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 →A, →B and →C as shown in fig.4.10 below:
• They are 'any three' vectors.
• If we add →B and →C, we will not get →A. This is shown in fig.4.10(b)
2. Now isolate →A. Let it's tail be P and tip be Q
• Draw a line parallel to →B through P. This is shown in green colour in fig.c
• Draw a line parallel to →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 →PR. It is drawn over the green line.
• But the green line is parallel to →B
• So →PR can be obtained by multiplying →B by a scalar λ
■ Thus we get: →PR = λ →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 →QR. It is drawn over the yellow line.
• But the yellow line is parallel to →C
• So →QR can be obtained by multiplying →C by a scalar μ
■ Thus we get: →QR = μ →C.
5. In the fig.e, we see that: →A = λ →B + μ →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 →B in terms of →A and →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
7. In the final fig.d, we see that: →B = λ →A + μ →C
• But direction of μ →C is opposite to that of →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: ˆi
• 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:
(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: ˆj
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 ˆi,ˆj and ˆk respectively
5. Since the magnitude of unit vectors is one, we can write:
|ˆi| = |ˆj| = |ˆk| = 1
■ In our present chapter, we are discussing motion in two dimensions. So in this chapter, we will need ˆi and ˆj only.
6. We can specify unit vectors in any required direction we want.
• If we multiply a unit vector ˆn by a scalar λ, we will get a new vector: λˆn
• The magnitude of this new vector will be λ
• The direction of this new vector will be same as that of ˆn
7. In general, any →A can be written as:
→A = |→A| ˆn
• Where n is a unit vector which has the same direction as A
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 →A is shown
2. We draw a green line (parallel to ˆi ) through P. This is shown in fig.b
• But ˆi is parallel to the x axis.
• So the green line is parallel to the x axis
3. We draw a yellow line (parallel to ˆj) through Q
• But ˆj 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
• →PR is parallel to ˆi. So we can write:
→PR = λˆi
• Where λ is a real number
6. →QR is parallel to ˆj. So we can write:
→QR = μˆj
• Where μ is a real number
7. From fig.4.12(b), we can see that [λˆi + μˆj] = →A
■ That is., λˆi and μˆj are the components of →A
8. Since ˆi is parallel to x axis, λˆi is called the horizontal component of →A
• Another name for horizontal component is x component
• The x component of →A is denoted as →Ax
■ So we can write: →Ax = λˆi
This is shown in fig.4.12(c) above.
9. Since ˆj is parallel to y axis, μˆy is called the vertical component of →A
• Another name for vertical component is y component
• The y component of →A is denoted as →Ay
[The x and y components together are called rectangular components of a vector]
■ So we can write: →Ay = μˆj
10. If we can find the values of λ and μ, we can calculate the components Ax and Ay
• 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 = |→A|
12. Length of the base PR will be equal to the magnitude of λˆi.
That is., PR = λ
13. Length of the altitude QR will be equal to the magnitude of μˆi.
That is., QR = μ.
14, Note the angle 'θ' shown in fig.c
It is the angle which →A makes with the x axis
So it is the direction of the →A. It will be given to us.
15. Now we can apply the trigonometric ratios
• We have cos θ = λmagnitudeof→A
⟹ λ = |→A| cos θ
• We have sin θ = μmagnitudeof→A
⟹ μ = |→A| sin θ
16. We can write the final results:
Eq.4.1:
(i) Magnitude of the x component of →A = |→Ax| = Length of PR = λ = |→A| cos θ
(ii) Magnitude of the y component of →A = |→Ay| = Length of QR = μ = |→A| sin θ
An example:
• Fig.4.13(a) below shows →A
♦ It has a magnitude of 5 units
♦ It makes 30o with the x axis
• Find the horizontal component →Ax and the vertical component →Ay of →A
Solution:
1. Using Eq.4.1, we have:
(i) |→Ax| = |→A| cos θ
• Substituting the values, we get:
|→Ax| = 5 cos30 = 5 × 0.8660 = 4.33 units
• Direction of →Ax is towards the positive side of the x axis
• So we can write: →Ax = 4.33ˆi. This is shown in fig.4.13(b)
(ii) |→Ay| = |→A| sin θ
• Substituting the values, we get:
|→Ay| = 5 sin30 = 5 × 0.5 = 2.5 units
• Direction of →Ay is towards the positive side of the y axis
• So we can write: →Ay = 2.5ˆj
2. Let u see a practical application:
• Since we are dealing with free vectors in this chapter, we will shift →Ay
• We will shift it so that it's tail coincides with the tails of →A and →Ax
• This is shown in fig.c
3. If a force of 5 N acts at an angle of 30o 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 (4.33ˆi + 2.5ˆj)
• 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 30o with the x axis
Method 2:
A force vector 4.33ˆi + 2.5ˆj acts at a point
17. Thus we successfully calculated the x and y components of the given →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:
|→A|2 = λ2 + μ2 .
