25 kg
(iv) Thus we get:
$\mathbf\small{\vec{r}_C=\frac{1.75\hat{i}+3.0\hat{j}+1.75\hat{k}}{1.25}=1.4\hat{i}+2.4\hat{j}+1.4\hat{k}}$
(v) So the coordinates of C are: (1.4,2.4,1.4)
• This is the same result that we obtained before

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15. We will generalize this method:

• P, Q, R, S, . . . are various particles in a system
• There are a total of 'n' particles in the system
    ♦ P is the first particle
    ♦ Q is the second particle
    ♦ R is the third particle . . . so on . . .
• Then:
    ♦ m_P is the first mass
    ♦ m_Q is the second mass
    ♦ m_R is the third mass . . . so on . . .
16. We can write:
    ♦ m₁ which is equal to m_P is the first mass
    ♦ m₂ which is equal to m_Q is the second mass
    ♦ m₃ which is equal to m_R is the third mass . . . so on . . .
• Then:
    ♦ m_i is the i^th mass
    ♦ m_n is the last mass
17. Also we can write:
    ♦ $\mathbf\small{\vec{r}_1}$ which is equal to $\mathbf\small{\vec{r}_P}$ is the first position vector
    ♦ $\mathbf\small{\vec{r}_2}$ which is equal to $\mathbf\small{\vec{r}_Q}$ is the second position vector
    ♦ $\mathbf\small{\vec{r}_3}$ which is equal to $\mathbf\small{\vec{r}_R}$ is the third position vector . . . so on . . .
• Then:
    ♦ $\mathbf\small{\vec{r}_i}$ is the i^th position vector
    ♦ $\mathbf\small{\vec{r}_n}$ is the last position vector
18. So the equation in (13) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{m_1+m_2\,+\,.\,.\,.\,+\,m_i+\,.\,.\,.\,+\,m_n}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{M}}$
19. Thus we get 
Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• This equation is applicable to 2- dimensional and 1-dimensional problems as well

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• Let us apply it to a two dimensional problem that we solved in the previous section

• We will consider solved example 7.3. We will do it as a new solved example:

Solved example 7.7
Three particles P, Q and R are situated at the vertices of an equilateral triangle of side 0.5 m. Their masses are 100 g, 150 g and 200 g respectively. Find the location of the 'C' of this system of 3 particles
Solution:
1. We have three particles so situated in space that, they are at the vertices of an equilateral triangle
• For ease in calculations, we arrange the frame of reference in the following way:
    ♦ The plane of the triangle lies in the xy-plane
    ♦ One of the sides (say PQ), lies on the x-axis
    ♦ The left vertex (P) of that side coincides with O
• This arrangement is shown in fig.7.35 below:

Fig.7.35

2. Once this arrangement is fixed, we can easily write the coordinates of the bottom vertices
• The coordinates of P will be (0,0)
• The coordinates of Q will be (0.5,0)
• The x-coordinate of R will be 0.25
3. To find the y-coordinate of R, we use the following steps:
• Drop a perpendicular from R. This is shown in fig.b
• The foot of this perpendicular is R'
• In the right triangle PR'R, we have:
$\mathbf\small{\sin 60=\frac{RR'}{PR}=\frac{RR'}{0.5}}$
$\mathbf\small{\Rightarrow RR'=\sin 60 \times 0.5=0.433}$
So the y-coordinate of R is 0.433
4. Once we obtain the coordinates, we can write the position vectors:
(i) $\mathbf\small{\vec{r}_P=0\hat{i}+0\hat{j}}$
• This is a null vector
(ii) $\mathbf\small{\vec{r}_Q=0.5\hat{i}+0\hat{j}}$
(iii) $\mathbf\small{\vec{r}_R=0.25\hat{i}+0.433\hat{j}}$
5. We have: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• The numerator becomes: $\mathbf\small{0.1 \times 0+0.15 \times 0.5\hat{i}+0.2 \times 0.25\hat{i}+0.2 \times 0.433\hat{j}}$
= $\mathbf\small{0.125\hat{i}+0.0866\hat{j}}$ 
• The denominator is the total mass M = 0.450 kg
6. Thus we get: $\mathbf\small{\vec{r}_C=\frac{0.125\hat{i}+0.0866\hat{j}}{0.45}=0.278 \hat{i}+0.192 \hat{j}}$
7. So the coordinates of 'C' are: (0.278,0.192)
• This is the same result that we obtained earlier
• Note that, the unit vector $\mathbf\small{\hat{k}}$ does not come in the calculations because, this is a 2-dimensional problem

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So now we know how to find the location of the 'C' of any given system. In the next section, we will see the significance of 'C'

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Chapter 6.1 - Properties of Scalar product

In the previous section, we saw that scalar product of vectors obeys distributive law. In this section, we will see a practical application of the distributive law. We will also see a few more properties of the scalar product.

