Scalar Product of Vectors

Two vectors or a vector and a scalar can be multiplied. There are mainly two kinds of products of vectors in physics, scalar multiplication of vectors and Vector Product (Cross Product) of two vectors. The result of the scalar product of two vectors is a number (a scalar). The common use of the scalar product is to define work and energy relations like to find the work performed on an object by a force (vector) while causing the displacement of the object, this work can be defined as a scalar product of the force vector with the displacement vector.

What is Scalar Product?

Scalar product or dot product of two vectors is an algebraic operation that takes two equal-length sequences of numbers and returns a single number as result. In geometrical terms, scalar products can be found by taking the component of one vector in the direction of the other vector and multiplying it with the magnitude of the other vector.

The scalar product of two vectors A and B is defined by,

A.B = |A| × |B| cos ∅ 

where 

• |A| is magnitude of vector A,
• |B| is magnitude of vector B and
• ∅  is the angle between vector A and B.

As the scalar product is denoted by dot (.) so it is also called the dot product.

• Scalar Product in terms of Unit Vector Representation

In unit vector representation of vectors, where i, j, k are along the x-axis, y-axis, and z-axis respectively. The scalar product can be calculated as: 

A.B = A_x × B_x + A_y × B_y + A_z × B_z

where,

A = A_xi +A_yj +A_zk

B = B_xi +B_yj + B_zk

• Geometrical interpretation of Scalar Product of Vectors

The product of two non-zero vectors can be visualized as multiplying the magnitude of any one of the vectors, by the magnitude of the projection of the other vector upon it.

Case 1: When the angle between two vectors is greater than 0 degrees and lesser than 90 degrees then the result of the scalar product is positive. 

Case 2: When the angle between two vectors is greater than 90 degrees and lesser than 180 degrees then the result of the scalar product is negative. 

Case 3: When the angle between two vectors is 90 degrees then the result of scalar product is 0. 

Special Cases of Scalar Product

(1) Scalar product of Two parallel Vectors: Scalar product of two parallel vectors is simply the product of magnitudes of two vectors. As the angle between vectors when they are parallel is 0 degree, and cos 0= 1. 

Therefore, 

a.b = |a| × |b| cos 0  

      = |a| × |b|

(2) Scalar Product of Two Antiparallel Vectors: Scalar product of two antiparallel vectors is negative of the product of magnitudes of two vectors.

a.b cos 180 = −|a| × |b|                         (Since, cos 180 = -1)

(3) Scalar product of Two Orthogonal Vectors: Scalar product of two orthogonal vectors is 0.

a.b cos 90 = 0                                          (Since, cos  90 = 0)

Check: Dot and Cross Products on Vectors

Matrix Representation of Scalar Product of Vectors

In matrix form, vectors can be represented as row matrices or column matrices. If we treat vectors as row matrices of their x, y, and z components, then the transposes of these vectors would be column matrices containing x, y, and z components.

Therefore, the vectors A and B  can be seen as follows :

A =

B = 

So, 

A.B = A_xB_x +A_yB_y + A_zB_z

Check: Product of Vectors

Properties of Scalar Multiplication of Vectors

1. Commutative Properties of Scalar Multiplication:

   If a and b are two vectors then:

   a.b = b.a

   As,

   a.b = |a||b| cos θ , and

   b.a = |b||a| cos θ

   So  a.b = b.a

2. Distributive Over Vector Addition:

    If a, b ,c are vector then,    

    a.(b+c) = a.b + a.c

3. Associative Property:

    If a is a vectors and  c, d are scalars then,

    c (da) = (cd) a

4. Identity Property:

    If a is a vector then,

    1⋅a = a

5. Multiplicative Property of 0:

   If a is a vector then,

   0 (a)=0

Sample Problems on Scalar Product of Vectors

Problem 1: Find the scalar product of vector A= 2i + 5j +3k and vector B= 3i + j +2k.

