Section Formula – Vector Algebra

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 addition.

As, 

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 PQ

then 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 internally

m = 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 externally

m = 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 P₁ be the foot of the perpendicular from P on the XY plane.

Hence we see that P₁P 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 OQP₁, we have

Using triangle law, from triangle OP₁P, we have

Hence, the position vector of P with reference to O is as follows:

Length

The length of vector  , 

OP₁² = OQ² + QP₁²    (Using Pythagoras theorem)

OP1² = x² + y²

and in the right-angled triangle OP₁ P, we have

 OP² = OP₁² + PP₁²

OP² = x² + y² + z² 

OP = 

Hence, the length of vector 

If  and  are two vectors as  and  then,

Sum

Difference

The vectors , iff

a₁ = a₂, b₁ = b₂ and c₁ = c₂

The multiplication of vector  by any scalar k is given by

Vector joining two points

If P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) 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

a₁ = a₂ , b₁ = b₂ and c₁ = c₂

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.