We will write the steps:
1. Consider the scalar product $\mathbf\small{\vec{A}.\vec{B}}$
2. Let the rectangular components of  $\mathbf\small{\vec{B}}$ be $\mathbf\small{B_x\,\hat{i}\;\; \text{and}\;\;B_y\,\hat{j}}$
3. Then we can write: $\mathbf\small{\vec{A}.\vec{B}=\vec{A}.[B_x\,\hat{i}+B_y\,\hat{j}]}$
• Applying distributive law, we get: $\mathbf\small{\vec{A}.\vec{B}=\vec{A}.(B_x\,\hat{i})+\vec{A}.(B_y\,\hat{j})}$
4. Now, let the rectangular components of  $\mathbf\small{\vec{A}}$ be $\mathbf\small{A_x\,\hat{i}\;\; \text{and}\;\;A_y\,\hat{j}}$
• Then the result in (3) becomes: $\mathbf\small{\vec{A}.\vec{B}=[A_x\,\hat{i}+A_y\,\hat{j}].(B_x\,\hat{i})+[A_x\,\hat{i}+A_y\,\hat{j}].(B_y\,\hat{j})}$
5. Applying distributive law, we get:
$\mathbf\small{\vec{A}.\vec{B}=(A_x\,\hat{i}.B_x\,\hat{i})+(A_y\,\hat{j}.B_x\,\hat{i})+(A_x\,\hat{i}.B_y\,\hat{j})+(A_y\,\hat{j}.B_y\,\hat{j})}$
6. There are 4 terms on the right side. We will analyse each of them separately.
First term is $\mathbf\small{(A_x\,\hat{i}.B_x\,\hat{i})}$  
• This is a dot product of two vectors $\mathbf\small{A_x\,\hat{i}\;\; \text{and}\;\;B_x\,\hat{i}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is $\mathbf\small{A_x}$
(ii) The magnitude of the second vector
• In this case, it is $\mathbf\small{B_x}$
(iii) Cosine of the angle θ between the two vectors
• In this case, both the vectors lie along the x axis. That means they are parallel
• The angle θ between them will be 0
• So cos θ = cos 0 = 1
■ Thus we get: $\mathbf\small{(A_x\,\hat{i}.B_x\,\hat{i})=A_x \times B_x \times \cos 0 =A_x \times B_x \times 1 = A_x \times B_x }$

Second term is $\mathbf\small{(A_y\,\hat{j}.B_x\,\hat{i})}$  
• This is a dot product of two vectors $\mathbf\small{A_y\,\hat{j}\;\; \text{and}\;\;B_x\,\hat{i}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is $\mathbf\small{A_y}$
(ii) The magnitude of the second vector
• In this case, it is $\mathbf\small{B_x}$
(iii) Cosine of the angle θ between the two vectors
• In this case, the first vector lies along the y axis and the second vector lies along the x axis. That means they are perpendicular
• The angle θ between them will be 90^o
• So cos θ = cos 90 = 0
(This is obvious because, we cannot project a vector onto another perpendicular vector)
■ Thus we get: $\mathbf\small{(A_y\,\hat{j}.B_x\,\hat{i})=A_y \times B_x \times \cos 90 =A_y \times B_x \times 0 = 0}$

Third term is $\mathbf\small{(A_x\,\hat{i}.B_y\,\hat{j})}$  
• This is a dot product of two vectors $\mathbf\small{A_x\,\hat{i}\;\; \text{and}\;\;B_y\,\hat{j}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is $\mathbf\small{A_x}$
(ii) The magnitude of the second vector
• In this case, it is $\mathbf\small{B_y}$
(iii) Cosine of the angle θ between the two vectors
• In this case, the first vector lies along the x axis and the second vector lies along the y axis. That means they are perpendicular
• The angle θ between them will be 90^o
• So cos θ = cos 90 = 0
(This is obvious because, we cannot project a vector onto another perpendicular vector)
■ Thus we get: $\mathbf\small{(A_x\,\hat{i}.B_y\,\hat{j})=A_x \times B_y \times \cos 90 =A_x \times B_y \times 0 = 0}$