Solution:

A. B = (2 * 3) + (5 * 1) +(3 * 2)

        = 6 + 5 + 6

        = 17 

Problem 2: A particle covers a displacement from position 2i + j + 2k to 3i + 2j +5k due to uniform force of ( 7i + 5j +2k) N. If the displacement is in mitres calculate the work done. 

Solution: 

Given,

Final position P2 = 3i + 2j +5k

Initial position P1 = 2i +j +2k

Force F = ( 7 i + 5 j + 2 k ) N

So  

The displacement D  = P2 – P1

D = ( 3 – 2 ) i + ( 2 -1 ) j + ( 5 – 2 ) k

   = i + j + 3k

Work done = F . D

= ( 7 i + 5 j + 2 k ) . ( i + j  +  3k )

= ( 7*1 ) + ( 5*1 ) + ( 2*3 )

= 7 + 5 + 6

= 18  J

Problem 3: Find the value of m such that  vectors A = 2 i + 3j + k and B = 3 i + 2 j + mk may be perpendicular.

Solution:

Given,

A and B are perpendicular

so A . B = 0 

( 2 i + 3j + k ) . (  3i + 2j + mk ) = 0

( 2 * 3 ) + ( 3 * 2 ) + ( 1 * m ) = 0

6 + 6 + m = 0

12 + m = 0

m = -12 

Problem 4: Prove that vectors U = 2i + 3j + k and V = 4i - 2j + 2k are perpendicular to each other.

Solution:

Given,

U = 2i + 3j + k

V = 4i - 2j + 2k 

to prove U and V are perpendicular to each other

U . V = ( 2i + 3j + k ) . ( 4i - 2j - 2k )

         = ( 2 * 4 ) + ( 3 * -2 ) + ( 1 * -2 )

         = 8 - 6 - 2

         = 0

we know that if scalar product of two vectors is 0 then they are perpendicular to each other.

U . V = 0 so, U and V are perpendicular to each other

Hence proved  

Problem 5: Prove that vectors A = 4ax -2ay - az and B = ax + 4ay  -az are perpendicular to each other.

Solution:

|A|=√(4)^2 +(-2)^2 + (-1)^2 =√21

|B| = √(1)^2+(4)^2+(-1)^2 = √18

A . B = (4ax -2ay - az) . (ax + 4ay  -az)

        = 4 -8 + 4 

        =0

|A| |B| cos ∅ = 0

cos ∅ = 0

so ∅ = 90 °

Hence proved.

Applications of Scalar Product

The scalar product, also known as the dot product, serves multiple purposes in vector theory, including:

1. Projection of a Vector: The scalar product is essential for computing the projection of one vector onto another. Specifically, the projection of vector a onto vector b can be expressed as (a·b)/|b|, while the projection of vector b onto vector a is (a·b)/|a|.
2. Scalar Triple Product: The scalar product also plays a role in determining the scalar triple product of three vectors. The formula for this is a·(b × c) = b·(c × a) = c·(a × b), illustrating its cyclic nature.
3. Angle Between Two Vectors: The scalar product is used to calculate the angle between two vectors. This is done using the cosine formula: cos(θ) = (a·b)/( |a| |b| ), which relates the dot product of the vectors to the product of their magnitudes and the cosine of the angle between them.

Scalar Product Properties

Having grasped the concept of scalar product, let's examine some key properties of the scalar product of vectors ? and ? that are useful in problem-solving:

1. Commutative Property: The scalar product is commutative, meaning ?⋅?=?⋅?.
2. Distributive Property: The scalar product obeys the distributive law:
   • ?⋅(?+?)=?⋅?+?⋅?
   • (?+?)⋅?=?⋅?+?⋅?
   • ?⋅(?−?)=?⋅?−?⋅?
   • (?−?)⋅?=?⋅?−?⋅?
3. Scalar Multiplication: The scalar product is consistent with scalar multiplication:
   • (?1?)⋅(?2?)=?1?2(?⋅?)
4. Orthogonality: Two vectors are orthogonal if their scalar product is zero, i.e., vectors ? and ? are orthogonal if ?⋅?=0.