Physical quantities are divided into two categories – scalar and vector quantities. The quantities which have only magnitude and not any fixed direction are called **Scalar quantities**. Eg. Mass, volume, density, etc. Quantities that have both magnitude and direction. Such quantities are called **vector quantities or vectors. **Eg. Displacement, velocity, acceleration, momentum, etc.

Vectors are represented by line segments that are directed such as, the direction of an arrow marked at one end, which denotes direction and the length of the line segment is the magnitude of the vector. Eg.

It denotes two points A and B, such that the magnitude of the vector is the length of the straight line AB and its direction is from A to B. Here, point A is called the initial point of vector and point B is called the terminal point(or endpoint).

The magnitude or modulus of a vector \overrightarrow{AB} is a positive number(>0) which is the measure of its length and is denoted by |\overrightarrow{AB}|.

### Types of Vectors

**Zero Vector **

A vector whose initial and terminal(end) points coincide with each other, is called a zero vector (or null vector), and denoted as . Zero vector can not be assigned a defined direction as it has zero magnitude. Or, in other words, it may be defined as having any direction.

The vectors , represent the zero vector.

**Unit Vector**

A vector whose magnitude is unity (=1) is called as **unit vector**. The unit vector is denoted by

and = 1

**Coinitial Vectors **

Two or more vectors having the same initial point or starting point are called coinitial vectors.

**Collinear Vectors **

If two or more vectors are parallel to each other, irrespective of their magnitudes and directions. Then they will be called collinear vectors.

**Equal Vectors **

If two vectors and have the same magnitude and direction regardless of the positions of their initial points will be called as equal vectors, and it can be represented as

**Negative of a Vector **

A vector whose magnitude(length) is equal to of given vector (say ) but the direction is opposite to that of it(initial and terminal points are reversed), is called negative of the given vector.

For example, vector is negative of the vector and written as,

**Position Vector**

Consider a point P in 3D space, having coordinates as (x, y, z) with respect to the origin O(0, 0, 0). And, the vector is having O as the initial point and P as its terminal point is called the **position vector** of the point P with respect to O.

Using the distance formula, the magnitude(or length) of is

=

Note:The vector’s definition defined above are such type of vectors that can be subjected to its parallel displacement without changing its magnitude and direction. Such vectors are called free vectors.

### Vector Addition

A vector simply denotes the displacement of anything from point A to point B.

**Triangle law of vector addition**

Consider a situation that a person moves from A to B and then from B to C. The net displacement made by the person from point A to point C, is given by the vector and expressed as

This is known as the

.triangle law of vector additionAs,

**When the sides of a triangle are taken in order, then it leads to zero resultant(no displacement result) as the initial and terminal points get coincided with each other.**

### Parallelogram Law of Vector Addition

If two vectors and are represented in magnitude and direction by the two adjacent sides of the parallelogram, then their sum() is represented by the diagonals of the given parallelogram which is coinitial with the given vectors and .

Now, let’s consider the parallelogram ABCD

where, and

then by using the triangle law of vector addition, from triangle ABC, we have

………………….(1)

Now, as the opposite sides of a parallelogram are equal and parallel

and

Again using triangle law of vector addition, from triangle ADC, we have

…………………………….(2)

From (1) and (2), we get

commutative property

Note:The magnitude is not equal to the sum of the magnitude(length) of and .

**(ii) Associative property**

Using triangle law, from triangle PQR, we have

Using triangle law, from triangle QRS, we have

Using triangle law, from triangle PRS, we have

Using triangle law, from triangle PQS, we have

Hence,

**(iii) Additive identity**

For any vector

**(iv) Additive inverse**

For any vector

**Problem 1: If** , **show that the points P, Q and R are collinear.**

**Solution:**

We have,

Using, converse of triangle law of addition of vectors, we get

and ae either parallel or collinear. But, Q is a point common to them.

So, and are collinear. Hence, points P, Q, R are collinear.

**Problem 2: B, P, Q, R and A are five points in a plane. Show that the sum of vectors** and in 3 .

**Solution:**

Using triangle law addition of vectors in △APB

……………(1)

Using triangle law addition of vectors in △AQB

…………………..(2)

Using triangle law addition of vectors in △ARB

…………….(3)

Adding (1), (2) and (3), we get

Hence, the sum of the vectors is 3

**Problem 3: Let O be the centre of a regular hexagon CDEFAB. Find the sum of the vectors** ** and** .

**Solution:**

As hexagon property states that, the centre of a regular hexagon bisects all the diagonals passing through it.

So,

and

………………..(1)

…………………..(2)

…………………..(3)

By adding (1), (2) and (3), we get

### Section formula

Here, the points P and Q are the two points represented by the position vectors and , respectively, with respect to the origin O. Then the line segment joining the points P and Q can be divided by a third point, here we say R, in two ways as follows:

Here, we intend to find the position vector for the point R with respect to the origin O. We take the two cases one by one.