Fourth term is $\mathbf\small{(A_y\,\hat{j}.B_y\,\hat{j})}$  
• This is a dot product of two vectors $\mathbf\small{A_y\,\hat{j}\;\; \text{and}\;\;B_y\,\hat{j}}$ 
• To calculate the dot product, we want three items:
(i) The magnitude of the first vector
• In this case, it is $\mathbf\small{A_y}$
(ii) The magnitude of the second vector
• In this case, it is $\mathbf\small{B_y}$
(iii) Cosine of the angle θ between the two vectors
• In this case, both the vectors lie along the y axis. That means they are parallel
• The angle θ between them will be 0
• So cos θ = cos 0 = 1
■ Thus we get: $\mathbf\small{(A_y\,\hat{j}.B_y\,\hat{j})=A_y \times B_y \times \cos 0 =A_y \times B_y \times 1 = A_y \times B_y }$
7. Out of the four terms, the second and third terms become zero. So the final result is:
$\mathbf\small{\vec{A}.\vec{B}=(A_x\,\hat{i}.B_x\,\hat{i})+(A_y\,\hat{j}.B_y\,\hat{j})}$
That is: $\mathbf\small{\vec{A}.\vec{B}=A_xB_x+A_yB_y}$
■ We will write it as a result for easy reference
Eq.6.5:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}}$
$\mathbf\small{\text{And}\;\;\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=A_xB_x+A_yB_y}$
8. From the above steps, we get two important results:
(i) The scalar product of two parallel vectors is simply the product of their magnitudes
• This is clear from the analysis of first and fourth terms
(ii) The scalar product of two perpendicular vectors is zero
• This is clear from the analysis of second and third terms
■ We will write them as results for easy reference
Eq.6.6:
$\mathbf\small{\text{If}\;\;\vec{A}\;\;\text{and}\;\;\vec{B}\;\;\text{are parallel}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=|\vec{A}|\times |\vec{B}|}$
Eq.6.7:
$\mathbf\small{\text{If}\;\;\vec{A}\;\;\text{and}\;\;\vec{B}\;\;\text{are perpendicular}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=0}$
9. The result in (7) can be extended to three dimensions also. We get:
Eq.6.8:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}+A_z\,\hat{k}}$
$\mathbf\small{\text{And}\;\;\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}+B_z\,\hat{k}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{B}=A_xB_x+A_yB_y+A_zB_z}$

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Just as we calculated $\mathbf\small{\vec{A}.\vec{B}}$ above, we can calculate $\mathbf\small{\vec{A}.\vec{A}}$ also
We will write the steps:
1. Consider the scalar product $\mathbf\small{\vec{A}.\vec{A}}$
2. Let the rectangular components of  $\mathbf\small{\vec{A}}$ be $\mathbf\small{A_x\,\hat{i}\;\; \text{and}\;\;A_y\,\hat{j}}$
3. Then we can write: $\mathbf\small{\vec{A}.\vec{A}=\vec{A}.[A_x\,\hat{i}+A_y\,\hat{j}]}$
Applying distributive law, we get: $\mathbf\small{\vec{A}.\vec{A}=\vec{A}.(A_x\,\hat{i})+\vec{A}.(A_y\,\hat{j})}$
4. Now, split the other $\mathbf\small{\vec{A}}$ also
Then the result in (3) becomes: $\mathbf\small{\vec{A}.\vec{A}=[A_x\,\hat{i}+A_y\,\hat{j}].(A_x\,\hat{i})+[A_x\,\hat{i}+A_y\,\hat{j}].(A_y\,\hat{j})}$
5. Applying distributive law, we get:
$\mathbf\small{\vec{A}.\vec{A}=(A_x\,\hat{i}.A_x\,\hat{i})+(A_y\,\hat{j}.A_x\,\hat{i})+(A_x\,\hat{i}.A_y\,\hat{j})+(A_y\,\hat{j}.A_y\,\hat{j})}$
6. There are 4 terms on the right side
• Each term is a dot product of two vectors
• If the two vectors in a term are perpendicular to each other, that term will become zero
• Thus the second and third terms will become zero 
7. So the final result is:
$\mathbf\small{\vec{A}.\vec{A}=(A_x\,\hat{i}.A_x\,\hat{i})+(A_y\,\hat{j}.A_y\,\hat{j})}$
That is: $\mathbf\small{\vec{A}.\vec{A}=A_xA_x+A_yA_y}$
That is: $\mathbf\small{\vec{A}.\vec{A}=A_x^2+A_y^2}$
■ We will write it as a result for easy reference
Eq.6.9:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{A}=A_x^2+A_y^2}$
8. The result in (7) can be extended to three dimensions also. We get:
$\mathbf\small{\vec{A}.\vec{A}=(A_x\,\hat{i}+A_y\,\hat{j}+A_z\,\hat{k}).(A_x\,\hat{i}+A_y\,\hat{j}+A_z\,\hat{k})=A_x^2+A_y^2+A_z^2}$
■ We will write it as a result for easy reference
Eq.6.10:
$\mathbf\small{\text{If}\;\;\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}+A_z\,\hat{k}}$
$\mathbf\small{\text{Then}\;\;\vec{A}.\vec{A}=A_x^2+A_y^2+A_z^2}$

