Multiplication of a vector, A, with another scalar quantity, a, results in another vector, B. The magnitude of the resulting vector is equal to the product of the magnitude of vector with the scalar quantity. The direction of the resulting vector, however, is same as that of the original vector (See Figures below).

Figure 2

Multiplication with scalar

We have already made use of this type of multiplication intuitively in expressing a vector in component form.

A = A x i + A y j + A z k A = A x i + A y j + A z k

In this vector representation, each component vector is obtained by multiplying the scalar component with the unit vector. As the unit vector has the magnitude of 1 with a specific direction, the resulting component vector retains the magnitude of the scalar component, but acquires the direction of unit vector.

Some physical quantities are themselves a scalar quantity, but are composed from the product of vector quantities. One such example is “work”. On the other hand, there are physical quantities like torque and magnetic force on a moving charge, which are themselves vectors and are also composed from vector quantities.

Thus, products of vectors are defined in two distinct manner – one resulting in a scalar quantity and the other resulting in a vector quantity. The product that results in scalar value is scalar product, also known as dot product as a "dot" (.) is the symbol of operator for this product. On the other hand, the product that results in vector value is vector product, also known as cross product as a "cross" (x) is the symbol of operator for this product. We shall discuss scalar product only in this module. We shall cover vector product in a separate module.

Scalar product of two vectors a and b is a scalar quantity defined as :

where “a” and “b” are the magnitudes of two vectors and “θ” is the angle between the direction of two vectors. It is important to note that vectors have two angles θ and 2π - θ. We can use either of them as cosine of both “θ” and “2π - θ” are same. However, it is suggested to use the smaller of the enclosed angles to be consistent with cross product in which it is required to use the smaller of the enclosed angles. This approach will maintain consistency with regard to enclosed angle in two types of vector multiplications.

The notation “a.b” is important and should be mentally noted to represent a scalar quantity – even though it involves bold faced vectors. It should be noted that the quantity on the right hand side of the equation is a scalar.

Angle between vectors

The angle between vectors is measured with precaution. The direction of vectors may sometimes be misleading. The basic consideration is that it is the angle between vectors at the common point of intersection. This intersection point, however, should be the common tail of vectors. If required, we may be required to shift the vector parallel to it or along its line of action to obtain common point at which tails of vectors meet.

Figure 3: Angle between vectors

Angle between vectors

See the steps shown in the figure. First, we need to shift one of two vectors say, a so that it touches the tail of vector b. Second, we move vector a along its line of action till tails of two vectors meet at the common point. Finally, we measure the angle θ such that 0≤ θ≤π.

Meaning of scalar product

We can read the definition of scalar product in either of the following manners :

a . b = a ( b cos θ ) a . b = b ( a cos θ ) a . b = a ( b cos θ ) a . b = b ( a cos θ )

(4)

Recall that “bcos θ” is the scalar component of vector b along the direction of vector a and “a cos θ” is the scalar component of vector a along the direction of vector b. Thus, we may consider the scalar product of vectors a and b as the product of the magnitude of one vector and the scalar component of other vector along the first vector.

The figure below shows drawing of scalar components. The scalar component of vector in figure (i) is obtained by drawing perpendicular from the tip of the vector, b, on the direction of vector, a. Similarly, the scalar component of vector in figure (ii) is obtained by drawing perpendicular from the tip of the vector, a, on the direction of vector, b.

Figure 4

Scalar product

The two alternate ways of evaluating dot product of two vectors indicate that the product is commutative i.e. independent of the order of two vectors :

a . b = b . a a . b = b . a

(5)

Exercise 1

A block of mass “m” moves from point A to B along a smooth plane surface under the action of force as shown in the figure. Find the work done if it is defined as :

W = F. Δx

Figure 5

Work done

where F and Δx are force and displacement vectors.

Solution

Expanding the expression of work, we have :

Figure 6

Work done

W = F . Δ x = F Δ x cos θ W = F . Δ x = F Δ x cos θ

Here, F = 10 N, Δx = 10 m and cosθ = cos60° = 0.5.

⇒ W = 10 x 10 x 0.5 = 50 J ⇒ W = 10 x 10 x 0.5 = 50 J

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A closer look at the expansion of dot product of two vectors reveals that the expression is very similar to the expression for a component of a vector. The expression of the dot product is :

a . b = a b cos θ a . b = a b cos θ

On the other hand, the component of a vector in a given direction is :

a x = a cos θ a x = a cos θ

Comparing two equations, we can define component of a vector in a direction given by unit vector "n" as :

This is a very general and useful relation to determine component of a vector in any direction. Only requirement is that we should know the unit vector in the direction in which component is to be determined.

