Chapter 4.2 - Addition and Subtraction of Vectors

In the previous section we saw multiplication of a vector by a scalar. In this section we will see addition and subtraction of vectors.

Addition of two vectors

We will see the steps using an example:
• Given two displacement vectors: $\small{\vec{A}}$ and $\small{\vec{B}}$
    ♦ $\small{\vec{A}}$ has a magnitude of 200 km. It makes an angle of 50^o with the x axis
    ♦ $\small{\vec{B}}$ has a magnitude of 300 km. It makes an angle of 105^o with the x axis
• We want to add the vectors $\small{\vec{A}}$ and $\small{\vec{B}}$
Step 1: Represent the vectors on a piece of paper using a suitable scale
• A scale of 1 cm = 50 km will be appropriate for this problem
• So we draw the following on a sheet of paper:
    ♦ A vector with length 4 cm, inclined at an angle of 50^o with the x axis. This represents $\small{\vec{A}}$  
    ♦ A vector with length 6 cm, inclined at an angle of 105^o with the x axis. This represents $\small{\vec{B}}$
• This is shown in fig.4.6(a) below

Fig.4.6

• Note that, in the fig., the coordinate axes are not shown. This is to save space. The reader may draw the axes and measure the angles from x axis with extreme precision.
• Also note that, the positions of the two vectors are not important for this step. 
Step 2: Shift $\small{\vec{B}}$ in such a way that, it's tail coincides with the tip of $\small{\vec{A}}$
This is shown in fig.b
Step 3: Join the following two points using a straight line:
(i) The tail of $\small{\vec{A}}$  (ii) The tip of $\small{\vec{B}}$
• This is shown in fig.c
■ This new line gives the sum of the two vectors
Step 4: Measure the length of this line in cm
• Multiply this length by 50
• The product thus obtained is the magnitude of ($\small{\vec{A}}$ + $\small{\vec{B}}$)
■ ($\small{\vec{A}}$ + $\small{\vec{B}}$) is called the resultant of the two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$
■ The resultant vector is represented as $\small{\vec{r}}$
■ We say: $\small{\vec{r}}$ = ($\small{\vec{A}}$ + $\small{\vec{B}}$)
Step 5: Measure the angle which $\small{\vec{r}}$ makes with the x axis
■ This angle can be used to specify the direction of $\small{\vec{r}}$

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■ What we saw above is a graphical method for addition of two vectors. Because, we have to plot the vectors precisely on a sheet of paper
• In this method, vectors are arranged head to tail
    ♦ So this method is called head-to-tail method for vector addition
• The two vectors and their resultant forms the three sides of a triangle.
    ♦ So this method is also known as the triangle method for vector addition.

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• We successfully calculated ($\small{\vec{A}}$ + $\small{\vec{B}}$). 
• What about ($\small{\vec{B}}$ + $\small{\vec{A}}$)?
• For that, we shift $\small{\vec{A}}$ in such a way that, the tail of $\small{\vec{A}}$ coincide with the tip of $\small{\vec{B}}$
• Then we join the following two points:
(i) The tail of $\small{\vec{B}}$  (ii) The tip of $\small{\vec{A}}$
This is shown in fig.4.6(d) above. 
• Thus we obtain ($\small{\vec{B}}$ + $\small{\vec{A}}$)
■ We find that ($\small{\vec{B}}$ + $\small{\vec{A}}$) has the same magnitude and direction as ($\small{\vec{A}}$ + $\small{\vec{B}}$)
• That is., ($\small{\vec{A}}$ + $\small{\vec{B}}$) = ($\small{\vec{B}}$ + $\small{\vec{A}}$)
■ So we can write: Vector addition is commutative

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Now we will see sum of three vectors
1. Consider fig.4.7(a) below:

