In the previous section, we completed a discussion on centripetal force. In this chapter, we will see work, energy and power.
■ When we hear the word ‘work’, different situations come to our minds:
♦ A farmer working in the field
♦ A student working to pass an exam
♦ An artist working on a sculpture
• In this chapter, we try to give a precise meaning for the word ‘work’.
• Once we do that, we will be able to do mathematical calculations to obtain ‘amount of work’
■ When we hear the word ‘energy’, different situations come to our minds:
♦ A person having the ability to work tirelessly for 14 hours surely has greater energy than a person who gets tired after 8 hours of work
♦ A long distance runner surely has greater energy than a short distance runner
♦ A large pile of firewood can supply a larger quantity of energy than a small pile
• In this chapter, we try to give a precise meaning for the word ‘energy’.
• Once we do that, we will be able to do mathematical calculations to obtain ‘amount of energy’
■ When we hear the word ‘power’, different situations come to our minds
♦ A large crane can unload a truck in a shorter time than a small crane. The amount of load is the same. But time requirements are different. We tend to say: The larger crane is more powerful than the smaller crane
♦ A large hotplate can cook food within a lesser time than a small hotplate. The amount of cooked food may be the same. But time requirements are different. We tend to say: The larger hotplate is more powerful than the small hotplate
• In this chapter, we try to give a precise meaning for the word ‘power’.
• Once we do that, we will be able to do mathematical calculations to obtain ‘amount of power’
• We have seen the details about vectors in a previous chapter
• Physical quantities like velocity, displacement, acceleration etc., are vectors
• We have seen how to add or subtract vectors
• Now we will see their multiplication
■ There are two ways for multiplying vectors
(i) The scalar product
This gives a scalar from two vectors
(ii) The vector product
This gives a new vector from two vectors
3. In fig.6.1(a), OP represents $\mathbf\small{\vec{A}}$ and OQ represents $\mathbf\small{\vec{B}}$
• Drop a perpendicular from P to OQ. Let R be the foot of the perpendicular
• Then obviously, $\mathbf\small{OR=|\vec{A}|\cos \theta}$
• That is., OR is the projection of $\mathbf\small{\vec{A}}$ onto $\mathbf\small{\vec{B}}$
• In other words, OR is the 'magnitude of the component of $\mathbf\small{\vec{A}}$' along $\mathbf\small{\vec{B}}$
4. We defined the scalar product as: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta}$
• On the right side, there are 3 items: $\mathbf\small{|\vec{A}|,\,|\vec{B}|,\,\cos \theta}$
• Let us group them into two:
(i) $\mathbf\small{|\vec{B}|}$
(ii) $\mathbf\small{|\vec{A}|\cos \theta}$
5. The scalar product is actually the product of 4(i) and 4(ii)
• So we can write:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{B}}$
(ii) 'Magnitude of the component of $\mathbf\small{\vec{A}}$' along $\mathbf\small{\vec{B}}$
• Alternate definition:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{B}}$
(ii) 'Projection of $\mathbf\small{\vec{A}}$' along $\mathbf\small{\vec{B}}$
(i) $\mathbf\small{|\vec{A}|}$
(ii) $\mathbf\small{|\vec{B}|\cos \theta}$
7. The scalar product is actually the product of 5(i) and 5(ii)
(This is because, three scalars can be multiplied in any order. The final product will not change)
• So we can write in this way also:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{A}}$
(ii) 'Magnitude of the component of $\mathbf\small{\vec{B}}$' along $\mathbf\small{\vec{A}}$
• This is shown in fig.c
• Alternate definition:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{A}}$
(ii) 'Projection of $\mathbf\small{\vec{B}}$' along $\mathbf\small{\vec{A}}$
• OS is the projection of $\mathbf\small{\vec{B}}$ onto $\mathbf\small{\vec{A}}$
8. To obtain the projection, the tail ends of the two vectors need not be at the same point
• If the two vectors are at some distance apart, two perpendiculars should be dropped. This is shown in fig.d
• The green dashed line is parallel to $\mathbf\small{\vec{B}}$
• Details about 'angle between vectors' can be seen here.
