Chapter 4.5 - Addition and Subtraction of Vectors using Formula

In the previous section we saw an analytical method to find the resultant vector. In this section, we will see another analytical method which uses a formula.
The steps are given below:

1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.18(a) below:

Fig.4.18

2. Shift them so that, their tails coincide at a point O. This is shown in fig.(b) 
    ♦ Let the tip of $\small{\vec{A}}$ be P
    ♦ Let the tip of $\small{\vec{B}}$ be Q
• Draw a green line parallel to OP
• Draw a yellow line parallel to OQ
• The green and yellow lines meet at S
3. Place a vector between O and S. 
• We know that this vector between O and S is $\small{\vec{R}}$, which is the resultant of $\small{\vec{A}}$ and $\small{\vec{B}}$ (Details in a the previous section)
• The length of OS will give the magnitude of $\small{\vec{R}}$
4. So our first aim is to find the length OS
• For that, extend OP 
• Drop a perpendicular SN on to that extension
5. Now we have a new right triangle: ⊿ONS
• The base ON of this triangle = (OP + PN)
    ♦ Instead of OP, we can write $\left | \vec{A} \right |$

    ♦ Instead PN, we can write PS cos $\theta$

    ♦ But PS =  $\left | \vec{B} \right |$

    ♦ So instead of PN, we can write $\left | \vec{B} \right |$ cos $\theta$

• So the base ON = (OP+PN) = $\left | \vec{A} \right |$  +  $\left | \vec{B} \right |$ cos $\theta$

6. Now, altitude of⊿ONS = SN = PS sin $\theta$ =  $\left | \vec{B} \right |$ sin $\theta$

• Applying Pythagoras theorem, we get: 

