Higher Secondary Mathematics

Saturday, May 2, 2026

26.2 - Addition of Vectors

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 $\small{\vec{a}~\text{and}~\vec{b}}$

Shift the vector to be added without changing magnitude or direction.

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 $\small{\vec{a}~\text{and}~\vec{b}}$

The two vectors to be added, form the adjacent sides of a parallelogram. The vectors have the same initial point.

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, $\small{\vec{PQ}=\vec{a}, \vec{QR}=\vec{b}~\text{and}~\vec{RS}=\vec{c}}$

Associative property is valid when three vectors are added together

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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Sunday, April 26, 2026

26.1 - Types of Vectors

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, 30^o 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 30^o 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

Checking whether the given vectors are equal or collinear

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, 30^o east of north
Solution:

1. An object undergoes a displacement of 40 km in a specified direction. We are asked to draw the displacement vector.

Fig.26.10

3. Next step is to draw a dashed line which satisfies two conditions:
♦ The line must make 30^o 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.

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

Solution:
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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Saturday, April 25, 2026

Chapter 26 - Vector Algebra

In the previous section, we completed a discussion on differential equations. In this chapter, we will see vector algebra.

Some basics about vectors can be written in 2 steps:
1. Consider the length of a room. We can write:
Length of the room is 3.2 m
• Length involves only one value, which is the magnitude.
• Quantities which involve only the magnitude, are called scalars.
• A scalar quantity can be represented by a real number and a proper unit.
• Some examples are:
    ♦ A mass of 32 kg
    ♦ An area of 90 sq.cm
    ♦ A speed of 7 m/s

2. A quantities having both magnitude and direction are called vectors.
• Some examples are:
♦ A car moving with a velocity. We can write: velocity is 35 km/h towards east. Here, velocity is a vector quantity.
♦ A force applied on a body. We can write: Force is 25 N along a line which makes 30 degrees with the x-axis. Here, force is a vector quantity.

Let us see vectors in some more detail. It can be written in 8 steps:

1. Fig.26.1(a) below, shows a line $\small{l}$. It extents to infinity in both ways.

A vector has both magnitude and direction

Fig.26.1

2. In fig.b, there is an arrowhead attached to the same line $\small{l}$. This arrowhead gives a sense of direction to the observer.

3. In fig.c the arrowhead is attached in the opposite direction. This arrowhead gives the sense of opposite direction to the observer.

4. In fig.d, two points A and B are marked on the same line $\small{l}$. AB is a line segment.
• Any line segment will have a definite length. That means, any line segment will have a magnitude.
• An arrowhead is also attached to the segment AB. This arrowhead gives the sense of direction.
• So the line segment AB in fig.d has both magnitude and direction. It is a vector.

5. The vector in fig.d can be denoted as: $\small{\vec{AB}}$ or as $\small{\vec{a}}$.
♦ $\small{\vec{AB}}$ is read as “vector AB”
♦ $\small{\vec{a}}$ is read as “vector a”

6. $\small{\vec{AB}}$ starts at A. So the point A is called the initial point of $\small{\vec{AB}}$

7. $\small{\vec{AB}}$ ends at B. So the point B is called the terminal point of $\small{\vec{AB}}$.

8. The distance between points A and B is called the magnitude of $\small{\vec{AB}}$. It is denoted as $\small{\left|{\vec{AB}}\right|}$
Note that, a length cannot be −ve. So we never write: $\small{\left|{\vec{AB}}\right| < 0}$

Position Vector

This can be explained in 4 steps:
1. In class 11, we have seen the three dimensional right handed rectangular coordinate system. Details here.
In that coordinate system, consider a point P. This is shown in fig.26.2 below:

Fig.26.2

♦ The coordinates of origin O are (0,0,0)
♦ The coordinates of P are (x,y,z)

2. In such a situation, we can draw $\small{\vec{OP}}$
• For $\small{\vec{OP}}$, O is the initial point and P is the terminal point.
• This $\small{\vec{OP}}$ is called the position vector of point P with respect to O.

3. Based on the discussions that we had in class 11, we are able to find the distance OP. Details here.
We get: $\small{OP = \sqrt{x^2 + y^2 + z^2}}$.
• So we can write: $\small{\left|{\vec{OP}}\right| = \sqrt{x^2 + y^2 + z^2}}$

4. Position vector is an important concept in science and engineering. For points A, B, C, . . . , we denote the position vectors as:
$\small{\vec{a},~\vec{b},~\vec{c},~.~.~.~,}$.
This is shown in fig.26.3 below:

Position vector of a point, starts from the origin and ends at that point.

Fig.26.3

Direction Cosines

There are three direction cosines. First we will see the direction cosine related to the x-axis. It can be written in 4 steps:

1. Fig.26.4 below shows a point P(x,y,z).

Fig.26.4

• The position vector $\small{\vec{OP}}$ makes an angle $\small{\alpha}$ with the +ve direction of the x-axis.
• For better clarity, the observer is looking at the coordinate system in the direction marked as I in fig.26.5 below.

