In the previous section, we saw the examples which demonstrates the uniform motion of 'C'. Next we have to learn about angular velocity. For that, first we must be familiar with vector cross products. So we will see that in this section
1. Fig.7.53(a) below shows a point P in space
• There can be infinite number of planes on which 'P' can lie. This is shown in the fig.b
2. Those planes in fig.b come in all possible orientations
• That means, we can take a plane of any orientation. That plane can be positioned so that 'P' will lie on it
• We get this freedom because there is only one point 'P'
3. What if there is a second point 'Q'?
• This is shown in fig.7.54(a) below:
• Note that, a straight line can be drawn between any two points in space
• If the 'line connecting P and Q' lies on a plane, we can say that, both P and Q lies on that plane
4. How many such planes are possible?
Ans: Infinite number of such planes are possible. Fig.7.54(b) shows three of them
5. However, the following 6 facts should be noted:
(i) Though infinite number of planes are possible in this case also, 'any orientation' cannot be allowed
(ii) We first fix up one plane which contains the line PQ
(iii) Then we rotate that plane by considering the line PQ as the 'axis of that rotation'
(iv) All planes thus obtained by rotating through various angles will contain both P and Q
(v) Thus infinite number of planes are possible
(vi) But 'all possible orientations' that we saw in fig.7.53(b) is not possible
6. Thus we find that, 2 points also give as a large amount of freedom with regard to the 'choice of planes'
• What if there is a third point R?
• This is shown in fig.7.55(a) below:
(i) We first fix up a plane containing line PQ
(ii) Then we rotate that plane by considering the line PQ as the axis
(iii) Just when the third point R falls on that plane, we stop the rotation
(iv) That is the plane we want
• No other plane will contain all the three point P, Q and R
• That means, there is one and only one plane which contains 3 given points in space
• This is shown in fig.7.55(b)
7. What if there is a fourth point 'S'?
Ans: If there are a total of four given points, there is no guarantee that, we will ever find a plane which contains all four of them
• It depends on the positions of those points in space
• We can randomly take any three of them. There sure will be a plane containing those three
• But accommodating the fourth point also in that plane may not be possible
• We have seen scalar product of two vectors in chapter 6
■ The result that we obtain when performing 'scalar multiplication' of two given vectors will be a scalar
♦ Eg: We multiply force vector and displacement vector to obtain 'work', which is a scalar
■ The result that we obtain when performing 'vector multiplication' of two given vectors will be a vector
• We will see examples in later sections
• First we will see the features of vector products. We will write them in steps:
1. Consider the two vectors $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ shown in fig.7.56(a) below:
• To find their vector product, both of them must lie on a plane
2. Now, $\mathbf\small{\vec{a}}$ has two well defined points: It's tail end and head end
• Similarly, $\mathbf\small{\vec{b}}$ has two well defined points: It's tail end and head end
• So there are a total of four distinct points
3. If both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ are to lie on a single plane, all those four point must lie on that plane
• But we have seen that any four points need not lie on a single plane
4. So we make an adjustment:
• We shift one of the vectors so that, the tail ends of both the vectors coincide
• Then there will only be three points to consider. We can sure find the plane which contains those three points
• In fig.7.56(b), $\mathbf\small{\vec{b}}$ has been shifted so that it's tail coincide with that of $\mathbf\small{\vec{a}}$
5. Now, we are going to work on the 'plane which contains both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• In our present case, it is convenient to make that plane coincide with our familiar xy-plane
• For that, we rearrange the frame of reference
• The rearrangement is a simple process. Only 3 steps are required:
(i) Take the reference frame and put it in such a way that, the 'O' is in the 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
(ii) With the 'O' as pivot, rotate the frame in a suitable direction so that, the x-axis falls on that 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
(iii) With the x-axis as the 'axis of rotation', rotate the frame in a suitable direction so that, the y-axis also falls on that 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• The final position is shown in fig.7.57(a) below:
6. Now, according to the 'definition of vector product', the 'product vector' that we are seeking, will be perpendicular to both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$
• That means: The 'product vector' will be perpendicular to the 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• For convenience, we will draw a line (perpendicular to the plane) through the tail ends of the vectors. It is shown in magenta color in fig.7.57(b)
• We can assume that, the product vector that we are seeking, lies along that magenta line
7. We see that the magenta line is parallel to the z axis.
• Now, the next question arises:
■ What is the direction of the vector that we are seeking?
