## 1. Concepts of Vectors and scalars ,magnitude and direction of a vector

There are two types of physical quantities vector quantity and scalar quantity

Scalar quantity

The quantities which are the only magnitude are called scalar quantities for example time

Vector quantity

The physical quantities which have both direction and magnitude are called vector quantities for example velocity

How can we find the magnitude of a vector

Let A = 2i+3j+4k

Magnitude of vector = ✓2²+3²+4²

= ✓ 4+9+16 = ✓ 29

## 1. Concepts of Vectors and scalars ,magnitude and direction of a vector

Chapter 10

Concepts of Vectors and scalars ,magnitude and direction of a vector :

The Physical quantity is either a vector or a scalar. These two categories can be distinguished from one another by their distinct definitions:

Scalars are Physical quantities that has magnitude but no particular direction is described as scalar.

Vectors are Physical quantities that are fully described by both a magnitude and a direction and also satisfy the triangle law of vector addition.

Examples of Scalar Quantities

Some examples of scalar include:

• Mass
• Speed
• Distance
• Time
• Area
• Volume
• Density
• Temperature

### Examples of Vector Quantities

Examples of vector quantity include:

• Linear momentum
• Acceleration
• Displacement
• Momentum
• Angular velocity
• Force
• Electric field
• Polarization

Initial Points – The point A where from the vector starts is known as initial point.

Terminal Point – The point B, where it ends is said to be the terminal point.

Magnitude The distance between initial point and terminal point of a vector is the

Position Vector – Consider a point p (x, y, z) in space. The vector with initial point, origin O and terminal point P, is called the position vector of P.

## 2. Direction cosines and direction ratios of a vector

Three dimensional Geometry

i ,j and k 3 unit vectors in direction of x axis y axis and z axis

Direction cosine give the angle with which align makes an angle with the axis formula for the direction cosine is

l = a cos ¢

m = b cos ¢

n = c cos ¢

Where a b and c are the direction ratios

## 2. Direction cosines and direction ratios of a vector

Direction cosines and direction ratios of a vector

Consider the position vector of a point P(x, y, z) The angles α, β, γ made by the vector  with the positive directions of x, y and z-axes respectively,are called its direction angles. The cosine values of these angles, i.e., cos α, cos β and cos γ are called direction cosines of the vector , and usually denoted by l, m and n, respectively.

i.e. l = cosα, m = cosβ and n = cosγ. The direction of a line cannot be fixed in space by knowing anyone or any two angles.

one may note that the triangle OAP is right angled, and in it, we have l = cosα=x/r, Similarly, from the right angled triangles OBP and OCP, we may write  m = cosβ=y/r and

• n = cosγ=z/r
• x=lr,y=mr,z=nr.

Thus, the coordinates of the point P may also be expressed as (lr, mr,nr). The numbers lr, mr and nr, proportional to the direction cosines are called as direction ratios of vector , and denoted as a, b and c, respectively. Where r  denotes the magnitude of the vector and it is given by,

Example : Find the direction ratios and direction cosines of a line joining the points (3, -4, 6) and (5, 2, 5).

Solution:

Given points are A(3, -4, 6) and B(5, 2, 5)

The direction ratios of the line joining AB is

a = x– x1 = 5 – 3 = 2

b = y– y1 = 2 + 4 = 6

c = z– z1 = 5 – 6 = -1

AB =

So direction cosines of the line = 2/√41, 6/√41, -1/√41.

Example : Find the direction cosines of the line joining the points (2,1,2) and (4,2,0).

Solution:

Let the points are A(2,1,2) and B(4,2,0).

x2-x1 = 4-2 = 2

y2-y1 = 2-1 = 1

z2-z1 = 0-2 = -2

AB = √(22+12+(-2)2)= 3

Hence the direction cosines are ⅔, ⅓, -⅔.

Example : Find the direction cosines of the line joining the points (2,3,-1) and (3,-2,1).

Solution:

Let the points are A(2,3,-1) and B(3,-2,1).

x2-x1 = 3-2 = 1

y2-y1 = -2-3 = -5

z2-z1 = 1- (-1) = 2

AB = √(12+(-5)2+22)= √30

Hence the direction cosines are 1/√30, -5/√30, 2/√30.

## 4. Algebra of Vectors( Addition and Multiplication of Vectors)

Associative property:

[1] Multiplication of Vector by a Scalar

[2] Components of Vector

[3] Vector joining two points

Let P1(x1, y1, z1) and P2(x2, y2 z2) be the two points. Then vector joining the

points P1 and P2 is P1P2 . Join P1, P2 with O. Now OP2 = OP1 + P1P2   (by triangle law)

## Section Formula

To begin with, take a look at the figure given below:

As shown above, P and Q are two points represented by position vectors OP and OQ, respectively, with respect to origin O. We can divide the line segment joining the points P and Q by a third point R in two ways:

• Internally
• Externally

If we want to find the position vector OR for the point R with respect to the origin O, then we should take both the cases one by one.