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Next we will prove that, if $\mathbf\small{\vec{B}=B_x\:\hat{i}+B_y\:\hat{j}}$, 
Then: $\mathbf\small{\lambda\vec{B}=\lambda(B_x\:\hat{i}+B_y\:\hat{j})=(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$
Where $\mathbf\small{\lambda}$  is a scalar quantity
• A graphic representation of this property is shown in fig.6.5 below:

Fig.6.5

• $\mathbf\small{B_x\,\hat{i}\; \text{and}\;B_y\,\hat{j}}$ are the rectangular components of $\mathbf\small{\vec{B}}$
• $\mathbf\small{B_x\,\hat{i}}$ is multiplied by a scalar $\mathbf\small{\lambda}$
    ♦ This gives a new vector
• $\mathbf\small{B_y\,\hat{j}}$ is multiplied by the same scalar $\mathbf\small{\lambda}$
    ♦ This gives another new vector
• There will be a resultant for these two new vectors
• This new resultant is $\mathbf\small{\vec{B}}$ multiplied by $\mathbf\small{\lambda}$
■ That means:
    ♦ The magnitude of the new resultant is: $\mathbf\small{\lambda |\vec{B}|}$
    ♦ The direction of the new resultant is same as that of $\mathbf\small{\vec{B}}$   
We will write the steps to prove it:
1. On the left side, we are multiplying a vector by a scalar $\mathbf\small{\lambda}$
• Then the magnitude of the new vector will become ${\mathbf\small{\lambda |\vec{B}|}}$
• The direction of the new vector will be same as the original
• We have to arrive at the above results from the right side also. Let us try:
2. On the right side, we have a vector: $\mathbf\small{(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$
• The magnitude of this vector will be: $\mathbf\small{\sqrt{(\lambda B_x)^2+(\lambda B_y)^2}=\sqrt{\lambda^2 B_x^2+\lambda^2 B_y^2}}$ (Using Eq.4.2)
$\mathbf\small{=\sqrt{\lambda^2 (B_x^2+ B_y^2)}=\lambda \sqrt{(B_x^2+ B_y^2)}=\lambda |\vec{B}|}$
• So magnitude of  $\mathbf\small{(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$ is same as the magnitude of $\mathbf\small{\lambda\vec{B}}$
3. The direction of $\mathbf\small{(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$ is given by:
$\mathbf\small{\theta =\tan ^{-1}\frac{\lambda B_y}{\lambda B_x}=\tan ^{-1}\frac{B_y}{B_x}}$ 
• So direction of  $\mathbf\small{(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$ is same as the direction of $\mathbf\small{\lambda\vec{B}}$
■Thus we proved that: $\mathbf\small{\lambda\vec{B}=\lambda(B_x\:\hat{i}+B_y\:\hat{j})=(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$
■ We will write it as a result for easy reference
Eq.6.11:
$\mathbf\small{\text{If}\;\;\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}}$
$\mathbf\small{\text{Then}\;\;\lambda\vec{B}=(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$