Fig.4.7

• Three vectors $\small{\vec{A}}$, $\small{\vec{B}}$ and $\small{\vec{C}}$ are shown
2. We want their resultant $\small{\vec{r}}$ which is given by:
$\small{\vec{A}}$ + $\small{\vec{B}}$ + $\small{\vec{C}}$
3. For that, we first find ($\small{\vec{A}}$ + $\small{\vec{B}}$). This is shown in fig.b
• To that, we add $\small{\vec{C}}$. This is shown in fig.c
So we successfully calculated ( $\small{\vec{A}}$ + $\small{\vec{B}}$ ) + $\small{\vec{C}}$
4. Is there any other method to find ( $\small{\vec{A}}$ + $\small{\vec{B}}$ + $\small{\vec{C}}$ )?
Let us try:
• In fig.d, ($\small{\vec{B}}$ + $\small{\vec{C}}$) is calculated first.
• This sum is added to $\small{\vec{A}}$
• We thus get $\small{\vec{A}}$ + ( $\small{\vec{B}}$ + $\small{\vec{C}}$ )
• We can see that this result is same as ( $\small{\vec{A}}$ + $\small{\vec{B}}$ ) + $\small{\vec{C}}$ in fig.c
5. In fig.e, ($\small{\vec{A}}$ + $\small{\vec{C}}$) is calculated first.
• Then $\small{\vec{B}}$ is added to this sum
• We thus get ($\small{\vec{A}}$ + $\small{\vec{C}}$) + $\small{\vec{B}}$
We can see that this result is same as $\small{\vec{A}}$ + ( $\small{\vec{B}}$ + $\small{\vec{C}}$ ) in fig.d
6. Based on the above results, we can write:
($\small{\vec{A}}$ + $\small{\vec{B}}$) + $\small{\vec{C}}$
= $\small{\vec{A}}$ + ($\small{\vec{B}}$ + $\small{\vec{C}}$)
= ($\small{\vec{A}}$ + $\small{\vec{C}}$) + $\small{\vec{B}}$
■ So we can write: Vector addition is associative

Null vector

1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{-A}}$ shown in fig.4.8(a) below:

Fig.4.8

• Let us add them. That is., we want to find ($\small{\vec{A}}$ + $\small{\vec{-A}}$)
2. For that, we shift $\small{\vec{-A}}$
• We shift $\small{\vec{-A}}$ until it's tail coincides with the tip of $\small{\vec{A}}$
3. Fig.b shows the position just before the tail of $\small{\vec{-A}}$ meets tip of $\small{\vec{A}}$
• We can see that, when they actually meet, the tip of $\small{\vec{-A}}$ will coincide with tail of $\small{\vec{A}}$
4. So it is clear:
If we start from the tail of $\small{\vec{A}}$, we will reach back at the tail of $\small{\vec{A}}$
5. That means, this addition gives us a vector of zero magnitude
■ A vector having zero magnitude is called null vector or a zero vector
• It is represented as $\small{\vec{0}}$ 
• We can write: ($\small{\vec{A}}$ + $\small{\vec{-A}}$) = $\small{\vec{0}}$
■ Since the magnitude of a null vector is zero, it's direction cannot be specified
6. We will get a null vector under the following circumstances:
(i) When we add two vectors which are equal in magnitude but opposite in directions
• That is., ($\small{\vec{A}}$ + $\small{\vec{-A}}$) = $\small{\vec{0}}$
(ii) When we multiply a vector by a scalar, if the value of the scalar is 0, we will get a null vector
• That is., 0 $\small{\vec{A}}$ = $\small{\vec{0}}$  
(iii) When we multiply a vector by a scalar, if the vector is a null vector, we will get a null vector
• That is., λ$\small{\vec{0}}$ = $\small{\vec{0}}$

Subtraction of vectors

1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.8(c) above
• We want to subtract B from A. That is., we want to find ($\small{\vec{A}}$ - $\small{\vec{B}}$)
2. For that, we add $\small{\vec{-B}}$ to $\small{\vec{A}}$. That is.,
($\small{\vec{A}}$ - $\small{\vec{B}}$) = [$\small{\vec{A}}$ + ($\small{\vec{-B}}$)]
3. So first we draw $\small{\vec{-B}}$. This is shown in fig.4.8(c)
• Then we shift it
• We shift it until it's tail coincide with the tip of $\small{\vec{A}}$
• Then we join tail of $\small{\vec{A}}$ and tip of $\small{\vec{B}}$
4. The result is different from ($\small{\vec{A}}$ + $\small{\vec{B}}$) in both magnitude and direction
• For a comparison, ($\small{\vec{A}}$ + $\small{\vec{B}}$) is shown in fig.e