### Case 1 – R Divides Segment PQ Internally

Take a look at Fig. 1 again. In this figure, if the point R divides such that,

where ‘m’ and ‘n’ are positive scalars, then we can say that R divides PQ internally in the ratio m:n. Now, from the triangles ORQ and OPR, we have

Therefore, replacing the values of and  in equation (1) above, we get

Hence, the position vector formula of the point R which divides PQ internally in the ratio m:n is,

### Case 2 – R Divides Segment PQ Externally

Look at the figure given below:

In Fig. 2, point R divides the segment PQ externally in the ratio m:n. Hence, we can say that point Q divides PR internally in the ratio: (m – n) : n. Therefore,

PQQR = (mn)/n

Now, by using equation (2), we have

### Note: If R is the Mid-Point of PQ

If R is the mid-point of PQ, then m = n. Therefore, from equation (2) above, we have

Therefore, r  = (b +a)/ 2. Hence, the position vector formula of mid-point R of PQ is,

OR= (b +a )/2 … (4)

Let’s look a solved example now:

Example 1

Consider two points P and Q with position vectors

Find the position vector formula of a point R which divides the line joining P and Q in the ratio 2:1, (i) internally, and (ii) externally.

Solution: Since point R divides PQ in the ratio 2:1. we have, m = 2 and n = 1

(i) R divides PQ internally

From equation (2), we have

Example:

Consider two points A and B with position vectors and

And

Find the position vector of a point C which divides the line joining A and B in the ratio 3 : 2,

(i) internally

(ii) externally.

Solution:

(i) The position vector of the point C dividing the join of A and B internally in the ratio 3 : 2 is:

Expanding the terms in the numerator,

(ii) The position vector of the point C dividing the join of A and B externally in the ratio 3 : 2 is:

Expanding the terms in the numerator,

Projection of one vector on a line:

If a vector  makes an angle θ with a given directed line l, in the anticlockwise direction, then the projection of  on l is a vector p  with magnitude |cosθ.

Also, the direction of p  is the same (or opposite) to that of the line l, depending upon whether cosθ is positive or negative. The vector p  is the projection vector and has magnitude |p |. It is also called the projection of vector   on the directed line l.In each of the figures shown above, the projection vector of  along the line l is the vector: .

The vector projection of one vector over another vector is the length of the shadow of the given vector over another vector. It is obtained by multiplying the magnitude of the given vectors with the cosecant of the angle between the two vectors. The resultant of a vector projection formula is a scalar value.

Problem:-

Find the projection of the vector î – ĵ  on the vector î + ĵ.

Let a  = ( î – ĵ) and b  = ( î + ĵ)

Now, the projection of vector a  on b  is given by,

(1)/ (I b I) (a . b ) = (1)/ (√1+1) ({1.1 + (-1) (1)})

= (1/√2) (1-1) =0

Hence, the projection of vector a   on b  is  0.

Question: Represent graphically a displacement of 40 km, 30° east of north.

Solution:

The vector represents the displacement of 40 km, 30o east of north.

Question:  Find the unit vector in the direction of vector , where P and Q are the points

(1, 2, 3) and (4, 5, 6), respectively

Solution:

We know that,

QUESTION: Find a vector in the direction of vector which has magnitude 8 units.

Solution:

Firstly,

## 6. Product of Two Vectors and Scalar (or dot) product of two vectors

Product of Two Vectors and Scalar (or dot) product of two vectors:

1. scalar product of vectors or dot product
2. vector product of vectors or cross product

## Dot Product of Vectors:

The scalar or dot product of two non-zero vectors a  and b , denoted by a .b  is

a .b  = |a | |b cosθ

where θ is the angle between a  and b  and 0 ≤ θ ≤ π as shown in the figure below.

It is important to note that if either a  = 0  or b  = 0 , then θ is not defined, and in this case

a .b  = 0

We can express the scalar product as:

where || and || represent the magnitude of the vectors  and  while cos θ denotes the cosine of the angle between both the vectors and .  indicate the dot product of the two vectors.

### Projection of Vectors:

BP is known to be the projection of a vector a on vector b in the direction of vector b given by |a| cos θ.

Similarly, the projection of vector b on a vector a in the direction of the vector a is given by |b| cos θ.

The projection of vector    on a vector

Question: Find the angle between two vectors a  and b  with magnitudes 3 and 2 ,
respectively having
a .b  = 6.