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Next we will prove that $\mathbf\small{\vec{A}.(\lambda \vec{B})=\lambda(\vec{A}.\vec{B})}$

We will write the steps:
1. First we will work on the left side
$\mathbf\small{\lambda\vec{B}=(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}}$
[Using Eq.6.11]
• Now, multiplying both sides by $\mathbf\small{\vec{A}}$, we get:
$\mathbf\small{\vec{A}.(\lambda\vec{B})=\vec{A}[(\lambda B_x)\:\hat{i}+(\lambda B_y)\:\hat{j}]}$
• Applying distributive law, we get:
$\mathbf\small{\vec{A}.(\lambda\vec{B})=\vec{A}.[(\lambda B_x)\:\hat{i}]+\vec{A.}[(\lambda B_y)\:\hat{j}}]$
• Now we put $\mathbf\small{\vec{A}=(A_x\:\hat{i}+A_y\:\hat{j})}$. We get: 
$\mathbf\small{\vec{A}.(\lambda \vec{B})=(A_x\:\hat{i}+A_y\:\hat{j}).[(\lambda B_x)\:\hat{i}]+(A_x\:\hat{i}+A_y\:\hat{j}).[(\lambda B_y)\:\hat{j}}]$
• Applying distributive law, we get:
$\mathbf\small{\vec{A}.(\lambda \vec{B})=A_x\:\hat{i}.[(\lambda B_x)\:\hat{i}]+A_y\:\hat{j}.[(\lambda B_x)\:\hat{i}]+A_x\:\hat{i}.[(\lambda B_y)\:\hat{j}]+A_y\:\hat{j}.[(\lambda B_y)\:\hat{j}]}$
2. There are 4 terms in the above result
• Each term is a dot product of two vectors
• If the two vectors in a term are perpendicular to each other, that term will become zero
• Thus the second and third terms will become zero
• We get: $\mathbf\small{\vec{A}.(\lambda \vec{B})=A_x\:\hat{i}.[(\lambda B_x)\:\hat{i}]+A_y\:\hat{j}.[(\lambda B_y)\:\hat{j}]}$
$\mathbf\small{=A_x(\lambda B_x)+A_y(\lambda B_y)}$
$\mathbf\small{=\lambda A_xB_x+\lambda A_yB_y}$
3. Now we work on the right side
• On the right side, we have $\mathbf\small{\lambda(\vec{A}.\vec{B})}$
• But using Eq6.5, $\mathbf\small{\vec{A}.\vec{B}=A_xB_x+A_yB_y}$
• So the right side becomes: $\mathbf\small{\lambda (\vec{A}.\vec{B})=\lambda (A_xB_x+A_yB_y)=\lambda A_xB_x+\lambda A_yB_y}$
• This is same as the result in (2)
• So L.H.S = R.H.S
• Thus we proved that $\mathbf\small{\vec{A}.(\lambda \vec{B})=\lambda(\vec{A}.\vec{B})}$
■ We will write it as a result for easy reference
Eq.6.12:
$\mathbf\small{\vec{A}.(\lambda \vec{B})=\lambda(\vec{A}.\vec{B})}$

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In the next section, we will continue this discussion and see some properties related to unit vectors. We will also see some solved examples

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Chapter 4.3 - Resolution of Vector

In the previous section we saw addition and subtraction of vectors. In this section we will see resolution of vectors.
1. Consider any three vectors $\small{\vec{A}}$, $\small{\vec{B}}$ and $\small{\vec{C}}$ as shown in fig.4.10 below:

Fig.4.10

• They are 'any three' vectors. 
• If we add $\small{\vec{B}}$ and $\small{\vec{C}}$, we will not get $\small{\vec{A}}$. This is shown in fig.4.10(b)
2. Now isolate $\small{\vec{A}}$. Let it's tail be P and tip be Q
• Draw a line parallel to $\small{\vec{B}}$ through P. This is shown in green colour in fig.c  
• Draw a line parallel to $\small{\vec{C}}$ through Q. This is shown in yellow colour in fig.d
• Let the green and yellow lines intersect at R
3. Consider the two points: P and R
• A vector can fit between those two points. This is shown in fig.e
• So we have a $\small{\vec{PR}}$. It is drawn over the green line.
• But the green line is parallel to $\small{\vec{B}}$  
• So $\small{\vec{PR}}$ can be obtained by multiplying $\small{\vec{B}}$ by a scalar λ
■ Thus we get: $\small{\vec{PR}}$ = λ $\small{\vec{B}}$
4. Consider the two points: Q and R
• A vector can fit between those two points. This is shown in fig.e
• So we have a $\small{\vec{QR}}$. It is drawn over the yellow line.
• But the yellow line is parallel to $\small{\vec{C}}$  
• So $\small{\vec{QR}}$ can be obtained by multiplying $\small{\vec{C}}$ by a scalar μ 
■ Thus we get: $\small{\vec{QR}}$ = μ $\small{\vec{C}}$.
5. In the fig.e, we see that: $\small{\vec{A}}$ = λ $\small{\vec{B}}$ + μ $\small{\vec{C}}$
■ Even if A, B and C are 'any three' vectors, we are able to express A in terms of B and C 
6. Can we express $\small{\vec{B}}$ in terms of $\small{\vec{A}}$ and $\small{\vec{C}}$?
Let us try. We will write only the minimum required steps:
(i) Isolate B and draw the green and yellow lines through it's ends. See fig.4.11 below:
    ♦ Green line through P is parallel to A
    ♦ Yellow line through Q is parallel to C

Fig.4.11

7. In the final fig.d, we see that: $\small{\vec{B}}$ = λ $\small{\vec{A}}$ + μ $\small{\vec{C}}$
• But direction of μ $\small{\vec{C}}$ is opposite to that of $\small{\vec{C}}$
• This indicates that, the scalar μ in this case is a negative real number
8. Anyway, we are able to express B in terms of A and C
• Following a similar procedure, we can express C in terms of A and B also
• Reader may draw different sets of 'any three' vectors and try all the three combinations for each set

---

The above discussion leads us to write this:
■ Any vector lying in a plane can be expressed in terms of 'any other two vectors' lying in the same plane
• Concentrate on the words: 'any other two vectors'
• Those words tell us that, the two vectors can have any magnitude and any direction
■ So why not make them like this:
(i) One of them have a magnitude of one unit
• It is directed towards the positive side of the x axis
• We will call it unit vector i. In vector notation, it is written as: $\small{\hat{i}}$ 
• It is read as 'i cap' 
[A cap sign ('^') is given above all unit vectors. This is to distinguish them other vectors]
See fig.4.12(a) below:

Fig.4.12

(ii) The other also have a magnitude of one unit
• But this one is directed towards the positive side of the y axis
• We will call it unit vector j. In vector notation, it is written as: $\small{\hat{j}}$

---

Let us see some properties of unit vectors:
1. A unit vector is a vector having a magnitude of one unit. It points in a particular direction.
2. It has no dimension (like mass, length, time etc.,)
• So it has no unit (like kg, m, s etc.,)
3. It is used to specify a direction only
4. Unit vectors along the x-, y- and z-axes are denoted by $\small{\hat{i},\hat{j}}$ and $\small{\hat{k}}$ respectively
5. Since the magnitude of unit vectors is one, we can write:
$\small{\left | \hat{i} \right |}$ = $\small{\left | \hat{j} \right |}$ = $\small{\left | \hat{k} \right |}$ = 1
■ In our present chapter, we are discussing motion in two dimensions. So in this chapter, we will need $\small{\hat{i}}$ and $\small{\hat{j}}$ only.
6. We can specify unit vectors in any required direction we want. 
• If we multiply a unit vector $\small{\hat{n}}$ by a scalar λ, we will get a new vector: λ$\small{\hat{n}}$
• The magnitude of this new vector will be λ
• The direction of this new vector will be same as that of $\small{\hat{n}}$
7. In general, any $\small{\vec{A}}$ can be written as:
$\small{\vec{A}}$ = $\small{\left | \vec{A} \right |}$ $\small{\hat{n}}$  
• Where n is a unit vector which has the same direction as A