An example:
• $\mathbf\small{\vec{F}}$ has magnitude 3 units. It makes 60o with the x axis
• $\mathbf\small{\vec{d}}$ has magnitude 5 units. It makes 20o with the x axis
• Find $\mathbf\small{\vec{F}.\vec{d}}$
Solution:
The two vectors are shown in fig.6.2(a) below:
We will find the answer using 3 different methods
Method 1:
1. Shift $\mathbf\small{\vec{F}}$ so that it's tail end coincide with the tail end of $\mathbf\small{\vec{d}}$
• This is shown in fig.6.2(b)
2. The angle between the two vectors is obviously (60-20) = 40o.
3. The 'projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$ is $\mathbf\small{|\vec{F}|\cos 40=3 \cos 40=2.298}$
• This is marked as OP
4. We have:
The scalar product of two vectors $\mathbf\small{\vec{d}}$and $\mathbf\small{\vec{F}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{d}}$
♦ This magnitude is given as 5 units
(ii) 'Projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$
♦ We obtained this as 2.298
■ Thus the required product is: (5 × 2.298) = 11.49 units.
• This product is a scalar quantity.
Method 1 (alternate):
1. In this method, we do not shift $\mathbf\small{\vec{F}}$. It is kept in it's original position
• The projection is obtained by drawing two perpendiculars from the ends of $\mathbf\small{\vec{F}}$
• This is shown in fig.6.2(c)
2. The 'projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$ is $\mathbf\small{|\vec{F}|\cos 40=3 \cos 40=2.298}$
• This is marked as MN
3. We have:
The scalar product of two vectors $\mathbf\small{\vec{d}}$and $\mathbf\small{\vec{F}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{d}}$
♦ This magnitude is given as 5 units
(ii) 'Projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$
♦ We obtained this as 2.298
■ Thus the required product is: (5 × 2.298) = 11.49 units.
• This product is a scalar quantity.
Method 2:
1. Shift $\mathbf\small{\vec{F}}$ so that it's tail end coincide with the tail end of $\mathbf\small{\vec{d}}$
• This is shown in fig.6.2(b) above
2. The angle between the two vectors is obviously (60-20) = 40o.
3. The 'projection of $\mathbf\small{\vec{d}}$' along $\mathbf\small{\vec{F}}$ is $\mathbf\small{|\vec{d}|\cos 40=5 \cos 40=3.83}$
• This is marked as OQ in fig.6.3(a) below:
4. We have:
The scalar product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{F}}$
♦ This magnitude is given as 3 units
(ii) 'Projection of $\mathbf\small{\vec{d}}$' along $\mathbf\small{\vec{F}}$
♦ We obtained this as 3.83
■ Thus the required product is: (3 × 3.83) = 11.49 units.
• This product is a scalar quantity.
Method 2 (alternate):
1. In this method, we do not shift $\mathbf\small{\vec{F}}$. It is kept in it's original position
• The projection is obtained by drawing two perpendiculars from the ends of $\mathbf\small{\vec{d}}$
• This is shown in fig.6.3(b)
2. The 'projection of $\mathbf\small{\vec{d}}$' along $\mathbf\small{\vec{F}}$ is $\mathbf\small{|\vec{F}|\cos 40=5 \cos 40=3.83}$
• This is marked as UV
3. We have:
The scalar product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{F}}$
♦ This magnitude is given as 3 units
(ii) 'Projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$
♦ We obtained this as 3.83
■ Thus the required product is: (3 × 3.83) = 11.49 units.
• This product is a scalar quantity.
Method 3:
• This time, we simply use the formula: $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}|\times|\vec{d}|\times\cos \theta}$
♦ Where θ is the angle between the two vectors
• Substituting the values, we get: $\mathbf\small{\vec{F}.\vec{d}=3\times5\times\cos 40=3\times5\times 0.7660=11.49\,\text{units}}$
■ So in all the three methods, we get the same result. We find that, method 3 is the simplest
■ We will write it as a result for easy reference
Eq.6.1: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta}$
• Where θ is the angle between $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$
Based on this, we get another interesting result:
■ What is the scalar product of a vector with itself?
■ In other words, What is $\mathbf\small{\vec{A}.\vec{A}}$?