OS² = ON² + SN²

$\small{\Rightarrow {OS}^2\, =\, \left ( \left | \vec{A} \right |\, +\, \left | \vec{B} \right |\, cos\, \theta  \right )^2\: +\: \left ( \left | \vec{B} \right |\, sin\, \theta  \right )^2}$
• The right side can be expanded and simplified:
$\small{\Rightarrow {OS}^2\, =\, \left ( \left | \vec{A} \right |^2\, +\,2\left | \vec{A} \right |\, \left | \vec{B} \right |\, cos\,\theta \, +\,   \left | \vec{B} \right |^2 \, {cos}^2\, \theta  \right )\: +\: \left ( \left | \vec{B} \right |^2\, {sin}^2\, \theta  \right )}$
• Group the last two terms together:
$\small{\Rightarrow {OS}^2\, =\, \left | \vec{A} \right |^2\, +\,2\left | \vec{A} \right |\, \left | \vec{B} \right |\, cos\,\theta \, +\,   \left ( \left | \vec{B} \right |^2 \, {cos}^2\, \theta  \: +\:  \left | \vec{B} \right |^2\, {sin}^2\, \theta  \right )}$
$\small{\Rightarrow\, {OS}^2\: =\: \left | \vec{A} \right |^2 \, +\, 2\left | \vec{A} \right | \left | \vec{B} \right |cos\, \theta \, +\, \left | \vec{B} \right |^2\, \left ( {cos}^2\, \theta \, +\, {sin}^2\, \theta  \right )}$
• But $\small{ \left ( {cos}^2\, \theta \, +\, {sin}^2\, \theta  \right ) = 1}$
• So we get:
$\small{{OS}^2\: =\: \left | \vec{A} \right |^2 \, +\, 2\left | \vec{A} \right | \left | \vec{B} \right |cos\, \theta \, +\, \left | \vec{B} \right |^2}$
Thus we can write:
Eq.4.4:
OS = $\small{\left | \vec{R} \right |=\left ( \left | \vec{A}+\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 + 2\left | \vec{A} \right |\left | \vec{B} \right |cos \theta}}$
■ Thus we get the magnitude of $\small{\vec{R}}$
■ Eq.4.4 is known as the Law of cosines
7. Next we want the direction of $\small{\vec{R}}$
• For that, we consider 𝜟ONS
• We have: $tan\,  \alpha \, =\, \frac{SN}{ON}$ = $\frac{SN}{OP\, +\, PN}$
• We have already obtained the expressions for SN, OP and PN 
• Thus we get:
Eq.4.5:
$tan\,  \alpha \, =\, \frac{\left | \vec{B} \right |\, sin\, \theta }{\left | \vec{A} \right |\, +\,\left | \vec{B} \right |\, cos\, \theta }$
• From this expression, we get 'α', which is the angle made by $\small{\vec{R}}$ with one of the vectors $\small{\vec{A}}$
■ We may not be always able to denote the vectors as $\small{\vec{A}}$ and $\small{\vec{B}}$. So remember that, α in Eq.4.5 is the angle between $\small{\vec{R}}$ and the first term (the term without the cosine ratio) in the denominator 
8. Another method to find 'α':
We have two right triangles: ⊿ONS and ⊿PNS
• Consider ⊿ONS. We have:
SN = OS sin α = |$\small{\vec{R}}$| sin α
• Consider ⊿PNS. We have:
SN = SP sin θ = |$\small{\vec{B}}$| sin θ.
• Equating the two, we get: |$\small{\vec{R}}$| sin α = |$\small{\vec{B}}$| sin θ. 
So we can write: 
Eq.4.6:
$\large{\frac{\left | \vec{R} \right |}{sin\, \theta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• So, after calculating the magnitude of $\small{\vec{R}}$, we can use Eq.4.6 to calculate 'α'
9. Now we want the angle 'β', which $\small{\vec{R}}$ makes with the other vector $\small{\vec{B}}$
• For that, we drop a perpendicular from P onto OS. This is shown in fig.4.18(c)
• Now we have two new right triangles: ⊿OPM and ⊿SPM
■ Note that, in ⊿SPM, the angle at S is the same $\small{\beta}$ that we are trying to find. This is because, ∠QOS and ∠OSP are alternate angles 
10. Consider ⊿OPM. We have:
PM = OP sin α = |$\small{\vec{A}}$| sin α
• Consider ⊿SPM. We have:
PM = SP sin β = |$\small{\vec{B}}$| sin β.
• Equating the two, we get: |$\small{\vec{A}}$| sin α = |$\small{\vec{B}}$| sin β.
$\large{\Rightarrow \, \frac{\left | \vec{A} \right |}{sin\, \beta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• But Eq.4.6 above gives another expression for $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• So we can equate the three items. We get:
Eq.4.7:
$\large{\frac{\left | \vec{R} \right |}{sin\, \theta}}$ = $\large{\frac{\left | \vec{A} \right |}{sin\, \beta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
■ Eq.4.7 gives the relation between the three vectors and the angles. It is known as the Law of sines.
• We can see a pattern:
(i) When $\small{\left | \vec{A} \right |}$ is in the numerator
    ♦ Angle 'β', which is related to $\small{\vec{B}}$ is in the denominator
(ii) When $\small{\left | \vec{B} \right |}$ is in the numerator
    ♦ Angle 'α', which is related to $\small{\vec{A}}$ is in the denominator
(iii) When $\small{\left | \vec{R} \right |}$ is in the numerator
    ♦ Angle 'θ', which is related to both $\small{\vec{A}}$ and $\small{\vec{B}}$ is in the denominator
• If we select the first two ratios from Eq.4.7, we will be using $\small{\left | \vec{R} \right |}$, $\small{\left | \vec{A} \right |}$ and θ to calculate β
• If we select the last two ratios from Eq.4.7, we will be using $\small{\left | \vec{A} \right |}$, $\small{\left | \vec{B} \right |}$ and α to calculate β
11. Special cases:
(i) Maximum magnitude for resultant
• Consider Eq.4.4 that we saw above. It gives the magnitude of the resultant. We want to know the 'condition for obtaining the maximum magnitude' 
• From our math classes, we know that, the range of cosine function varies from -1 to +1 both inclusive
• In most cases, the cosine value that we will be using in Eq.4.4 will be a fraction
• We know that, when a quantity is multiplied by a fraction, it's value will decrease
• The only 'non-fraction values' that cos θ can take are: -1 and +1
• So, maximum value of $\small{\left | \vec{R} \right |}$ will be obtained when cos θ = 1
• For cos θ to be 1, θ should be equal to zero
• That is., for maximum magnitude, the two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ should have the same direction  
(ii) Minimum magnitude for resultant
• We want to know the 'condition for obtaining the least possible magnitude'
• When we add two vectors, we may get a resultant which has a negative direction
• But even if the direction is negative, the resultant will have a magnitude.
• So the least possible magnitude is zero. 
• That is., for the least possible magnitude, the two vectors should have the same magnitude but opposite directions
• Resultant of such a pair will have a zero magnitude
• Resultant of all other pairs will have a certain magnitude