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Now we will see how unit vectors can be helpful to find the components of a given vector
1. In fig.4.12(a) above, a vector $\small{\vec{A}}$ is shown
2. We draw a green line (parallel to $\small{\hat{i}}$ ) through P. This is shown in fig.b
• But $\small{\hat{i}}$ is parallel to the x axis.
• So the green line is parallel to the x axis
3. We draw a yellow line (parallel to $\small{\hat{j}}$) through Q
• But $\small{\hat{j}}$ is parallel to the y axis.
• So the yellow line is parallel to the y axis
4. The green and yellow lines meet at R
• $\small{\vec{PR}}$ is parallel to $\small{\hat{i}}$. So we can write:
$\small{\vec{PR}}$ = λ$\small{\hat{i}}$
• Where λ is a real number
6. $\small{\vec{QR}}$ is parallel to $\small{\hat{j}}$. So we can write:
$\small{\vec{QR}}$ = μ$\small{\hat{j}}$
• Where μ is a real number 
7. From fig.4.12(b), we can see that  [λ$\small{\hat{i}}$ + μ$\small{\hat{j}}$] = $\small{\vec{A}}$
■ That is., λ$\small{\hat{i}}$ and μ$\small{\hat{j}}$ are the components of $\small{\vec{A}}$
8. Since $\small{\hat{i}}$ is parallel to x axis, λ$\small{\hat{i}}$ is called the horizontal component of $\small{\vec{A}}$
• Another name for horizontal component is x component
• The x component of $\small{\vec{A}}$ is denoted as $\small{\vec{A_x}}$
■ So we can write: $\small{\vec{A_x}}$ = λ$\small{\hat{i}}$
This is shown in fig.4.12(c) above.
9. Since $\small{\hat{j}}$ is parallel to y axis, μ$\small{\hat{y}}$ is called the vertical component of $\small{\vec{A}}$
• Another name for vertical component is y component
• The y component of $\small{\vec{A}}$ is denoted as $\small{\vec{A_y}}$
[The x and y components together are called rectangular components of a vector] 
■ So we can write: $\small{\vec{A_y}}$ = μ$\small{\hat{j}}$ 
10. If we can find the values of λ and μ, we can calculate the components A_x and A_y
• So our next aim is to find λ and μ. 
11. For that, consider PQR as a right angled triangle. This is shown in fig.d
Length of PQ will be equal to the magnitude of A. 
That is., PQ = $\small{\left | \vec{A} \right |}$  
12. Length of the base PR will be equal to the magnitude of λ$\small{\hat{i}}$. 
That is., PR = λ 
13. Length of the altitude QR will be equal to the magnitude of μ$\small{\hat{i}}$. 
That is., QR = μ.
14, Note the angle 'θ' shown in fig.c
It is the angle which $\small{\vec{A}}$ makes with the x axis
So it is the direction of the $\small{\vec{A}}$. It will be given to us.
15. Now we can apply the trigonometric ratios
• We have cos $\theta$ = $\frac{\lambda}{magnitude\:of\: \vec{A}}$
⟹ $\lambda$ = ${\left | \vec{A} \right |}$ cos $\theta$
• We have sin $\theta$ = $\frac{\mu}{magnitude\:of\: \vec{A}}$
⟹ $\mu$ = ${\left | \vec{A} \right |}$ sin $\theta$
16. We can write the final results:
Eq.4.1:
(i) Magnitude of the x component of $\small{\vec{A}}$ = $\small{\left | \vec{A_x} \right |}$ = Length of PR = $\lambda$ = ${\left | \vec{A} \right |}$ cos $\theta$
(ii) Magnitude of the y component of $\small{\vec{A}}$ = $\small{\left | \vec{A_y} \right |}$ = Length of QR = $\mu$ = ${\left | \vec{A} \right |}$ sin $\theta$
An example:
• Fig.4.13(a) below shows $\small{\vec{A}}$
    ♦ It has a magnitude of 5 units
    ♦ It makes 30^o with the x axis