Answer:
• The projection of a vector $\mathbf\small{\vec{A}}$ on itself is obviously $\mathbf\small{|\vec{A}|}$
• The angle θ between a vector and itself is obviously 0
♦ So cos θ = cos 0 = 1
• Thus we get: $\mathbf\small{\vec{A}.\vec{A}=|\vec{A}|\times |\vec{A}|\times \cos 0=|\vec{A}|\times |\vec{A}|\times 1=|\vec{A}|^2}$
■ We will write it as a result for easy reference
Eq.6.2: $\mathbf\small{\vec{A}.\vec{A}=|\vec{A}|^2}$
• In the above example, the magnitude and direction (in the form of angle with x axis) of the vectors were given. We were able to find the scalar product easily
• But some times vectors will be given in the form:
$\mathbf\small{\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}}$
$\mathbf\small{\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}}$
■ How do we find the scalar products in such cases?
• We will see the method in the next section
■ From the above discussions, it is obvious that $\mathbf\small{\vec{A}.\vec{B}=\vec{B}.\vec{A}}$
■ That is., scalar product obeys commutative law
■ We will write it as a result for easy reference
Eq.6.3: $\mathbf\small{\vec{A}.\vec{B}=\vec{B}.\vec{A}}$
• That is., p(q+r) = pq+pr
An example:
• 3×(5+9) = 3 × 14 = 42
• 3×(5+9) = (3×5)+(3×9) = 15 + 27 =42
■ Now we want to prove that scalar product obeys distributive law
Let us write the steps:
1. In fig.6.4(a) below, $\mathbf\small{\vec{A},\,\vec{B}\,\text{and}\,\vec{C}}$ are 3 vectors
• We want to prove that: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=\vec{A}.\vec{C}+\vec{B}.\vec{C}}$
2. The resultant of $\mathbf\small{\vec{A}\,\text{and}\,\vec{B}}$ is $\mathbf\small{(\vec{A}+\vec{B})}$. It is drawn in fig.b
• It is obtained using the 'head to tail method'
3. The resultant $\mathbf\small{(\vec{A}+\vec{B})}$ makes an angle θ with $\mathbf\small{\vec{C}}$
• So projection of $\mathbf\small{(\vec{A}+\vec{B})}$ on $\mathbf\small{\vec{C}}$ is $\mathbf\small{|\vec{R}|\cos \theta}$
• Where $\mathbf\small{\vec{R}=\vec{A}+\vec{B}}$
• This projection is shown as OP in fig.c
4. So we can write: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=\vec{R}.\vec{C}}$
$\mathbf\small{=\text{Projection of}\,\,\vec{R}\,\,\text{on}\,\,\vec{C}\times| \vec{C}|}$
$\mathbf\small{=OP\times| \vec{C}|}$
5. Now consider fig.d. We see that OP = (OQ+QP)
• That is.,$\mathbf\small{OP=\text{Projection of}\,\,\vec{A}\,\,\text{on}\,\,\vec{C}+\text{Projection of}\,\,\vec{B}\,\,\text{on}\,\,\vec{C}}$
• We can write this in the place of 'OP' in the result in step (4). We get:
$\mathbf\small{OP\times| \vec{C}|=(\text{Projection of}\,\,\vec{A}\,\,\text{on}\,\,\vec{C}+\text{Projection of}\,\,\vec{B}\,\,\text{on}\,\,\vec{C})\times|\vec{C}|}$
$\mathbf\small{\Longrightarrow OP\times| \vec{C}|=(\text{Projection of}\,\,\vec{A}\,\,\text{on}\,\,\vec{C}\times|\vec{C}|+\text{Projection of}\,\,\vec{B}\,\,\text{on}\,\,\vec{C}\times|\vec{C}|)}$
$\mathbf\small{\Longrightarrow OP\times| \vec{C}|=(\vec{A}.\vec{C}+\vec{B}.\vec{C})}$
6. But from (4), we have: $\mathbf\small{OP\times| \vec{C}|=(\vec{A}+\vec{B}).\vec{C}}$
Thus we get: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=(\vec{A}.\vec{C}+\vec{B}.\vec{C})}$
■ We will write it as a result for easy reference
Eq.6.4: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=(\vec{A}.\vec{C}+\vec{B}.\vec{C})}$
■ When we hear the word ‘work’, different situations come to our minds:
♦ A farmer working in the field
♦ A student working to pass an exam
♦ An artist working on a sculpture
• In this chapter, we try to give a precise meaning for the word ‘work’.
• Once we do that, we will be able to do mathematical calculations to obtain ‘amount of work’
■ When we hear the word ‘energy’, different situations come to our minds:
♦ A person having the ability to work tirelessly for 14 hours surely has greater energy than a person who gets tired after 8 hours of work
♦ A long distance runner surely has greater energy than a short distance runner
♦ A large pile of firewood can supply a larger quantity of energy than a small pile
• In this chapter, we try to give a precise meaning for the word ‘energy’.