Fig.26.5
Top view of rectangular coordinate system

2. A perpendicular is dropped from P onto the xy-plane. Foot of this perpendicular is marked as P₁.
• From P₁, a perpendicular is dropped onto the x-axis. Foot of this perpendicular is marked as A.

3. Consider the plane containing the three point P, P₁ and A.
• This plane will be perpendicular to the x-axis because, AP₁ is perpendicular to the x-axis.
• Since this plane is perpendicular to the x-axis, any line in the plane will be perpendicular to the x-axis.
• So AP is perpendicular to the x-axis.
• Therefore, $\small{\angle{OAP} = 90^\circ}$

4. So triangle OAP is a right triangle with OP as hypotenuse. We get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \alpha} & {~=~} &{\frac{OA}{\left|\vec{OP} \right|} = \frac{x}{\left|\vec{OP} \right|}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{x} & {~=~} &{\cos \alpha\,\left|\vec{OP} \right|}
\\ \end{array}}$

Next we will see the direction cosine related to the y-axis. It can be written in 4 steps:

1. Fig.26.6 below shows the same point P(x,y,z).

Fig.26.6

• The position vector $\small{\vec{OP}}$ makes an angle $\small{\beta}$ with the +ve direction of the y-axis.
• For better clarity, the observer is looking at the coordinate system in the direction marked as II in fig.26.5 above.

2. We have the same point P₁ as before.
• This time, the perpendicular from P₁, is dropped onto the y-axis. Foot of this perpendicular is marked as B.

3. Consider the plane containing the three point P, P₁ and B.
• This plane will be perpendicular to the y-axis because, BP₁ is perpendicular to the y-axis.
• Since this plane is perpendicular to the y-axis, any line in the plane will be perpendicular to the y-axis.
• So BP is perpendicular to the y-axis.
• Therefore, $\small{\angle{OBP} = 90^\circ}$

4. So triangle OBP is a right triangle with OP as hypotenuse. We get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \beta} & {~=~} &{\frac{OB}{\left|\vec{OP} \right|} = \frac{y}{\left|\vec{OP} \right|}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{y} & {~=~} &{\cos \beta\,\left|\vec{OP} \right|}
\\ \end{array}}$

Finally we will see the direction cosine related to the z-axis. It can be written in 3 steps:

1. Fig.26.7 below shows the same point P(x,y,z).

Fig.26.7

• The position vector $\small{\vec{OP}}$ makes an angle $\small{\gamma}$ with the +ve direction of the z-axis.
• For better clarity, the observer is looking at the coordinate system in the direction marked as I in fig.26.5 above.

2. This time, we do not need the point P₁.
• This time, we directly drop the perpendicular from P onto the z-axis. Foot of this perpendicular is marked as C.

3. Since PC is perpendicular to the z-axis, triangle OPC is a right triangle with OP as hypotenuse. We get:

$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \gamma} & {~=~} &{\frac{OC}{\left|\vec{OP} \right|} = \frac{z}{\left|\vec{OP} \right|}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{z} & {~=~} &{\cos \gamma\,\left|\vec{OP} \right|}
\\ \end{array}}$

Now we can write the definitions. It can be written in 7 steps:

1. The angles $\small{\alpha, \beta~\text{and}~\gamma}$ that we saw above, are angles made with the +ve direction of the x, y and z axes respectively.
• They are called direction angles of the position vector $\small{\vec{OP}}$

2. We can take cosines of the above angles. We get:
$\small{\cos\alpha, \cos\beta~\text{and}~\cos\gamma}$
• These are called direction cosines of the position vector $\small{\vec{OP}}$

3. Usually we use short forms for denoting the direction cosines.
    ♦ $\small{\cos\alpha}$ is denoted by $\small{l}$
    ♦ $\small{\cos\beta}$ is denoted by $\small{m}$
    ♦ $\small{\cos\gamma}$ is denoted by $\small{n}$

4. There are short forms for the position vector also:
♦ $\small{\vec{OP}}$ is denoted by $\small{\vec{r}}$
♦ $\small{\left|\vec{OP} \right|}$ is denoted by $\small{r}$

5. Based on the above short forms, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{\cos \alpha\,\left|\vec{OP} \right|}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{x}    & {~=~}    &{lr}
\\ {~\color{magenta}    3    }    &{}    &{y}    & {~=~}    &{\cos \beta\,\left|\vec{OP} \right|}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{y}    & {~=~}    &{mr}
\\ {~\color{magenta}    5    }    &{}    &{z}    & {~=~}    &{\cos \gamma\,\left|\vec{OP} \right|}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{z}    & {~=~}    &{nr}
\\ \end{array}}$

6. So the coordinates (x,y,z) of the point P, can be written as: $\small{(lr,mr,nr)}$

7. $\small{lr,mr~\text{and}~nr}$ are mere numbers.
• They are called direction ratios of $\small{\vec{r}}$. We will see more details about this in later sections.

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

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