⟹ |→A|2 = |→Ax|2 + |→Ay|2
From this we get:
Eq.4.2:
|→A| = ⎷[|→Ax|2 + |→Ay|2]
(ii) Now we want the direction of →A
• From the right triangle PQR in fig.4.12(d), we have:
tan θ = μλ
• From this we get:
Eq.4.3:
tan θ = |→Ay||→Ax|
• We can write: θ = tan-1|→Ay||→Ax|
An example:
Given that:
• x component →Ax of a vector →A is 9ˆi
• y component →Ay of a vector →A is 11ˆi
■ Find →A
Solution:
1. →Ax and →Ay are shown in fig.4.14(a) below:
2. Shift →Ay until it's tail coicide with the tip of →Ax
This is shown in fig.b
3. Join the tail of →Ax and tip of →Ay. This gives →A. It is shown in fig.c
4. In the resulting right angled triangle,
Base = |→Ax| = 9 units
Altitude = |→Ay| = 11 units
5. So hypotenuse = |→A| = √[92 + 112] = 14.213 units
6. Now we want the direction.
• We have: tan θ = |→Ay||→Ax| = 119 = 1.222
• So θ = tan-1 1.222 = 50.71o.
7. So →A has a magnitude of 14.213 units and it makes an angle of 50.71o with the x axis
1. Consider any three vectors →A, →B and →C as shown in fig.4.10 below:
 |
| Fig.4.10 |
• If we add →B and →C, we will not get →A. This is shown in fig.4.10(b)
2. Now isolate →A. Let it's tail be P and tip be Q
• Draw a line parallel to →B through P. This is shown in green colour in fig.c
• Draw a line parallel to →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 →PR. It is drawn over the green line.
• But the green line is parallel to →B
• So →PR can be obtained by multiplying →B by a scalar λ
■ Thus we get: →PR = λ →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 →QR. It is drawn over the yellow line.
• But the yellow line is parallel to →C
• So →QR can be obtained by multiplying →C by a scalar μ
■ Thus we get: →QR = μ →C.
5. In the fig.e, we see that: →A = λ →B + μ →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 →B in terms of →A and →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 |
• But direction of μ →C is opposite to that of →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: ˆi
• 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 |
• 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: ˆj
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 ˆi,ˆj and ˆk respectively
5. Since the magnitude of unit vectors is one, we can write:
|ˆi| = |ˆj| = |ˆk| = 1
■ In our present chapter, we are discussing motion in two dimensions. So in this chapter, we will need ˆi and ˆj only.
6. We can specify unit vectors in any required direction we want.
• If we multiply a unit vector ˆn by a scalar λ, we will get a new vector: λˆn
• The magnitude of this new vector will be λ
• The direction of this new vector will be same as that of ˆn
7. In general, any →A can be written as:
→A = |→A| ˆn
• Where n is a unit vector which has the same direction as A
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 →A is shown
2. We draw a green line (parallel to ˆi ) through P. This is shown in fig.b
• But ˆi is parallel to the x axis.
