In the previous section,
we saw types of vectors. In this section, we will see addition of vectors.
Some basic details about addition can be written in 4 steps:
1. Suppose that, a person moves from A to B. This displacement can be represented by $\small{\vec{AB}}$ in fig.26.12 below:
 |
| Fig.26.12 |
2. If after reaching B, the person moves to C, we will get a new vector $\small{\vec{BC}}$
3. So the net effect is that, the person has moved from A to C. This net displacement can be represented by a third vector $\small{\vec{AC}}$.
• The three vectors $\small{\vec{AB}, \vec{BC}~\text{and}~\vec{AC}}$ form the sides of a triangle ABC.
4. We say that, $\small{\vec{AC}}$ is the resultant of $\small{\vec{AB}~\text{and}~\vec{BC}}$
• Mathematically, it is written as:$\small{\vec{AC} = \vec{AB} + \vec{BC}}$
• This is known as the
triangle law of vector addition.
Now we will see the general method of vector addition. It can be written in 3 steps:
1. Fig.26.13 (i) below shows two vectors $\small{\vec{a}~\text{and}~\vec{b}}$
 |
| Fig.26.13 |
• We want to find $\small{\vec{a} + \vec{b}}$
2. For that, we shift $\small{\vec{b}}$ in such a way that, the initial point of $\small{\vec{b}}$ coincides with the terminal point of $\small{\vec{a}}$. It is important not to change the magnitude or direction of $\small{\vec{b}}$ while making the shift. Fig(ii) shows the position after the shift. We see that, $\small{\vec{a}~\text{and}~\vec{b}}$ now forms the sides AB and BC of the triangle ABC
3. The resultant vector $\small{\vec{a} + \vec{b}}$ is represented by the third side AC of triangle ABC.
We can write: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$
From the above discussion, we get an interesting result. It can be written in 4 steps:
1. Suppose that the person traveled three segments:
• First from A to B, then from B to C and finally from C to A
• We want to know the resultant displacement.
• That is., we want to find $\small{\vec{AB} + \vec{BC} + \vec{CA}}$
2. We have obtained: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$
Substituting this in the above result, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Resultant}} & {~=~} &{\vec{AB} + \vec{BC} + \vec{CA}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\vec{AC}+ \vec{CA}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{-\vec{CA}+ \vec{CA}}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\vec{0}}
\\ \end{array}}$
◼ Remarks
3 (magenta color): Here we apply the fact that, $\small{-\vec{CA}}$ is same as $\small{\vec{AC}}$
3. So the resultant is a null vector. That means, the resultant does not have magnitude. That means, the net displacement is zero.
4. We can write:
When sides of a triangle are taken in order, the resultant will be a null vector. This is because, the initial and terminal points will be the same.
Now we will see subtraction. It can be written in 4 steps:
1. We want $\small{\vec{a} - \vec{b}}$. That is., we want the resultant of $\small{\vec{a}~\text{and}~-\vec{b}}$
2. For that, we draw $\small{-\vec{b}}$ at a convenient place and shift it. We shift it in such a way that, the initial point of $\small{-\vec{b}}$ coincides with the terminal point of $\small{\vec{a}}$. It is important not to change the magnitude or direction of $\small{-\vec{b}}$ while making the shift. Fig.26.13(iii) above, shows the position after the shift. We see that, $\small{\vec{a}~\text{and}~-\vec{b}}$ forms the sides AB and BC' of the triangle ABC'
3. Applying the triangle law of vector addition, we get:
$\small{\vec{a} + \left(-\vec{b} \right) = \vec{AB} + \vec{BC'} = \vec{AC'}}$
• $\small{\vec{AC'}}$ is said to represent the difference of $\small{\vec{a}}$ and $\small{\vec{b}}$
4. So, for subtraction, we make use of the negative of the vector to be subtracted.
Now we will see the
parallelogram law of vector addition. It can be explained in 7 steps:
1. In fig.26.14(i) below, we have two vectors $\small{\vec{a}~\text{and}~\vec{b}}$
 |
| Fig.26.14 |
• We want to find $\small{\vec{a} + \vec{b}}$
2. For that, we shift $\small{\vec{b}}$ in such a way that, the initial point of $\small{\vec{b}}$ coincides with the initial point of $\small{\vec{a}}$. This is shown in fig(ii). [Recall that in the triangle method, initial point of $\small{\vec{b}}$ must coincides with the terminal point of $\small{\vec{a}}$. This is shown in fig(iii)] It is important not to change the magnitude or direction of $\small{\vec{b}}$ while making the shift.
3. Next step is to draw two parallel lines:
♦ Through the terminal point of $\small{\vec{b}}$, draw a dashed line parallel to $\small{\vec{a}}$
♦ Through the terminal point of $\small{\vec{a}}$, draw a dashed line parallel to $\small{\vec{b}}$
4. Let the two dashed lines intersect at C. Then the resultant of $\small{\vec{a}~\text{and}~\vec{b}}$ is given by $\small{\vec{AC}}$
• Note that AC is the diagonal of the parallelogram ABCD
• Note also that, the diagonal AC, which is drawn through the initial points of $\small{\vec{a}~\text{and}~\vec{b}}$ has to be considered. We must not consider the other diagonal BD
5. Now we can write the parallelogram law of vector addition:
If two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ can be represented in magnitude and direction by two adjacent sides of a parallelogram, then their sum $\small{\vec{a} + \vec{b}}$ is represented in magnitude and direction by the diagonal of the parallelogram through their common point.