Parallelogram method of vector addition

• Two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ are shown in fig.4.9(a) below:

Fig.4.9

• We want to find ($\small{\vec{A}}$ + $\small{\vec{B}}$)
Step 1: Shift $\small{\vec{B}}$ in such a way that, tail of $\small{\vec{B}}$ coincides with tail of $\small{\vec{A}}$
[Note that, in triangle method of addition, we make the tail of $\small{\vec{B}}$ to coincide with head of $\small{\vec{A}}$]
• In our present method, we want the two tails to be at the same point
• This is shown in fig.4.9(b)
Step 2: Draw a line parallel to $\small{\vec{A}}$
• It must pass through the tip of $\small{\vec{B}}$
• It is shown in green colour in fig.c
Step 3: Draw a line parallel to $\small{\vec{B}}$
• It must pass through the tip of $\small{\vec{A}}$
• It is shown in yellow colour in fig.d
• The green and yellow lines intersect at a point. We get a parallelogram 
Step 4: Consider the following two points
(i) The point at which tails of $\small{\vec{A}}$ and $\small{\vec{B}}$ coincide
(ii) The point of intersection of green and yellow lines
• Draw a vector connecting the above two points
• This vector represents ($\small{\vec{A}}$ + $\small{\vec{B}}$)
• This is shown in fig.e
■ Proof: In fig.e, shift $\small{\vec{B}}$ in such a way that, it's tail coincide with the tip of $\small{\vec{A}}$
• This is shown in fig.f
• We can see that, ($\small{\vec{A}}$ + $\small{\vec{B}}$) now forms the third side of a triangle.
• It is the triangle method. 
■ So parallelogram method is equivalent to triangle method
• Fig.g shows triangle method used independently for the same problem
• We can see that ($\small{\vec{A}}$ + $\small{\vec{B}}$) in figs. e, f and g are the same

Solved example 4.1
Rain is falling vertically with a speed of 35 ms^-1 . Winds starts blowing after sometime with a speed of 12 ms^-1 in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella ?
Solution:
1. Rain is falling vertically. So it can be represented by a vertical vector directed downwards.
• The actual length of that vector is 35 units
• When we draw it on paper, we can use a scale of 1 cm = 5 ms^-1
• So the vector on paper will have a length of $\frac{35}{5}$ = 7 cm
• This is shown in the fig.a below:

2. The standard notation for geographic directions is shown in red colour. 
• Based on that, we have a horizontal vector (representing wind) pointing towards the left   
• Actual length of that vector is 12 units
• When we draw it on paper, we must use the same scale of 1 cm = 5 ms^-1.
• So the vector on paper will have a length of $\frac{12}{5}$ = 2.4 cm 
3. Now we shift the wind vector. We shift it so that, it's tail coincides with the tip of the rain vector
This is shown in fig.b
4. Next we join the following two points with a vector:
(i) Tail of rain vector
(ii) Tip of wind vector
• This vector is the resultant $\small{\vec{R}}$
5. Measure the length (in cm) of this vector 
• Multiply that length by 5. The result is the magnitude of the $\small{\vec{R}}$ in ms^-1.
6. Measure the angle $\theta$ made by $\small{\vec{R}}$ with the vertical
• The boy must hold his umbrella in such a way that, the angle between the umbrella and the vertical is this $\theta$
7. Another method:
(i) To find the magnitude of the resultant: 
• Since the given vectors are vertical and horizontal, we get a right angled triangle in fig.c
Applying Pythagoras theorem, we get:
| $\small{\vec{A}}$ |² = 12² + 35² = 1369
⟹ | $\small{\vec{A}}$ | = √1369 = 37 ms^-1
(ii) To find the direction of the resultant:
• Since we have a right angled triangle, we can use trigonometric ratios to find $\theta$
• In the right angled triangle in fig.c, we have:
tan$\theta$ = $\frac{12}{35}$  
So we get: $\theta$ = tan^-1$\frac{12}{35}$ = tan^-1(0.343) = 19^o

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In the next section, we will see resolution of vectors.

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