• We know that, every vector has a direction
• So just saying that 'the product vector lies along the magenta line in fig.7.57(b)' is not good enough
• We want the answer for these questions as well:
Is the product vector directed towards the positive side of the z-axis?
OR, Is it directed towards the negative side of the z-axis?
8. To find the answer, we use the right hand screw rule. Let us see how this rule is applied:
• To learn about this rule, first we must know the difference between these two:
(i) A right handed screw (ii) A left handed screw
9. In fig.7.58(a) below, a screw is about to be driven into a wooden block
• The direction in which the head of the screw is turned is important
• In the fig.a, the direction of turning is shown by the brown curved arrow
10. Is this brown arrow showing clockwise direction?
OR, Is it showing anti-clockwise direction?
Ans: The answer depends on the direction in which we look at the screw
• If we look at the screw (along it's axis) from it's head towards the tail, the arrow can be said to be clockwise
• If we look at the screw (along it's axis) from it's tail towards the head, the arrow will become anti-clockwise
11. Normally, while turning a screw, we will be looking at it from the head towards the tail
• So we need not specify the direction in which we are looking
• All we need to mention is one of the two below (whichever is appropriate):
(i) Turning a screw in clockwise direction
(ii) Turning a screw in anti-clockwise direction
• Both will imply that, we are looking from the head towards the tail
12. So, the brown screw in fig.7.58(a) above, is being turned in the clockwise direction
• That screw will advance into the block
13. The cyan screw in fig.7.58(b) above, is also being turned in the clockwise direction
• That screw will not advance into the block
• In fact, the opposite will happen. That is., the cyan screw will back out from the block
14. To put it in simple words, we can write the following 3 statements:
(i) Both brown and cyan screws are turned in the clockwise direction
(ii) The brown screw will tighten into the block
(iii) The cyan screw will loosen away from the block
■ The brown screw in fig.a is a right handed screw
■ The cyan screw in fig.b is a left handed screw
• Normally, those we buy from hardware stores are right handed screws
• Left handed screws are manufactured only in small quantities. They are used for some special purposes
1. Fig.7.53(a) below shows a point P in space
 |
| Fig.7.53 |
2. Those planes in fig.b come in all possible orientations
• That means, we can take a plane of any orientation. That plane can be positioned so that 'P' will lie on it
• We get this freedom because there is only one point 'P'
3. What if there is a second point 'Q'?
• This is shown in fig.7.54(a) below:
 |
| Fig.7.54 |
• If the 'line connecting P and Q' lies on a plane, we can say that, both P and Q lies on that plane
4. How many such planes are possible?
Ans: Infinite number of such planes are possible. Fig.7.54(b) shows three of them
5. However, the following 6 facts should be noted:
(i) Though infinite number of planes are possible in this case also, 'any orientation' cannot be allowed
(ii) We first fix up one plane which contains the line PQ
(iii) Then we rotate that plane by considering the line PQ as the 'axis of that rotation'
(iv) All planes thus obtained by rotating through various angles will contain both P and Q
(v) Thus infinite number of planes are possible
(vi) But 'all possible orientations' that we saw in fig.7.53(b) is not possible
6. Thus we find that, 2 points also give as a large amount of freedom with regard to the 'choice of planes'
• What if there is a third point R?
• This is shown in fig.7.55(a) below:
 |
| Fig.7.55 |
(ii) Then we rotate that plane by considering the line PQ as the axis
(iii) Just when the third point R falls on that plane, we stop the rotation
(iv) That is the plane we want
• No other plane will contain all the three point P, Q and R
• That means, there is one and only one plane which contains 3 given points in space
• This is shown in fig.7.55(b)
