We have learnt about vectors and their use in earlier lesson ‘Motion in a Plane’ and we know that physical quantities like displacement, velocity, acceleration, force etc., are vectors and how they are added or subtracted. Let us find out how vectors are multiplied. There are two ways of multiplying vectors:

- The scalar product gives a scalar from two vectors.
- Vector product produces a new vector from two vectors.

We shall learn about vector product later on. The scalar product or dot product of any two vectors **A **and **B** denoted as **A.****B **(read as **A **dot **B**) is defined as

**A.****B **= *AB cos θ ………………………………………………………………………………………………*(1a)

where *θ* is the angle between the two vectors as shown in Fig. 1a. Since *A, B *and *cos θ *are scalars, the dot product of **A **and **B** is a scalar quantity. Each vector, **A **and **B,** has a direction but their scalar product does not have a direction.

From Eq. (1a), we have

**A.****B **= *A*(*B cos θ*)

= *B*(*A cos θ*)

Geometrically, *B cos θ *is the projection of **B **onto **A **Fig. 1(b) and *A cos θ* is the projection of **A **onto **B **in Fig. 1(c). So, **A.B **is the product of the magnitude of **A **and the component of **B** along **A**. Alternatively, it is the product of the magnitude of **B** and the component of **A** along **B**.

**Fig. 1** (a) The scalar product of two vectors A and B is a scalar: A.B = A B cos θ (b) B cos θ is the projection of B on to A (c) A cos θ is the projection of A on to B.

Equation (1a) shows that the scalar product follows the commutative law:

**A.****B **= **B.****A**

Scalar product obeys the **distributive law:**

**A.(****B+C) **= **A.B + ****A.C**

Further, **A.**(λ**B)**** = **λ (**A.****B**) where λ is a real number.

The proofs of the above equations are left to you as an exercise.

For unit vectors **i, j, k **we have,

**i.i **= **j.j **= **k.k = **1

**i.j **=** j.k** =** k.i** = 0

Given two vectors,

**A = ***A*_{x} **i + ***A*_{y}**j*** + A _{z}*

**k**

**B = ***B*_{x} **i + ***B*_{y}**j*** + B _{z}*

**k**

their scalar product is

**A.****B **= (*A*_{x} **i + ***A*_{y}**j*** + A _{z}*

**k**)

**.**(

*B*

_{x}**i +**

*B*

_{y}**j**

*+ B*

_{z}**k**)

= *A** _{x} B_{x}* +

*A*

_{y}*B*

_{y}*+ A*

_{z}*B*

_{z ………………………………………………………………………………………………}_{(1b)}

From the definition of scalar product and Eq. 1(b) we have:

(i) **A.****A = ***A** _{x}A_{x}* +

*A*

_{y}*A*

_{y}*+ A*

_{z}*A*

_{z}Or, *A*^{2 }= (*A** _{x})^{2}* + (

*A*

_{y}*)*

^{2}*+ (A*

_{z}*)*

^{2}

_{ ……………………………………………………………………………………}_{(1c)}

since **A.****A = **I **A **I **A **I cos 0 = *A*^{2}

(ii) **A.****B **= 0, if **A **and **B **are perpendicular.*Example 1* Find the angle between force **F = **(3i +4j – 5k) unit and displacement **d = **(5i + 4j +3k) unit. Also find the projection of **F **and **d.**

##### Answer

**F.d = ***F** _{x}d_{x}* +

*F*

_{y}*d*

_{y}*+ F*

_{z}*d*

_{z}= 3(5) + 4(4) + (-5)(3)

= 16

Hence **F.d =***F **d *cos *θ *= 16 unit

Now **F.F = ***F*^{2 }= (*F** _{x})^{2}* + (

*F*

_{y}*)*

^{2}*+ (F*

_{z}*)*

^{2}= 9 +16 +25

= 50 unit

and **d.d **= *d*^{2 }= (*d** _{x})^{2}* + (

*d*

_{y}*)*

^{2}*+ (d*

_{z}*)*

^{2}= 25 +16 +9

= 50 unit

Therefore, cos *θ *= 16/(√50 √50) = 16/50 = 0.32

*θ *= cos^{-1}0.32