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:
• 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
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:
(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
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 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 = $\small{\left | \vec{A} \right |}$
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 30o with the x axis
• 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 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 (4.33$\small{\hat{i}}$ + 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 30o with the x axis
Method 2:
A force vector 4.33$\small{\hat{i}}$ + 2.5$\small{\hat{j}}$ acts at a point
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:
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 = 50.71o.
7. So $\small{\vec{A}}$ has a magnitude of 14.213 units and it makes an angle of 50.71o with the x axis
1. Consider any three vectors $\small{\vec{A}}$, $\small{\vec{B}}$ and $\small{\vec{C}}$ as shown in fig.4.10 below:
 |
| Fig.4.10 |
• 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 |
• 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 |
• 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
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 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 = $\small{\left | \vec{A} \right |}$
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 30o with the x axis
 |
| Fig.4.13 |
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 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 (4.33$\small{\hat{i}}$ + 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 30o with the x axis
Method 2:
A force vector 4.33$\small{\hat{i}}$ + 2.5$\small{\hat{j}}$ acts at a point
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 |
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 = 50.71o.
7. So $\small{\vec{A}}$ 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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