Chapter 4.6 - Displacement vector from Position vectors

In the previous section we saw the formula to find the resultant of two vectors. In this section, we will see motion in two dimension.
First we will see Displacement. We will write it in steps:
1.Fig.4.20(a) shows the position 'P' of an object in the xy plane. 

Fig.4.20

• The object is at 'P' when the stop watch shows a reading of 't' s. 
2.We can draw the position vector $\small{\vec{r}}$. We have seen the details about position vector in a previous section. 
3. We have also learned about the 'rectangular components of any given vector'. So we can write:
$\small{\vec{r}=\text{x}\hat{i}+\text{y}\hat{j}}$. This is also shown in fig.a
• The horizontal component of  $\small{\vec{r}}$ is x$\small{\hat{i}}$ 
    ♦ Clearly, this horizontal component has a magnitude of x units
    ♦ So the distance OP' = x units 
• The horizontal component of  $\small{\vec{r}}$ is y$\small{\hat{j}}$ 
    ♦ Clearly, this vertical component has a magnitude of y units
    ♦ So the distance OP'' = y units 
So we can write: The coordinates of the object at time 't' s are (x,y)
4. Now consider fig.b
■ The object is moving along the path shown in green color
• When the stop watch showed t₁ s, it is at P₁
    ♦ So $\small{\vec{r_1}}$ is the position vector when t = t₁ s 
• When the stop watch showed t₂ s, it is at P₂
    ♦ So $\small{\vec{r_2}}$ is the position vector when t = t₂ s
5. Put a vector between P₁ and P₂, pointing from P₁ towards P₂. It is shown in fig.c
■ It is a vector joining the initial and final positions. It is shown in fig.c
• So it is a displacement vector. We will denote it as $\vec{\Delta r}$
• We can write:
$\vec{\Delta r}$ is the displacement vector corresponding to the 'motion in a time duration of (t₂-t₁) s'
6. The horizontal and vertical dashed lines in fig.c help us to find the rectangular components of $\vec{\Delta r}$. We find that:
    ♦ The horizontal component of $\small{\vec{\Delta r}}$  has a magnitude of Δx
    ♦ The vertical component of $\small{\vec{\Delta r}}$ has a magnitude of Δy
■ Thus we can write:
$\vec{\Delta r}={\left (\Delta x \right ) \hat i}+{\left (\Delta y \right ) \hat j}$
7. But from our math classes, we know how to find the actual values of Δx and Δy
• We have: Δx = (x₂-x₁) and Δy = (y₂-y₁)
Where, 
    ♦ (x₁,y₁) are the coordinates of P₁
    ♦ (x₂,y₂) are the coordinates of P₂.
■ Thus we get:
$\vec{\Delta r}={(x_2-x_1)\hat i}+{(y_2-y_1)\hat j}$

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8. Now we want the relation between $\small{\vec{r_1}}$ , $\small{\vec{r_2}}$ and $\small{\vec{\Delta r}}$
• Fig.4.21(a) below shows the three vectors

Fig.4.21

• We see that:
(i) The tail of $\small{\vec{\Delta r}}$ coincides with the head of $\small{\vec{r_1}}$
(ii) $\small{\vec{r_2}}$ starts from the tail of $\small{\vec{r_1}}$ and ends at the head of $\small{\vec{\Delta r}}$
■ Thus, by the triangle method of addition, what we have in fig.4.21(a) is:
$\small{\vec{r_1}}$ + $\small{\vec{\Delta r}}$ = $\small{\vec{r_2}}$
9. Let us find ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) 
• That is., we want: [$\small{\vec{r_2}}$ + (-$\small{\vec{r_1}}$) ]
10. So we must first find (-$\small{\vec{r_1}}$) 
For that, reverse ($\small{\vec{r_1}}$) to form (-$\small{\vec{r_1}}$). This is shown in fig.b
11. Then shift (-$\small{\vec{r_1}}$) so that it's tail coincide with the head of $\small{\vec{r_2}}$. This is shown in fig.c
12. Fig.d shows the result ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$)
Compare ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) in fig.d with $\small{\vec{\Delta r}}$ in fig.a. We see that, both are same
• So it is proved:
($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) = $\small{\vec{\Delta r}}$  
■ We can write a summary in the form of Question and Answer:
• How do we find the displacement vector analytically?
Ans: Find ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) 
• How do we find the displacement vector graphically?
Ans: Draw a new vector such that:
    ♦ It's tail coincides with the head of $\small{\vec{r_1}}$  
    ♦ It's head coincides with the head of $\small{\vec{r_2}}$  
This new vector is the required displacement vector

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So we have seen how to obtain displacement vectors from position vectors. In the next section, we will see how to obtain velocity vectors from displacement vectors.

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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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