• Once we do that, we will be able to do mathematical calculations to obtain ‘amount of energy’
■ When we hear the word ‘power’, different situations come to our minds
♦ A large crane can unload a truck in a shorter time than a small crane. The amount of load is the same. But time requirements are different. We tend to say: The larger crane is more powerful than the smaller crane
♦ A large hotplate can cook food within a lesser time than a small hotplate. The amount of cooked food may be the same. But time requirements are different. We tend to say: The larger hotplate is more powerful than the small hotplate
• In this chapter, we try to give a precise meaning for the word ‘power’.
• Once we do that, we will be able to do mathematical calculations to obtain ‘amount of power’
• Before we begin our discussion on work, energy and power, we have to learn about the ‘multiplication of vectors’
• Physical quantities like velocity, displacement, acceleration etc., are vectors
• We have seen how to add or subtract vectors
• Now we will see their multiplication
■ There are two ways for multiplying vectors
(i) The scalar product
This gives a scalar from two vectors
(ii) The vector product
This gives a new vector from two vectors
In this chapter we will be using scalar products only. We will see vector products in the next chapter. Following steps will help us to learn about scalar products:
1. Consider two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$
• Their scalar product is defined as: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta}$
♦ Where θ is the angle between A and B
• This is shown in fig.6.1(a) below:
2. $\mathbf\small{\vec{A}.\vec{B}}$ is read as: 'A dot B'
• So 'scalar product' is also known as 'dot product'
1. Consider two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$
• Their scalar product is defined as: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta}$
♦ Where θ is the angle between A and B
• This is shown in fig.6.1(a) below:
 |
| Fig.6.1 |
• So 'scalar product' is also known as 'dot product'
• Drop a perpendicular from P to OQ. Let R be the foot of the perpendicular
• Then obviously, $\mathbf\small{OR=|\vec{A}|\cos \theta}$
• That is., OR is the projection of $\mathbf\small{\vec{A}}$ onto $\mathbf\small{\vec{B}}$
• In other words, OR is the 'magnitude of the component of $\mathbf\small{\vec{A}}$' along $\mathbf\small{\vec{B}}$
4. We defined the scalar product as: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta}$
• On the right side, there are 3 items: $\mathbf\small{|\vec{A}|,\,|\vec{B}|,\,\cos \theta}$
• Let us group them into two:
(i) $\mathbf\small{|\vec{B}|}$
(ii) $\mathbf\small{|\vec{A}|\cos \theta}$
5. The scalar product is actually the product of 4(i) and 4(ii)
• So we can write:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{B}}$
(ii) 'Magnitude of the component of $\mathbf\small{\vec{A}}$' along $\mathbf\small{\vec{B}}$
• Alternate definition:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{B}}$
(ii) 'Projection of $\mathbf\small{\vec{A}}$' along $\mathbf\small{\vec{B}}$
(i) $\mathbf\small{|\vec{A}|}$
(ii) $\mathbf\small{|\vec{B}|\cos \theta}$
7. The scalar product is actually the product of 5(i) and 5(ii)
(This is because, three scalars can be multiplied in any order. The final product will not change)
• So we can write in this way also:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{A}}$
(ii) 'Magnitude of the component of $\mathbf\small{\vec{B}}$' along $\mathbf\small{\vec{A}}$
• This is shown in fig.c
• Alternate definition:
The scalar product of two vectors $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{A}}$
(ii) 'Projection of $\mathbf\small{\vec{B}}$' along $\mathbf\small{\vec{A}}$
• OS is the projection of $\mathbf\small{\vec{B}}$ onto $\mathbf\small{\vec{A}}$
8. To obtain the projection, the tail ends of the two vectors need not be at the same point
• If the two vectors are at some distance apart, two perpendiculars should be dropped. This is shown in fig.d
• The green dashed line is parallel to $\mathbf\small{\vec{B}}$
• Details about 'angle between vectors' can be seen here.