Solution: By definition of the scalar or dot product of vectors, we know that

a .b  = |a | |b cosθ

where θ is the angle between a  and b . In this question, we have,

|a | = 3, |b | = 2, and a .b  = 6

Replacing these values in the formula, we get

a .b  = |a | |b cosθ
6 = 3 x 2 x cosθ
3 x 232 x 2cosθ

Cancelling the common terms on both sides, we get

1 = 2 x cosθ
Or, cosθ = 12
Therefore, θ = π/4
Hence, the angle between the vectors a  and b  is π/4.

Dot Product Properties of Vector:

• Property 1: Dot product of two vectors is commutative i.e. a.b = b.a = ab cos θ.
• Property 2: If a.b = 0 then it can be clearly seen that either b or a is zero or cos θ = 0
•   q = π/2
• Property 3: Also we know that using scalar product of vectors (pa).(qb)=(pb).(qa)=pq (a.b)
• Property 4: The dot product of a vector to itself is the magnitude squared of the vector i.e. a.a = a.a  cos 0 = a2
• Property 5: The dot product follows the distributive law also i.e. a.(b + c) = a.b + a.c
• Property 6: In terms of orthogonal coordinates for mutually perpendicular vectors it is seen that

Example : Let there be two vectors [6, 2, -1] and [5, -8, 2]. Find the dot product of the vectors.

Solution:

Given vectors: [6, 2, -1] and [5, -8, 2] be a and b respectively.

a.b = (6)(5) + (2)(-8) + (-1)(2)

a.b = 30 – 16 – 2

a.b = 12

Example 2: Let there be two vectors |a|=4 and |b|=2 and θ = 60°. Find their dot product.

Solution:

a.b = |a||b|cos θ

a.b = 4.2 cos 60°

a.b = 4.2 × (1/2)

a.b = 4

Example :

Find the angle between two vectors

And

Solution:

Given:

And

The formula to find the angle between two vectors is given by:

Hence,

= 1-1-1

= -1

Therefore,

Now, substituting the value in the formula, we get

Cos θ = -⅓

Hence, the angle between two vectors is θ = cos-1(-⅓).

Question:  Ifare such thatis perpendicular to, then find the value of λ.

Solution:

We know that the

# Vector (or Cross) Product of Two Vectors

Vectors can be multiplied in two ways, a scalar product where the result is a scalar and cross or vector product where is the result is a vector. In this article, we will look at the cross or vector product of two vectors.

## Explanation

We have already studied the three-dimensional right-handed rectangular coordinate system. As shown in the figure below, when the positive x-axis is rotated counter-clockwise into the positive y-axis, then a right-handed standard screw moves in the direction of the positive z-axis.

As can be seen above, in a three-dimensional right-handed rectangular coordinate system, the thumb of the right-hand points in the direction of the positive z-axis when the fingers are curled from the positive x-axis towards the positive y-axis.

## Definition

The cross or vector product of two non-zero vectors a  and b , is

a  x b  = |a | |b sinθn^

Where θ is the angle between a  and b , 0 ≤ θ ≤ π. Also, n^ is a unit vector perpendicular to both a  and b  such that a b , and n^ form a right-handed system as shown below.

As can be seen above, when the system is rotated from a  to b , it moves in the direction of n^. Also, if either a  = 0 or b  = 0, then θ is not defined and we can say,

a  x b  = 0

## Important Observations

• a  x b  is a vector.
• If a  and b  are two non-zero vectors, then a  x b  = 0, if and only if a  and b  are parallel (or collinear) to each other, i.e.

a  x b  = 0  a   b

Hence, a  x a  = 0 and a  x (−a)→ = 0. This is because in the first case θ = 0. Also, in the second case θ = π, giving the value of sinθ = 0.

• If θ = π2, then a  x b  = |a | |b |
• Considering observations 2 and 3 above, for mutually perpendicular vectors i j , and k , we have

• The angle between the two vectors a  and b  is,

sinθ = |a ×b ||a ||b |

• A cross or vector product is not commutative. We know this because a  x b  = b x a . Now, we know that,

a  x b  = |a | |b sinθn^.

Where a b , and n^ form a right-handed system. Or, θ is traversed from a  to b . On the other hand,

b  x a  = |b | |a sinθn1^.

Where b a , and n1^ form a right-handed system. Or, θ is traversed from b  to a . So, if a  and b  lie on a plane of paper, then n^ and n1^ are both perpendicular to the plane of the paper. However, n^ is directed above the paper and n1^ is directed below it. Or, n^ = – n1^. Hence,

a  x b  = |a | |b sinθn^ = – |a ||b |sinθ n1^

= – b  x a

• From the observations 4 and 6 above, we have

j x i = – k
k x j^ = – i
i x k = – j^

• If a  and b  represent the two sides of a triangle, then its area is |a  x b |. To understand this, look at the figure given below.

By the definition of the area of a triangle, we have area of ΔABC =  (AB).(CD). We know that, AB = |b | and CD = |a |sinθ. Therefore,

• If a  and b  represent the two adjacent sides of a parallelogram, then its area is |a  x b |. To understand this, look at the figure given below.

