In the previous section, we saw direction cosines. In this section, we will see types of vectors.
We need to learn about seven types of vectors:
1. Zero Vector
For a vector, if the terminal point is same as the initial point, then that vector is called a zero vector.
• A zero vector is denoted as $\small{\vec{0}}$
• A zero vector is also known as a null vector.
• A zero vector cannot be assigned any definite direction because, it has no magnitude.
• Alternatively, a zero vector can be regarded as a vector having any of the infinite possible directions.
• Vectors $\small{\vec{AA}, \vec{BB},~.~.~}$ are examples of zero vector.
2. Unit Vector
If the magnitude of a vector is 1 unit, then it is called a unit vector.
•
Suppose that, we are given a vector $\small{\vec{a}}$. We can draw a
unit vector which has the same direction as $\small{\vec{a}}$. Such a
vector is denoted as $\small{\hat{a}}$
3. Coinitial Vectors
If two or more vectors have the same initial point, then they are called coinitial vectors.
• $\small{\vec{AB}}$ and $\small{\vec{AC}}$ are coinitial vectors
4. Collinear Vectors
If two or more vectors are parallel to the same line, then they are called collinear vectors.
• Collinear vectors need not lie along the same line.
• Collinear vectors can have only two possible directions. One direction exactly opposite to the other.
5. Equal Vectors
Two vectors are said to be equal if they have the same magnitude and direction.
• Equal vectors need not have the same initial point.
• If two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ are equal, then we write $\small{\vec{a} = \vec{b}}$
6. Negative of a Vector
A vector is said to be the negative of another vector, if they have the same magnitude but opposite directions.
• $\small{\vec{AB}}$ is the negative of $\small{\vec{BA}}$. We write: $\small{\vec{BA} = -\vec{AB}}$
7. Free Vectors
Free vectors are those which can be moved around. The only condition is that, the magnitude and direction should not be altered.
• In this chapter, we will be dealing with free vectors only.
Now we will see some solved examples
Solved example 26.1
Represent graphically a displacement of 40 km, 30o west of south.
Solution:
1. An object undergoes a displacement of 40 km in a specified direction. We are asked to draw the displacement vector.
2. We setup the rectangular coordinate system in such a way that, the origin O of the coordinate system coincides with the initial position of the object. Also, the following directions should be clearly marked:
♦ +ve x direction towards east
♦ +ve y direction towards north
♦ −ve x direction towards west
♦ −ve y direction towards south
This is shown in fig.26.8 below:
 |
| Fig.26.8 |
3. Next step is to draw a dashed line which satisfies two conditions:
♦ The line must make 30o with the south direction
♦ The line must be inclined towards west
This dashed line is shown in fig.a
4. Next step is to mark a point P on the dashed line in such a way that, OP represents 40 km. A suitable scale can be used based on the size of the paper available.
5. Now, the dashed portion between O and P should be changed to a continuous line. And an arrowhead can be placed at P. This is shown in fig.b
• Thus we get $\small{\vec{OP}}$ which is the displacement vector.
6. Scale of the drawing should be written on the top left corner.
Solved example 26.2
Classify the following measures as scalars and vectors.
(i) 5 seconds (ii) 1000 cm3 (iii) 10 newton (iv) 30 km/hr (v) 10 g/cm3 (vi) 20 m/s towards north
Solution:
(i) Time has magnitude only. It has no direction. So 5 seconds is a scalar quantity
(ii) Volume has magnitude only. It has no direction. So 1000 cm3 is a scalar quantity
(iii) Force has both magnitude and direction. So 10 newton is a vector quantity.
(iv) Speed has magnitude only. It has no direction. So 30 km/hr is a scalar quantity.
(v) Density has magnitude only. It has no direction. So 10 g/cm3 is a scalar quantity.
(vi) Velocity has both magnitude and direction. So "20 m/s towards north" is a vector quantity.
Solved example 26.3
In fig.26.9, which of the vectors are:
(i) Collinear (ii) Equal (iii) Coinitial
 |
| Fig.26.9 |
Solution:
1. We are given four vectors.
• Each of $\small{\vec{a}}$ and $\small{\vec{c}}$ has a small white perpendicular line. This line divides the vector into two equal parts. So these two vectors have a magnitude of two units.
• Each of $\small{\vec{b}}$ and $\small{\vec{d}}$ has two small white perpendicular lines. These lines divide the vectors into three equal parts. So these two vectors have a magnitude of three units.
2. We see that $\small{\vec{a}~\text{and}~\vec{c}}$ have the same direction. Also, both have a magnitude of 2 units. So they are equal vectors.
3. $\small{\vec{b}, \vec{c}~\text{and}~\vec{d}}$ have the same initial point. So they are coinitial vectors.
4. $\small{\vec{a}, \vec{c}~\text{and}~\vec{d}}$ are parallel. So they are collinear vectors
Solved example 26.4
Represent graphically a displacement of 40 km, 30o east of north
Solution:
2.
We setup the rectangular coordinate system in such a way that, the
origin O of the coordinate system coincides with the initial position of
the object. Also, the following directions should be clearly marked:
♦ +ve x direction towards east
♦ +ve y direction towards north
♦ −ve x direction towards west
♦ −ve y direction towards south
This is shown in fig.26.10 below:
 |
| Fig.26.10 |
3. Next step is to draw a dashed line which satisfies two conditions:
♦ The line must make 30o with the north direction
♦ The line must be inclined towards east
This dashed line is shown in fig.a
4. Next step is to mark a point P on the dashed line in such a way that, OP represents 40 km. A suitable scale can be used based on the size of the paper available.
5. Now, the dashed portion between O and P
should be changed to a continuous line. And an arrowhead can be placed
at P. This is shown in fig.b
• Thus we get $\small{\vec{OP}}$ which is the displacement vector.
6. Scale of the drawing should be written on the top left corner.
Solved example 26.5
In fig.26.11 (a square), identify the following vectors.
(i) Coinitial (ii) Equal (iii) Collinear but not equal
 |
| Fig.26.11 |
1. We are given four vectors. They form the sides of a square. So all four of them have the same magnitude.
2. $\small{\vec{a}~\text{and}~\vec{d}}$ have the same initial point. So they are coinitial vectors.
3. $\small{\vec{b}~\text{and}~\vec{d}}$ have the same magnitude and direction. So they are equal vectors.
4. $\small{\vec{a}~\text{and}~\vec{c}}$ have the same magnitude but opposite directions. So they are not equal vectors. However, since they are parallel, they are collinear.
The link below gives a few more solved examples:
Exercise 26.1
In the next section, we will see addition of vectors.
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