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Now we will see a solved example.

We saw Solved example 4.2 in the previous section. We will do the same problem by the new method. The link to the new file is given below:

Solved example 4.3
We find that, the results obtained for $\small{\left | \vec{R} \right |}$, α, β, and $\small{\theta_R}$ are the same in both methods
• One more solved example is given below:
Solved example 4.4

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Now we will see the analytical method for vector subtraction
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.19(a) below:

Fig.4.19

• We want to find ($\small{\vec{A}}$ - $\small{\vec{B}}$)
• That is., we want to find [$\small{\vec{A}}$ + (-$\small{\vec{B}}$)]
2. Draw a vector with the same magnitude of $\small{\vec{B}}$, but opposite in direction.
• This new vector is $\small{\vec{-B}}$. It is also shown in fig.a
3. Shift $\small{\vec{-B}}$ until it's tail coincide with the tail of $\small{\vec{A}}$. This is shown in fig.b
Clearly:
IF angle between $\small{\vec{A}}$ and $\small{\vec{B}}$ is θ,
THEN angle between $\small{\vec{A}}$ and $\small{\vec{-B}}$ is (180-θ).
4. Draw green line parallel to $\small{\vec{A}}$
• Draw yellow line parallel to $\small{\vec{B}}$  
• Thus we get the parallelogram
• Based on the parallelogram, we can derive expressions for the following quantities:
$\small{\left | \vec{R} \right |}$, α, β, and $\small{\theta_R}$
(i) Magnitude of resultant:

$\small{\left | \vec{R} \right |=\left ( \left | \vec{A}-\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 + 2\left | \vec{A} \right |\left | \vec{B} \right |cos(180-\theta)}}$
• But cos (180-θ) = -cos θ. So we get:
Eq.4.4(a):
$\small{\left | \vec{R} \right |=\left ( \left | \vec{A}-\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 - 2\left | \vec{A} \right |\left | \vec{B} \right |cos \theta}}$
(ii) Direction of resultant:
$\tan\alpha =\frac{\left | \vec{B} \right |\sin(180-\theta )}{\left | \vec{A} \right |+\left | \vec{B} \right |\cos(180-\theta )}$
• But sin (180-θ) = sin θ and 
• cos (180-θ) = -cos θ. So we get:
Eq.4.5(a):
$\tan\alpha =\frac{\left | \vec{B} \right |\sin \theta}{\left | \vec{A} \right |-\left | \vec{B} \right |\cos \theta}$
■ In the same way, the other equations can also be written. Readers are advised to write all steps in their own notebooks and derive Eqs.4.4(a), 4.5(a), 4.6(a) and 4.7(a) independently.

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■ A  note about 'angle between two vectors':

1. Consider fig.4.19(c) above.
• Two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ are placed in such a way that, the head of $\small{\vec{B}}$ coincide with the tail of $\small{\vec{A}}$
2. In this position, the angle between the two vectors is $\theta_1$.
• But this is not the correct way for finding the angle between two vectors
3. Both the tails must coincide at the same point. This is shown in fig.d
• In this position, the angle between the two vectors is $\theta_2$.
■ So we can write:
• The angle $\theta$ between the two vectors is $\theta_2$.