6. Note that $\small{\vec{a} + \vec{b}}$ obtained in both fig(ii) and fig(iii) is the same. So we can write:
• Triangle law and parallelogram law are equivalent to each other.
7. The equivalence can be proved analytically also. It can be done in 5 steps:
(i) In fig.26.14(ii), $\small{\vec{BC} = \vec{AD}}$ because, they have the same magnitude and direction.
(ii) In triangle ABC, we can apply triangle law. We get: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$
(iii) So using parallelogram law, we have: $\small{\vec{AB} + \vec{AD} = \vec{AC}}$
(iv) Using triangle law, we have: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$, which is same as: $\small{\vec{AB} + \vec{AD} = \vec{AC}}$
(v) The results in (iii) and (iv) are the same. Hence the two laws are equivalent to each other.
Properties of vector addition
Property I (commutative property):
For any two vectors $\small{\vec{a}~\text{and}~\vec{b}}$, $\small{\vec{a}+\vec{b} = \vec{b} + \vec{a}}$
Proof can be written in 5 steps:
1. In fig.26.15 below, $\small{\vec{AB}=\vec{a}~\text{and}~\vec{AD}=\vec{b}}$
 |
| Fig.26.15 |
$\small{\vec{a}~\text{and}~\vec{b}}$ form the adjacent sides of the parallelogram ABCD. Also, the initial points of both vectors are at A.
2. In the parallelogram ABCD,
• $\small{\vec{BC}=\vec{AD}=\vec{b}}$ because, they have the same magnitude and direction.
• $\small{\vec{DC}=\vec{AB}=\vec{a}}$ because, they have the same magnitude and direction.
3. In the triangle ABC, we can apply the triangle law. We get:
$\small{\vec{AB}+\vec{BC}=\vec{AC}}$
$\small{\Rightarrow \vec{a}+\vec{b}=\vec{AC}}$
4. In the triangle ADC also, we can apply the triangle law. We get:
$\small{\vec{AD}+\vec{DC}=\vec{AC}}$
$\small{\Rightarrow \vec{b}+\vec{a}=\vec{AC}}$
5. From (3) and (4), we get: $\small{\vec{a}+\vec{b} = \vec{b} + \vec{a}}$
Property II (Associative property):
For any three vectors $\small{\vec{a}, \vec{b}~\text{and}~\vec{c}}$,
$\small{\left(\vec{a}+\vec{b} \right) + \vec{c}~=~\vec{a} +
\left(\vec{b}+\vec{c} \right)~=~\left(\vec{a}+\vec{c} \right) +
\vec{b}}$
Proof can be written in 4 steps:
1. In fig.26.16(i) below, $\small{\vec{PQ}=\vec{a}, \vec{QR}=\vec{b}~\text{and}~\vec{RS}=\vec{c}}$
 |
| Fig.26.16 |
• Consider triangle PQR. Applying triangle law, we get:
$\small{\vec{a}+\vec{b} = \vec{PQ} + \vec{QR} = \vec{PR}}$
• Consider triangle PRS. Applying triangle law, we get:
$\small{\left(\vec{a}+\vec{b} \right) + \vec{c}= \vec{PR} + \vec{RS} = \vec{PS}}$
2. In fig.26.16(ii) above, $\small{\vec{PQ}=\vec{a}, \vec{QR}=\vec{b}~\text{and}~\vec{RS}=\vec{c}}$
• Consider triangle QRS. Applying triangle law, we get:
$\small{\vec{b}+\vec{c} = \vec{QR} + \vec{RS} = \vec{QS}}$
• Consider triangle PQS. Applying triangle law, we get:
$\small{\vec{a} + \left(\vec{b}+\vec{c} \right)= \vec{PQ} + \vec{QS} = \vec{PS}}$
3. In fig.26.16(iii) above, $\small{\vec{PQ}=\vec{a}, \vec{RS}=\vec{b}~\text{and}~\vec{QR}=\vec{c}}$
• Consider triangle PQR. Applying triangle law, we get:
$\small{\vec{a}+\vec{c} = \vec{PQ} + \vec{QR} = \vec{PR}}$
• Consider triangle PRS. Applying triangle law, we get:
$\small{\left(\vec{a}+\vec{c} \right) + \vec{b} = \vec{PR} + \vec{RS} = \vec{PS}}$
4. In all of the above three steps, we get the resultant as $\small{\vec{PS}}$. So we can write:
$\small{\left(\vec{a}+\vec{b} \right) + \vec{c}~=~\vec{a} + \left(\vec{b}+\vec{c} \right)~=~\left(\vec{a}+\vec{c} \right) + \vec{b}}$
◼ Based on property I, we can write:
$\small{\vec{a}+\vec{0}~=~\vec{0} + \vec{a}~=~\vec{a}}$
So $\small{\vec{0}}$ is called the
additive identity for vector addition
◼ Based on property II, we can add three vectors $\small{\vec{a}, \vec{b}, \vec{c}}$ in any order we like. There is no need to group any two of them.
In the next section, we will see scalar multiplication of vectors by scalars.
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