An example:
• $\mathbf\small{\vec{F}}$ has magnitude 3 units. It makes 60o with the x axis
• $\mathbf\small{\vec{d}}$ has magnitude 5 units. It makes 20o with the x axis
• Find $\mathbf\small{\vec{F}.\vec{d}}$
Solution:
The two vectors are shown in fig.6.2(a) below:
 |
| Fig.6.2 |
Method 1:
1. Shift $\mathbf\small{\vec{F}}$ so that it's tail end coincide with the tail end of $\mathbf\small{\vec{d}}$
• This is shown in fig.6.2(b)
2. The angle between the two vectors is obviously (60-20) = 40o.
3. The 'projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$ is $\mathbf\small{|\vec{F}|\cos 40=3 \cos 40=2.298}$
• This is marked as OP
4. We have:
The scalar product of two vectors $\mathbf\small{\vec{d}}$and $\mathbf\small{\vec{F}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{d}}$
♦ This magnitude is given as 5 units
(ii) 'Projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$
♦ We obtained this as 2.298
■ Thus the required product is: (5 × 2.298) = 11.49 units.
• This product is a scalar quantity.
Method 1 (alternate):
1. In this method, we do not shift $\mathbf\small{\vec{F}}$. It is kept in it's original position
• The projection is obtained by drawing two perpendiculars from the ends of $\mathbf\small{\vec{F}}$
• This is shown in fig.6.2(c)
2. The 'projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$ is $\mathbf\small{|\vec{F}|\cos 40=3 \cos 40=2.298}$
• This is marked as MN
3. We have:
The scalar product of two vectors $\mathbf\small{\vec{d}}$and $\mathbf\small{\vec{F}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{d}}$
♦ This magnitude is given as 5 units
(ii) 'Projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$
♦ We obtained this as 2.298
■ Thus the required product is: (5 × 2.298) = 11.49 units.
• This product is a scalar quantity.
Method 2:
1. Shift $\mathbf\small{\vec{F}}$ so that it's tail end coincide with the tail end of $\mathbf\small{\vec{d}}$
• This is shown in fig.6.2(b) above
2. The angle between the two vectors is obviously (60-20) = 40o.
3. The 'projection of $\mathbf\small{\vec{d}}$' along $\mathbf\small{\vec{F}}$ is $\mathbf\small{|\vec{d}|\cos 40=5 \cos 40=3.83}$
• This is marked as OQ in fig.6.3(a) below:
 |
| Fig.6.3 |
The scalar product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{F}}$
♦ This magnitude is given as 3 units
(ii) 'Projection of $\mathbf\small{\vec{d}}$' along $\mathbf\small{\vec{F}}$
♦ We obtained this as 3.83
■ Thus the required product is: (3 × 3.83) = 11.49 units.
• This product is a scalar quantity.
Method 2 (alternate):
1. In this method, we do not shift $\mathbf\small{\vec{F}}$. It is kept in it's original position
• The projection is obtained by drawing two perpendiculars from the ends of $\mathbf\small{\vec{d}}$
• This is shown in fig.6.3(b)
2. The 'projection of $\mathbf\small{\vec{d}}$' along $\mathbf\small{\vec{F}}$ is $\mathbf\small{|\vec{F}|\cos 40=5 \cos 40=3.83}$
• This is marked as UV
3. We have:
The scalar product of two vectors $\mathbf\small{\vec{F}}$ and $\mathbf\small{\vec{d}}$ is the product of two items:
(i) Magnitude of $\mathbf\small{\vec{F}}$
♦ This magnitude is given as 3 units
(ii) 'Projection of $\mathbf\small{\vec{F}}$' along $\mathbf\small{\vec{d}}$
♦ We obtained this as 3.83
■ Thus the required product is: (3 × 3.83) = 11.49 units.
• This product is a scalar quantity.
Method 3:
• This time, we simply use the formula: $\mathbf\small{\vec{F}.\vec{d}=|\vec{F}|\times|\vec{d}|\times\cos \theta}$
♦ Where θ is the angle between the two vectors
• Substituting the values, we get: $\mathbf\small{\vec{F}.\vec{d}=3\times5\times\cos 40=3\times5\times 0.7660=11.49\,\text{units}}$
■ So in all the three methods, we get the same result. We find that, method 3 is the simplest
■ We will write it as a result for easy reference
Eq.6.1: $\mathbf\small{\vec{A}.\vec{B}=|\vec{A}||\vec{B}|\cos \theta}$
• Where θ is the angle between $\mathbf\small{\vec{A}}$ and $\mathbf\small{\vec{B}}$
Based on this, we get another interesting result:
■ What is the scalar product of a vector with itself?