By the definition of the area of a parallelogram, we have area of parallelogram ABCD = (AB).(DE). We know that, AB = |b | and DE = |a |sinθ. Therefore,

Area of parallelogram ABCD = |a ||b |sinθ = |a  x b |

## Property: Distributivity of a cross or vector product over addition

If a b , and c  are any three vectors and λ is a scalar, then

• a  x (b  + c ) = a  x b  + a  x c
• λ(a  x b ) = (λa ) x b  = a  x (λb )

Question 1: Find the area of the parallelogram whose adjacent sides are determined by the following vectors,

• a  = i^ – j^ + 3k  and
• b  = 2i^ – 7j^ + k .

Answer : We know that if a  and b  represent the two adjacent sides of a parallelogram, then its area is |a  x b |. Also,

Substituting the values of a1,a2,a3,b1,b2,and b3, we get

Solving the determinant, we get

• a  x b  = {[(-1) x 1)] – [(-7) x 3]} – {[1 x 1)] – [2 x 3]} + {[1 x (-7))] – [2 x (-1)]}
= 20
i^ + 5j^ – 5k^.

Also, the magnitude of a  x b  is,

• |a  x b | = [20+5+(−5)]450 25×9×2 = 152.

Therefore, the area of the parallelogram is 152.

Question : Explain the characteristics of vector product?

Answer: The characteristics of vector product are as follows:

• Vector product two vectors always happen to be a vector.
• Vector product of two vectors happens to be noncommutative.
• Vector product is in accordance with the distributive law of multiplication.
• If a • b = 0 and a ≠ o, b ≠ o, then the two vectors shall be parallel to each other.

## 8. Scalar triple product and properties of scalar triple product

Scalar triple product and properties of scalar triple product

The scalar triple product of three vectors a, b, c is the scalar product of ector a with the

cross product of the vectors b and c, i.e., a · (b × c). Symbolically, it is also written as [a b c] = [a, b, c] = a · (b × c). The scalar triple product [a b c] gives the volume of a parallelepiped with adjacent sides a, b, and c. If we are given three vectors a, b, c, then their scalar triple products [a b c] are:

a · (b × c)

• a · (c × b)
• b · (a × c)
• b · (c × a)
• c · (b × a)
• c · (a × b)

Now, before moving to the formula of the scalar triple product, we need to note that:

• [a, b, c] = a · (b × c) = b · (c × a) = c · (a × b)
• a · (b × c) = - a · (c × b)
• b · (c × a) = - b · (a × c)
• c · (a × b) = - c · (b × a)
• a · (b × c) = (a × b) · c

the scalar triple product of vectors means the product of three vectors. It means taking the dot product of one of the vectors with the cross product of the remaining two. It is denoted as

[a b c ] = ( a × b) . c

The following conclusions can be drawn, by looking into the above formula:

i) The resultant is always a scalar quantity.

ii) Cross product of the vectors is calculated first, followed by the dot product which gives the scalar triple product.

iii) The physical significance of the scalar triple product formula represents the volume of the parallelepiped whose three coterminous edges represent the three

vectors a, b and c. The figure will make this point more clear.

According to this figure, the three vectors are represented by the coterminous edges as shown. The cross product of vectors a and b  gives the area of the base, and also, the direction of the cross product of vectors is perpendicular to both the vectors. As volume is the product of area and height, the height, in this case, is given by the component of vector c along the direction of the cross product of a and b. The component is given by c cos α.

Thus, we can conclude that for a Parallelepiped, if the coterminous edges are denoted by three vectors and a, b and c then,

Volume of parallelepiped = ( a × b) c cos α =  ( a × b) . c

Where α is the angle between  ( a × b)  and c.

We are familiar with the expansion of cross products of vectors. Keeping that in mind,

## Properties of Scalar Triple Product:

i) If the vectors are cyclically calculated, then

( a × b) . c = a.( b × c)

ii)  The product is cyclic in nature, i.e.,

a.(b × c) = b.(c × a) = c.(a × b)

Thus,

[ a b c ] = [ b c a ] = [ c a b ] = – [ b a c ] = – [ c b a ] = – [ a c b ]

Example : Evaluate the volume of a parallelepiped whose coterminous edges are i - j + k, 2i + 3j - k, and -i - j + 5k.

Solution: To determine the volume of the parallelepiped with edges i - j + k, 2i + 3j - k, and -i - j + 5k, we will determine its scalar triple product.

[ i - j + k, 2i + 3j - k, -i - j + 5k] =

= 1(15 - 1) + 1(10 - 1) + 1(-2 + 3)

= 14 + 9 + 1

= 24

Answer: The volume of the parallelepiped is 24 cubic units.