7. What if there is a fourth point 'S'?
Ans: If there are a total of four given points, there is no guarantee that, we will ever find a plane which contains all four of them
• It depends on the positions of those points in space
• We can randomly take any three of them. There sure will be a plane containing those three
• But accommodating the fourth point also in that plane may not be possible
Now we can start the discussion on vector products
■ The result that we obtain when performing 'scalar multiplication' of two given vectors will be a scalar
♦ Eg: We multiply force vector and displacement vector to obtain 'work', which is a scalar
■ The result that we obtain when performing 'vector multiplication' of two given vectors will be a vector
• We will see examples in later sections
• First we will see the features of vector products. We will write them in steps:
1. Consider the two vectors $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ shown in fig.7.56(a) below:
 |
| Fig.7.56 |
2. Now, $\mathbf\small{\vec{a}}$ has two well defined points: It's tail end and head end
• Similarly, $\mathbf\small{\vec{b}}$ has two well defined points: It's tail end and head end
• So there are a total of four distinct points
3. If both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ are to lie on a single plane, all those four point must lie on that plane
• But we have seen that any four points need not lie on a single plane
4. So we make an adjustment:
• We shift one of the vectors so that, the tail ends of both the vectors coincide
• Then there will only be three points to consider. We can sure find the plane which contains those three points
• In fig.7.56(b), $\mathbf\small{\vec{b}}$ has been shifted so that it's tail coincide with that of $\mathbf\small{\vec{a}}$
5. Now, we are going to work on the 'plane which contains both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• In our present case, it is convenient to make that plane coincide with our familiar xy-plane
• For that, we rearrange the frame of reference
• The rearrangement is a simple process. Only 3 steps are required:
(i) Take the reference frame and put it in such a way that, the 'O' is in the 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
(ii) With the 'O' as pivot, rotate the frame in a suitable direction so that, the x-axis falls on that 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
(iii) With the x-axis as the 'axis of rotation', rotate the frame in a suitable direction so that, the y-axis also falls on that 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• The final position is shown in fig.7.57(a) below:
 |
| Fig.7.57 |
• That means: The 'product vector' will be perpendicular to the 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• For convenience, we will draw a line (perpendicular to the plane) through the tail ends of the vectors. It is shown in magenta color in fig.7.57(b)
• We can assume that, the product vector that we are seeking, lies along that magenta line
7. We see that the magenta line is parallel to the z axis.
• Now, the next question arises:
■ What is the direction of the vector that we are seeking?
• We know that, every vector has a direction
• So just saying that 'the product vector lies along the magenta line in fig.7.57(b)' is not good enough
• We want the answer for these questions as well:
Is the product vector directed towards the positive side of the z-axis?
OR, Is it directed towards the negative side of the z-axis?
8. To find the answer, we use the right hand screw rule. Let us see how this rule is applied:
• To learn about this rule, first we must know the difference between these two:
(i) A right handed screw (ii) A left handed screw
9. In fig.7.58(a) below, a screw is about to be driven into a wooden block
 |
| Fig.7.58 |
• In the fig.a, the direction of turning is shown by the brown curved arrow
10. Is this brown arrow showing clockwise direction?
OR, Is it showing anti-clockwise direction?
Ans: The answer depends on the direction in which we look at the screw
• If we look at the screw (along it's axis) from it's head towards the tail, the arrow can be said to be clockwise
• If we look at the screw (along it's axis) from it's tail towards the head, the arrow will become anti-clockwise
11. Normally, while turning a screw, we will be looking at it from the head towards the tail
• So we need not specify the direction in which we are looking
• All we need to mention is one of the two below (whichever is appropriate):
(i) Turning a screw in clockwise direction
(ii) Turning a screw in anti-clockwise direction
• Both will imply that, we are looking from the head towards the tail
12. So, the brown screw in fig.7.58(a) above, is being turned in the clockwise direction
• That screw will advance into the block
13. The cyan screw in fig.7.58(b) above, is also being turned in the clockwise direction
• That screw will not advance into the block
• In fact, the opposite will happen. That is., the cyan screw will back out from the block
14. To put it in simple words, we can write the following 3 statements:
(i) Both brown and cyan screws are turned in the clockwise direction
(ii) The brown screw will tighten into the block
(iii) The cyan screw will loosen away from the block
■ The brown screw in fig.a is a right handed screw
■ The cyan screw in fig.b is a left handed screw
• Normally, those we buy from hardware stores are right handed screws
• Left handed screws are manufactured only in small quantities. They are used for some special purposes
Now that we know the difference between right handed and left handed screws, we can see how the right hand screw rule is applied in vector multiplication. We will see that in the next section
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