■ In other words, What is $\mathbf\small{\vec{A}.\vec{A}}$?
Answer:
• The projection of a vector $\mathbf\small{\vec{A}}$ on itself is obviously $\mathbf\small{|\vec{A}|}$
• The angle θ between a vector and itself is obviously 0
♦ So cos θ = cos 0 = 1
• Thus we get: $\mathbf\small{\vec{A}.\vec{A}=|\vec{A}|\times |\vec{A}|\times \cos 0=|\vec{A}|\times |\vec{A}|\times 1=|\vec{A}|^2}$
■ We will write it as a result for easy reference
Eq.6.2: $\mathbf\small{\vec{A}.\vec{A}=|\vec{A}|^2}$
• In the above example, the magnitude and direction (in the form of angle with x axis) of the vectors were given. We were able to find the scalar product easily
• But some times vectors will be given in the form:
$\mathbf\small{\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}}$
$\mathbf\small{\vec{B}=B_x\,\hat{i}+B_y\,\hat{j}}$
■ How do we find the scalar products in such cases?
• We will see the method in the next section
■ From the above discussions, it is obvious that $\mathbf\small{\vec{A}.\vec{B}=\vec{B}.\vec{A}}$
■ That is., scalar product obeys commutative law
■ We will write it as a result for easy reference
Eq.6.3: $\mathbf\small{\vec{A}.\vec{B}=\vec{B}.\vec{A}}$
The scalar product obeys distributive law
• We know that 'ordinary numbers' obey distributive law• That is., p(q+r) = pq+pr
An example:
• 3×(5+9) = 3 × 14 = 42
• 3×(5+9) = (3×5)+(3×9) = 15 + 27 =42
■ Now we want to prove that scalar product obeys distributive law
Let us write the steps:
 |
| Fig.6.4 |
2. The resultant of $\mathbf\small{\vec{A}\,\text{and}\,\vec{B}}$ is $\mathbf\small{(\vec{A}+\vec{B})}$. It is drawn in fig.b
• It is obtained using the 'head to tail method'
3. The resultant $\mathbf\small{(\vec{A}+\vec{B})}$ makes an angle θ with $\mathbf\small{\vec{C}}$
• So projection of $\mathbf\small{(\vec{A}+\vec{B})}$ on $\mathbf\small{\vec{C}}$ is $\mathbf\small{|\vec{R}|\cos \theta}$
• Where $\mathbf\small{\vec{R}=\vec{A}+\vec{B}}$
• This projection is shown as OP in fig.c
4. So we can write: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=\vec{R}.\vec{C}}$
$\mathbf\small{=\text{Projection of}\,\,\vec{R}\,\,\text{on}\,\,\vec{C}\times| \vec{C}|}$
$\mathbf\small{=OP\times| \vec{C}|}$
5. Now consider fig.d. We see that OP = (OQ+QP)
• That is.,$\mathbf\small{OP=\text{Projection of}\,\,\vec{A}\,\,\text{on}\,\,\vec{C}+\text{Projection of}\,\,\vec{B}\,\,\text{on}\,\,\vec{C}}$
• We can write this in the place of 'OP' in the result in step (4). We get:
$\mathbf\small{OP\times| \vec{C}|=(\text{Projection of}\,\,\vec{A}\,\,\text{on}\,\,\vec{C}+\text{Projection of}\,\,\vec{B}\,\,\text{on}\,\,\vec{C})\times|\vec{C}|}$
$\mathbf\small{\Longrightarrow OP\times| \vec{C}|=(\text{Projection of}\,\,\vec{A}\,\,\text{on}\,\,\vec{C}\times|\vec{C}|+\text{Projection of}\,\,\vec{B}\,\,\text{on}\,\,\vec{C}\times|\vec{C}|)}$
$\mathbf\small{\Longrightarrow OP\times| \vec{C}|=(\vec{A}.\vec{C}+\vec{B}.\vec{C})}$
6. But from (4), we have: $\mathbf\small{OP\times| \vec{C}|=(\vec{A}+\vec{B}).\vec{C}}$
Thus we get: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=(\vec{A}.\vec{C}+\vec{B}.\vec{C})}$
■ We will write it as a result for easy reference
Eq.6.4: $\mathbf\small{(\vec{A}+\vec{B}).\vec{C}=(\vec{A}.\vec{C}+\vec{B}.\vec{C})}$
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