Fig.4.13

• Find the horizontal component $\small{\vec{A_x}}$ and the vertical component $\small{\vec{A_y}}$ of $\small{\vec{A}}$
Solution:
1. Using Eq.4.1, we have:
(i) $\small{\left | \vec{A_x} \right |}$ = ${\left | \vec{A} \right |}$ cos $\theta$
• Substituting the values, we get: 
$\small{\left | \vec{A_x} \right |}$ = 5 cos30 = 5 × 0.8660 = 4.33 units
• Direction of $\small{\vec{A_x}}$ is towards the positive side of the x axis 
• So we can write: $\small{\vec{A_x}}$ = 4.33$\small{\hat{i}}$. This is shown in fig.4.13(b)
(ii) $\small{\left | \vec{A_y} \right |}$ = ${\left | \vec{A} \right |}$ sin $\theta$
• Substituting the values, we get: 
$\small{\left | \vec{A_y} \right |}$ = 5 sin30 = 5 × 0.5 = 2.5 units  
• Direction of $\small{\vec{A_y}}$ is towards the positive side of the y axis 
• So we can write: $\small{\vec{A_y}}$ = 2.5$\small{\hat{j}}$
2. Let u see a practical application:
• Since we are dealing with free vectors in this chapter, we will shift $\small{\vec{A_y}}$
• We will shift it so that it's tail coincides with the tails of $\small{\vec{A}}$ and $\small{\vec{A_x}}$
• This is shown in fig.c
3. If a force of 5 N acts at an angle of 30^o at a point, the effect will be the combination of the following two:
(i) A force of 4.33 N pushing the point towards the positive side of x axis 
(i) A force of 2.5 N pushing the point towards the positive side of y axis 
• The combined action is the vector sum (4.33$\small{\hat{i}}$ + 2.5$\small{\hat{j}}$) 
• This sum produces the same effect of 5 N acting at an angle 
■ So there are two methods to represent the vector.
Method 1:
• A force vector acts at a point
• It has a magnitude 5 N
• It has a direction which makes 30^o with the x axis
Method 2:
A force vector 4.33$\small{\hat{i}}$ + 2.5$\small{\hat{j}}$ acts at a point

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17. Thus we successfully calculated the x and y components of the given $\small{\vec{A}}$
• Can we do the reverse?
• That is., if we are given the x and y components of a vector, we must be able to find the original vector. Let us try:
(i) First we will find the magnitude.
• From the right triangle PQR in fig.4.12(d), we have:
|$\small{\vec{A}}$|² = λ² + μ² .
⟹ |$\small{\vec{A}}$|² = |$\small{\vec{A_x}}$|² + |$\small{\vec{A_y}}$|²
From this we get:
Eq.4.2:
|$\small{\vec{A}}$| = ⎷[|$\small{\vec{A_x}}$|² + |$\small{\vec{A_y}}$|²]
(ii) Now we want the direction of $\small{\vec{A}}$
• From the right triangle PQR in fig.4.12(d), we have:
tan $\theta$ = $\frac{\mu}{\lambda}$
• From this we get:
Eq.4.3:
 tan $\theta$ = $\frac{|\vec{A_y}|}{|\vec{A_x}|}$
• We can write: $\theta$ = tan^-1$\frac{|\vec{A_y}|}{|\vec{A_x}|}$
An example:
Given that:
• x component $\small{\vec{A_x}}$ of a vector $\small{\vec{A}}$ is 9$\small{\hat{i}}$
• y component $\small{\vec{A_y}}$ of a vector $\small{\vec{A}}$ is 11$\small{\hat{i}}$
■ Find $\small{\vec{A}}$
Solution:
1. $\small{\vec{A_x}}$ and $\small{\vec{A_y}}$ are shown in fig.4.14(a) below:

Fig.4.14

2. Shift $\small{\vec{A_y}}$ until it's tail coicide with the tip of $\small{\vec{A_x}}$
This is shown in fig.b
3. Join the tail of $\small{\vec{A_x}}$ and tip of $\small{\vec{A_y}}$. This gives $\small{\vec{A}}$. It is shown in fig.c
4. In the resulting right angled triangle, 
Base = |$\small{\vec{A_x}}$| = 9 units   
Altitude = |$\small{\vec{A_y}}$| = 11 units   
5. So hypotenuse = |$\small{\vec{A}}$| = √[9² + 11²] = 14.213 units   
6. Now we want the direction. 
• We have: tan $\theta$ = $\frac{|\vec{A_y}|}{|\vec{A_x}|}$ = $\frac{11}{9}$ = 1.222
• So $\theta$ = tan^-1 1.222 = 50.71^o.
7. So $\small{\vec{A}}$ has a magnitude of 14.213 units and it makes an angle of 50.71^o with the x axis

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In the next section, we will see the analytical method of vector addition.

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