#### Internally

When R divides PQ internally. If R divides such that

where m and n are positive values, we specify that the point R divides internally in the ratio of m : n.

Now from triangles ORQ and OPR, we have

Hence, we can conclude that,

m = n

On simplification, we get

When R is the midpoint PQthen m = n

So, we get

#### Externally

When R divides PQ externally. If R divides such that

where m and n are positive values, we say that the point R divides externally in the ratio of the m : n.

Now from triangles ORQ and OPR, we have

Hence, we can conclude that,

On simplification, we get

**Problem 1: Find the position vectors of the points which divide the join of the points** **and** ** internally and externally in the ratio 2 : 3.**

**Solution:**

Let A and B be the given points with the position vectors and respectively.

Let P divide the in the ratio 2 : 3

internallym = 2 and n = 3

Using internally section formula,

Position vector of P =

Position vector of P =

Position vector of P =

Position vector of P =

Now, Let P divide the in the ratio 2 : 3

externallym = 2 and n = 3

Using externally section formula,

Position vector of P =

Position vector of P =

Position vector of P =

Position vector of P =

**Problem 2: If **** and** **are position vectors of points A and B respectively, then find the position vector of points of trisection of AB.**

**Solution:**

Let P and Q be points of trisection. Then, AP = PQ = QB = k (constant variable)

PB = PQ + QB = k + k = 2k

P divides AB in the ratio 1 : 2

Using internally section formula, where m=1 and n=2

Position vector of P =

Position vector of P =

Position vector of P =

Now, we can clearly see that Q is the mid-point of PB.

Apply mid-point section formula we have,

Position vector of Q =

Position vector of Q =

Position vector of Q =

**Problem 3: If D is the mid-point of the side BC of a triangle ABC, prove that**

**Solution:**

Let A be the origin here and position vectors of B and C be and respectively.

As D is the mid-point of BC.

Applying mid-point section formula, we get

Position vector of D =

As,

Hence, Proved!!

### Components of a vector

Let’s take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and z-axis, respectively. Then,

= 1, = 1 and = 1

The vectors, , and are having magnitude unity or 1, which are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by , and respectively.

Lets, consider the position vector , of a point P(x, y, z). Let P_{1} be the foot of the perpendicular from P on the XY plane.

Hence we see that P_{1}P is parallel to z-axis. As , and are the unit vectors along the axes x, y and z, respectively, and by the definition of the coordinates of P, we have

Using triangle law, from triangle OQP

_{1}, we have

Using triangle law, from triangle OP

_{1}P, we haveHence, the position vector of P with reference to O is as follows:

#### Length

The length of vector ,

OP

_{1}^{2}= OQ^{2}+ QP_{1}^{2}(Using Pythagoras theorem)OP1

^{2}= x^{2}+ y^{2}and in the right-angled triangle OP

_{1}P, we haveOP

^{2}= OP_{1}^{2}+ PP_{1}^{2}OP

^{2}= x^{2}+ y^{2 }+ z^{2 }OP =

Hence, the length of vector

If and are two vectors as and then,

**Sum**

**Difference**

The vectors , iff

a_{1}= a_{2}, b_{1}= b_{2}and c_{1}= c_{2}The

multiplicationof vector by any scalar k is given by

### Vector joining two points

If P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) are any two points, then the vector joining P and Q is the vector

Joining the points P and Q with the origin O, and applying triangle law, from the triangle OPQ, we have

Using the properties of vector addition, the above equation become

Hence, the magnitude of is as follows:

**Problem 1: Find the value of x, y and z so that the vectors** **and** **are equal.**

**Solution:**

Two vectors and are equal iff

a

_{1}= a_{2}, b_{1}= b_{2}and c_{1}= c_{2}Hence, the values of

x = 2, y = 2 and z =1.

**Problem 2: Find the magnitude of the vector **

**Solution:**

As, we have

= 7

**Problem 3: Find the unit vector of given vector as **

**Solution:**

Let,

= 7

So, the unit vector in the direction of is given by,

**Problem 4: Find the unit vector of given vector as ****, where the points P(1,2,3) and Q(4,5,6).**

**Solution:**

Position vector of P(1,2,3) =

Position vector of Q(4,5,6) =

= –

Now, magnitude of

= 3√3

So, the unit vector in the direction of is given by,

**Problem 5: Show that the vector **** and **** are collinear.**

**Solution:**

Let, and

As, we can see

This implies,

Hence, and are collinear.

**Problem 6: If A(2,0,0), B(0,1,0), C (0,0,2) have position vectors, show that △ABC is isosceles triangle.**

**Solution:**

Position vector of A(2,0,0) =

Position vector of B(0,1,0) =

Position vector of C(0,0,2) =

=

Now, magnitude of

=

Now, magnitude of

Clearly, AB = BC.

Hence,

△ABC is isosceles triangle.