In this section, we summarize the properties of dot product as discussed above. Besides, some additional derived attributes are included for reference.

1: Dot product is commutative

This means that the dot product of vectors is not dependent on the sequence of vectors :

a . b = b . a a . b = b . a

We must, however, be careful while writing sequence of dot product. For example, writing a sequence involving three vectors like a.b.c is incorrect. For, dot product of any two vectors is a scalar. As dot product is defined for two vectors (not one vector and one scalar), the resulting dot product of a scalar (a.b) and that of third vector c has no meaning.

2: Distributive property of dot product :

a . ( b + c ) = a . b + a . c a . ( b + c ) = a . b + a . c

3: The dot product of a vector with itself is equal to the square of the magnitude of the vector.

a . a = a x a cosθ = a 2 cos 0 ° = a 2 a . a = a x a cosθ = a 2 cos 0 ° = a 2

4: The magnitude of dot product of two vectors can be obtained in either of the following manner :

a . b = a b cos θ a . b = a b cos θ = a x ( b cos θ ) = a x component of b along a a . b = a b cos θ = ( a cos θ ) x b = b x component of a along b a . b = a b cos θ a . b = a b cos θ = a x ( b cos θ ) = a x component of b along a a . b = a b cos θ = ( a cos θ ) x b = b x component of a along b

The dot product of two vectors is equal to the algebraic product of magnitude of one vector and component of second vector in the direction of first vector.

5: The cosine of the angle between two vectors can be obtained in terms of dot product as :

a . b = a b cosθ a . b = a b cosθ

⇒ cosθ = a . b a b ⇒ cosθ = a . b a b

6: The condition of two perpendicular vectors in terms of dot product is given by :

a . b = a b cos 90 ° = 0 a . b = a b cos 90 ° = 0

7: Properties of dot product with respect to unit vectors along the axes of rectangular coordinate system are :

i . i = j . j = k . k = 1 i . j = j . k = k . i = 0 i . i = j . j = k . k = 1 i . j = j . k = k . i = 0

8: Dot product in component form is :

a . b = a x b x + a y b y + a z b z a . b = a x b x + a y b y + a z b z

9: The dot product does not yield to cancellation. For example, if a.b = a.c, then we can not conclude that b = c. Rearranging, we have :

a . b - a . c = 0 a . ( b - c ) = 0 a . b - a . c = 0 a . ( b - c ) = 0

This means that a and (b - c) are perpendicular to each other. In turn, this implies that (b - c) is not equal to zero (null vector). Hence, b is not equal to c as we would get after cancellation.

We can understand this difference with respect to cancellation more explicitly by working through the problem given here :

Law of cosine relates sides of a triangle with one included angle. We can determine this relationship using property of a dot product. Let three vectors are represented by sides of the triangle such that closing side is the sum of other two vectors. Then applying triangle law of addition :

Figure 7: Cosine law

Cosine law

c = ( a + b ) c = ( a + b )

We know that the dot product of a vector with itself is equal to the square of the magnitude of the vector. Hence,

c 2 = ( a + b ) . ( a + b ) = a . a + 2 a . b + b . b c 2 = a 2 + 2 a b cos θ + b 2 c 2 = a 2 + 2 a b cos ( π-φ ) + b 2 c 2 = a 2 - 2 a b cos φ + b 2 c 2 = ( a + b ) . ( a + b ) = a . a + 2 a . b + b . b c 2 = a 2 + 2 a b cos θ + b 2 c 2 = a 2 + 2 a b cos ( π-φ ) + b 2 c 2 = a 2 - 2 a b cos φ + b 2

This is known as cosine law of triangle. Curiously, we may pay attention to first two equations above. As a matter of fact, second equation gives the square of the magnitude of resultant of two vectors a and b.

Differentiation of a vector expression yields a vector. Consider a vector expression given as :

a = ( x 2 + 2 x + 3 ) i a = ( x 2 + 2 x + 3 ) i

The derivative of the vector with respect to x is :

a ' = ( 2 x + 2 ) i a ' = ( 2 x + 2 ) i

As the derivative is a vector, two vector expressions with dot product is differentiated in a manner so that dot product is retained in the final expression of derivative. For example,

d d x ( a . b ) = a ' b + a b ' d d x ( a . b ) = a ' b + a b '