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In the next section, we will see vectors applied to motion in a plane.

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Chapter 4.4 - Vector addition by Analytical method

In the previous section we saw resolution of vectors. Now we are in a position to use analytical method to find vector sum.  
• We have seen how to find the resultant using graphical method. 
• But graphical method is tedious and lacks accuracy. 
• So it is preferable to use analytical method. The steps are given below:

1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.15(a) below:

Fig.4.15

• Let the rectangular components of $\small{\vec{A}}$ be $\small{\vec{A_x}}$ and $\small{\vec{A_y}}$ respectively. This is shown in fig.4.15(b)
• Let the rectangular components of $\small{\vec{B}}$ be $\small{\vec{B_x}}$ and $\small{\vec{B_y}}$ respectively
2. Grouping of like-components:
• Group the horizontal components together. That is., group $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$ together. This is shown in fig.4.15(c)   

• Group the vertical components together. That is., group $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$ together
3. Add the like-components:
• Add the two horizontal components $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$.
    ♦ For that, shift $\small{\vec{B_x}}$ so that it's tail coincide with the tip of $\small{\vec{A_x}}$
    ♦ We thus get a new horizonatal vector: ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)    
• Add the two vertical components $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$.
    ♦ For that, shift $\small{\vec{B_y}}$ so that it's tail coincide with the tip of $\small{\vec{A_y}}$
    ♦ We thus get a new vertical vector: ($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)    
4. For that, we will add the two new vectors by the triangle method. 
• Let the resultant of this vector addition be $\small{\vec{R}}$.  This is shown in fig.d
• We can see that: $\small{\vec{R}}$ = [($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)] + [($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)]
• This is an easy analytical method for vector addition
5. Let us check the above result graphically:
For that we add the original vectors by triangle method. This is shown in fig.e
• We find that, the resultant in fig.e is same as the resultant in fig.d
■ So we can write:
$\small{(\vec{A}\: +\: \vec{B})\: =\: [(\vec{A_x}\: +\: \vec{B_x})]\: +\: [(\vec{A_y}\: +\: \vec{B_y})]}$
6. We proved the above result graphically. But once proved, we do not need to draw graphs for doing problems. That is., while adding two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$, we do not need to use graphical methods any more.
We can straight away write:
• The horizontal component of the required resultant $\small{\vec{R}}$ is: $\small{(\vec{A_x}\: +\: \vec{B_x})}$
• The vertical component of the required resultant $\small{\vec{R}}$ is: $\small{(\vec{A_y}\: +\: \vec{B_y})}$
7. Once we find the horizontal and vertical components mentioned in (6) above, we can easily find $\small{\vec{R}}$
But how do we find those components analytically?
• We have already seen the method in the previous section. We will write it again:
(i) If $\theta_A$ is the angle made by $\small{\vec{A}}$ with the x axis, then:
• $\small{\vec{A_x}}$ = $\small{\left [\left | \vec{A} \right |\, cos\, \theta_A \right ]\hat{i}}$  
    ♦ That means: magnitude of $\small{\vec{A_x}}$ is: $\small{\left [\left | \vec{A} \right |\, cos\, \theta_A \right ]}$    
    ♦ Direction of $\small{\vec{A_x}}$ is same as the direction of the unit vector $\small{\hat{i}}$
• $\small{\vec{A_y}}$ = $\small{\left [\left | \vec{A} \right |\, sin\, \theta_A \right ]\hat{j}}$  
    ♦ That means: magnitude of $\small{\vec{A_y}}$ is: $\small{\left [\left | \vec{A} \right |\, sin\, \theta_A \right ]}$    
    ♦ Direction of $\small{\vec{A_y}}$ is same as the direction of the unit vector $\small{\hat{j}}$
(ii) If $\theta_B$ is the angle made by $\small{\vec{B}}$ with the x axis, then:
• $\small{\vec{B_x}}$ = $\small{\left [\left | \vec{B} \right |\, cos\, \theta_B \right ]\hat{i}}$  
    ♦ That means: magnitude of $\small{\vec{B_x}}$ is: $\small{\left [\left | \vec{B} \right |\, cos\, \theta_B \right ]}$
    ♦ Direction of $\small{\vec{B_x}}$ is same as the direction of the unit vector $\small{\hat{i}}$