• So the green line is parallel to the x axis
3. We draw a yellow line (parallel to ˆj) through Q
• But ˆj 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
• →PR is parallel to ˆi. So we can write:
→PR = λˆi
• Where λ is a real number
6. →QR is parallel to ˆj. So we can write:
→QR = μˆj
• Where μ is a real number
7. From fig.4.12(b), we can see that [λˆi + μˆj] = →A
■ That is., λˆi and μˆj are the components of →A
8. Since ˆi is parallel to x axis, λˆi is called the horizontal component of →A
• Another name for horizontal component is x component
• The x component of →A is denoted as →Ax
■ So we can write: →Ax = λˆi
This is shown in fig.4.12(c) above.
9. Since ˆj is parallel to y axis, μˆy is called the vertical component of →A
• Another name for vertical component is y component
• The y component of →A is denoted as →Ay
[The x and y components together are called rectangular components of a vector]
■ So we can write: →Ay = μˆj
10. If we can find the values of λ and μ, we can calculate the components Ax and Ay
• 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 = |→A|
12. Length of the base PR will be equal to the magnitude of λˆi.
That is., PR = λ
13. Length of the altitude QR will be equal to the magnitude of μˆi.
That is., QR = μ.
14, Note the angle 'θ' shown in fig.c
It is the angle which →A makes with the x axis
So it is the direction of the →A. It will be given to us.
15. Now we can apply the trigonometric ratios
• We have cos θ = λmagnitudeof→A
⟹ λ = |→A| cos θ
• We have sin θ = μmagnitudeof→A
⟹ μ = |→A| sin θ
16. We can write the final results:
Eq.4.1:
(i) Magnitude of the x component of →A = |→Ax| = Length of PR = λ = |→A| cos θ
(ii) Magnitude of the y component of →A = |→Ay| = Length of QR = μ = |→A| sin θ
An example:
• Fig.4.13(a) below shows →A
♦ It has a magnitude of 5 units
♦ It makes 30o with the x axis
 |
| Fig.4.13 |
Solution:
1. Using Eq.4.1, we have:
(i) |→Ax| = |→A| cos θ
• Substituting the values, we get:
|→Ax| = 5 cos30 = 5 × 0.8660 = 4.33 units
• Direction of →Ax is towards the positive side of the x axis
• So we can write: →Ax = 4.33ˆi. This is shown in fig.4.13(b)
(ii) |→Ay| = |→A| sin θ
• Substituting the values, we get:
|→Ay| = 5 sin30 = 5 × 0.5 = 2.5 units
• Direction of →Ay is towards the positive side of the y axis
• So we can write: →Ay = 2.5ˆj
2. Let u see a practical application:
• Since we are dealing with free vectors in this chapter, we will shift →Ay
• We will shift it so that it's tail coincides with the tails of →A and →Ax
• This is shown in fig.c
3. If a force of 5 N acts at an angle of 30o 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 (4.33ˆi + 2.5ˆj)
• 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 30o with the x axis
Method 2:
A force vector 4.33ˆi + 2.5ˆj acts at a point
17. Thus we successfully calculated the x and y components of the given →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:
|→A|2 = λ2 + μ2 .
⟹ |→A|2 = |→Ax|2 + |→Ay|2
From this we get:
Eq.4.2:
|→A| = ⎷[|→Ax|2 + |→Ay|2]
(ii) Now we want the direction of →A
• From the right triangle PQR in fig.4.12(d), we have:
tan θ = μλ
• From this we get:
Eq.4.3:
tan θ = |→Ay||→Ax|
• We can write: θ = tan-1|→Ay||→Ax|
An example:
Given that:
• x component →Ax of a vector →A is 9ˆi
• y component →Ay of a vector →A is 11ˆi
■ Find →A
Solution:
1. →Ax and →Ay are shown in fig.4.14(a) below:
 |
| Fig.4.14 |
This is shown in fig.b
3. Join the tail of →Ax and tip of →Ay. This gives →A. It is shown in fig.c
4. In the resulting right angled triangle,
Base = |→Ax| = 9 units
Altitude = |→Ay| = 11 units
5. So hypotenuse = |→A| = √[92 + 112] = 14.213 units
6. Now we want the direction.
• We have: tan θ = |→Ay||→Ax| = 119 = 1.222
• So θ = tan-1 1.222 = 50.71o.
7. So →A has a magnitude of 14.213 units and it makes an angle of 50.71o with the x axis
In the next section, we will see the analytical method of vector addition.
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