• $\small{\vec{B_y}}$ = $\small{\left [\left | \vec{B} \right |\, sin\, \theta_B \right ]\hat{j}}$  

    ♦ That means: magnitude of $\small{\vec{B_y}}$ is: $\small{\left [\left | \vec{B} \right |\, sin\, \theta_B \right ]}$
    ♦ Direction of $\small{\vec{B_y}}$ is same as the direction of the unit vector $\small{\hat{j}}$

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• In the above example, we simply added the like-components
• We may encounter problems in which we will have to do subtraction. One such example is given below:
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.16(a) below:

Fig.4.16

• Let the rectangular components of $\small{\vec{A}}$ be $\small{\vec{A_x}}$ and $\small{\vec{A_y}}$ respectively. This is shown in fig.4.16(b)
• Let the rectangular components of $\small{\vec{B}}$ be $\small{\vec{B_x}}$ and $\small{\vec{B_y}}$ respectively
2. Grouping of like-components:
• Group the horizontal components together. That is., group $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$ together. This is shown in fig.4.15(c)   

• Group the vertical components together. That is., group $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$ together
3. Add the like-components:
• Add the two horizontal components $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$.
    ♦ For that, shift $\small{\vec{B_x}}$ so that it's tail coincide with the tip of $\small{\vec{A_x}}$
    ♦ We thus get a new horizonatal vector: ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)
    ♦ Note that, in the fig.d, $\small{\vec{B_x}}$ will come above $\small{\vec{A_x}}$. It is shown slightly out of position. This is for clarity only.
    ♦ From fig.d, we see that, ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$) is in effect, a subtraction
• Add the two vertical components $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$.
    ♦ For that, shift $\small{\vec{B_y}}$ so that it's tail coincide with the tip of $\small{\vec{A_y}}$
    ♦ We thus get a new vertical vector: ($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)    
4. Add the two new vectors by the triangle method. 
• Let the resultant of this vector addition be $\small{\vec{R}}$.  This is shown in fig.d
• We can see that: $\small{\vec{R}}$ = [($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)] + [($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)]
5. Addition of the original vectors by triangle method. This is shown in fig.e
• We find that, the resultant in fig.e is same as the resultant in fig.d

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■ The above points will be clear when we see a solved example. Link to the pdf file is given below:

Solved example 4.2

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Based on the above discussions, we can write a general form for finding the resultant of two vectors.

We will write it in steps:

1. Given two vectors:

(i) $\small{\vec{A}=\left | \vec{A_x} \right |\hat{i}+\left | \vec{A_y} \right |\hat{j}}$

(ii) $\small{\vec{B}=\left | \vec{B_x} \right |\hat{i}+\left | \vec{B_y} \right |\hat{j}}$

• We want ($\small{\vec{A}}$ + $\small{\vec{B}}$)

2. Add like-components to form new vectors:

(i) New vector in the x direction is: $\small{\left | \vec{A_x} \right |\hat{i}+\left | \vec{B_x}\right |\hat{i}}$
• This can be written as: $\small{\left ( \left | \vec{A_x} \right |+\left | \vec{B_x} \right | \right )\hat{i}}$
(ii) New vector in the y direction is: $\small{\left | \vec{A_y} \right |\hat{j}+\left | \vec{B_y}\right |\hat{j}}$
• This can be written as: $\small{\left ( \left | \vec{A_y} \right |+\left | \vec{B_y} \right | \right )\hat{j}}$

3. The vector in 2(i) is the x component of ($\small{\vec{A}}$ + $\small{\vec{B}}$) 

The vector in 2(ii) is the y component of ($\small{\vec{A}}$ + $\small{\vec{B}}$)
■ So we can write:

$\small{\left ( \vec{A}+\vec{B} \right )=\left ( \left | \vec{A_x} \right |+\left | \vec{B_x} \right | \right )\hat{i}+\left ( \left | \vec{A_y} \right |+\left | \vec{B_y} \right | \right )\hat{j}}$

An example:

Find the resultant of two vectors given below:

(i) -17 $\small{\hat{i}}$ + 29.44 $\small{\hat{j}}$

(ii) 37.28 $\small{\hat{i}}$ + 44.43 $\small{\hat{j}}$

Solution:

• Add like components to form new vectors:

    ♦ The new vector in the x direction is: (-17+37.28)$\small{\hat{i}}$ = 20.28$\small{\hat{i}}$ 

    ♦ The new vector in the y direction is: (29.44+44.43)$\small{\hat{j}}$ = 73.87$\small{\hat{j}}$

• So the resultant vector is: 20.28 $\small{\hat{i}}$ + 73.87$\small{\hat{j}}$. See fig.4.17 below.
• Note that, fig.4.17 shows rough sketches. For analytical method, we do not need accurate drawings

Fig.4.17

■ Magnitude of the resultant:

|$\small{\vec{R}}$| = √[20.28² + 73.87²] = 76.60 units

■ Direction of the resultant:
$\theta_R$ = $\tan^{-1}\frac{73.87}{20.28}$ = 74.61^o
■ So the resultant has a magnitude of 76.60 units. It makes an angle of 74.61^o with the x axis

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In the next section, we will see a formula to obtain resultant.

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Chapter 4 - Motion in a Plane

In the previous section we completed a discussion on rectilinear motion. In this chapter we will see motion in a plane.

We have already seen how to specify the position of an object which lies on a plane. (Details here)
For convenience, the fig.3.5 is shown again below:

Fig.3.5

1. Let in fig.a, a point object move on the xy plane
• To describe the motion of that object, we will need the help of two coordinates:
(i) the x coordinate
(ii) the y coordinate
2. Similarly, to describe the motion of an object on the xz plane, we will need the help of two coordinates:
(i) the x coordinate
(ii) the z coordinate
3. Similarly, to describe the motion of an object on the yz plane, we will need the help of two coordinates:
(i) the y coordinate
(ii) the z coordinate
■ Motion in a plane is called two dimensional motion

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• In one dimentional motion, we learned about four items:
(i) Position  (ii) Displacement  (iii) Velocity  (iv) Acceleration
• In two dimentional motion also we have to learn about the same four items. For that, first we have to learn about vectors.

Scalars and Vectors

■ If a physical quantity has magnitude but no direction, it is called a Scalar quantity
• Such a quantity can be completely specified by a single number and a proper unit
• Speed, area, mass, volume, temperature etc., are scalars
■ If a physical quantity has both magnitude and direction, it is called a Vector quantity
• To specify such a quantity completely, we need the following two items:
(i) The magnitude
It is given by a number and a proper unit
(ii) The direction
It is given by statements like:
    ♦ North direction
    ♦ East west direction
    ♦ South west direction
    ♦ 30^o east of north
    ♦ 25^o with x axis
etc.,
• Velocity, force, acceleration, weight, momentum etc., are vectors

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We can learn more details using an example:
1. Distance between two points A and B is 250 m
• Distance is a scalar quantity
• We use distance in those situations where direction is not required. Amount of fuel required for a journey will depend on the distance. Amount of fuel required will not depend on the direction
2. Displacement of an object is 350 m in the north west direction
• Displacement is a vector quantity
• We use displacement in those situations where direction is also necessary. An example is shown in fig.4.1(a) below:

Fig.4.1

(i) An object is displaced from A to B
• AB makes an angle 30^o with the horizontal  
■ So we can write:
Magnitude of the displacement = Length AB
Direction of the displacement = A Direction which makes 30^o with the horizontal
(ii) Consider another possibility:
• The displacement can be along AB' also
• Length AB = Length AB'
• But AB' makes an angle 60^o with the horizontal
• If the object moves along AB, it's final position will be at B
• If the object moves along AB', it's final position will be at B'
(iii) Now, a stationary point P is also present in the problem
• Distance PB' > Distance PB
• So, if the displacement is along AB', the object will end up being at a greater distance from P
• If the displacement is along AB, the object will end up being at a lesser distance from P
• Note that, the distance travelled is the same in both cases
■ So this is a situation where direction also has to be considered
• Like distance, speed, area, mass, volume, temperature etc., are scalars
• Like displacement, velocity, force, acceleration, weight, momentum etc., are vectors

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Let us learn more about vectors:
1. In print media, vectors are represented by boldface letters. For example: 
• v represents the velocity vector
• a represents the acceleration vector
2. But when wriiten by hand, boldface is impractical. So in hand writting, we place an arrow above the letter. For example:
• $\vec{v}$ represents the velocity vector
• $\vec{a}$ represents the acceleration vector
3. Now we need a method to represent the magnitude of a vector. It can be explained with the help of an example:
• Given a force vector $\vec{F}$. 
• Let it's magnitude be 15 N. 
■Then we write:
|$\vec{F}$|= 15 N

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Position vectors

1. Consider fig.4.1(b) above
• An object is moving on the xy plane
• The green curve is the path along which the object moves
2. At time t₁, the object is at P₁
• Join the origin O and P₁ using a straight line. This is shown in fig.4.1(c)  
• Mark an arrow on the line OP₁
• Then $\mathbf\small{\vec{OP_1}}$ is the position vector of the object at time t₁
[The direction of the arrow on the line OP₁ in the fig.4.1(c) is important
Based on the direction of that arrow, we say these:
    ♦ Tail of the position vector $\mathbf\small{\vec{OP_1}}$ is at O   
    ♦ Tip of the position vector $\mathbf\small{\vec{OP_1}}$ is at P₁]
■ We can write the following 3 points:
(i) Position vector of the object at time t₁ is $\mathbf\small{\vec{OP_1}}$
(ii) Magnitude of $\vec{OP_1}$ :
$\mathbf\small{|\vec{OP_1}|}$ = Length OP₁
(iii) Direction of $\vec{OP_1}$ :
The vector $\mathbf\small{\vec{OP_1}}$ makes an angle $\theta_1$ with the x axis
3. At time t₂, the object is at P₂
• Join the origin O and P₂. 
• Then $\mathbf\small{\vec{OP_2}}$ is the position vector of the object at time t₂
■ We can write the following 3 points:
(i) Position vector of the object at time t₂ is $\mathbf\small{\vec{OP_2}}$
(ii) Magnitude of $\mathbf\small{\vec{OP_2}}$:
$\mathbf\small{|\vec{OP_2}|}$ = Length OP₂
(iii) Direction of $\mathbf\small{\vec{OP_2}}$:
The vector $\mathbf\small{\vec{OP_2}}$  makes an angle $\theta_2$ with the x axis

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What is the practical application of position vectors?
Consider a situation:
1. A person gives us the following information:
(i) $\vec{OP_1}$ is the position vector of an object at time t₁
(ii) Magnitude of $\vec{OP_1}$ is 250 m. That is., |$\vec{OP_1}$|= 250 m 
(iii) Direction of $\vec{OP_1}$ is 70^o with the x axis
2. We can use the above information to represent the 'position of the object at time t₁' on a sheet of paper
Let us see how it is done:
(i) On a fresh sheet of paper (preferably a graph paper), draw x and y axis and mark distances to a suitable scale. (A scale of 1 cm = 40 m is appropriate for this problem)
(ii) From the origin O, draw a line OP₁ such that:
• Length of OP₁ = 250 m
• Angle which OP₁ makes with the x axis = 70^o
This is shown in fig.4.2 below:

Fig.4.2

■ Thus P₁ represents the position of the object at time t₁  
• In this way, if we have information about $\vec{OP_2}$, we can show the position P₂ also

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In the next section, we will see displacement vector.

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