Properties of scalar or dot products

(i) Dot product is commutative:- From definition, vector(A.B) = AB cosθ and vector(B.A) = BA cosθ = AB cosθ

Hence vector(A.B = B.A), which is commutative law.

(ii) Dot product is distributive over the addition of vectors:

i.e., vector{A.(B + C) = A.B + A.C}

(iii) Dot product of two parallel vectors:- The angle between two parallel vectors is zero,

i.e., θ = 0

Let vector vector(A + B) be two parallel vectors

Then vector(A.B) - AB cosθ = AB cos 0º = AB, (cos 0º = 1)

(iv) Dot product of two equal vectors:- The angle between two equal vectors is also zero

i.e., θ = 0, or cos 0º = 1

The dot product of two equal vectors is given by

vector(A.A) = AA cos0º = A²

Similarly, the dot product of unit vectors with itself is unity.

i.i = 1 x 1. cos0º = 1

j.j = 1.1. cos0º = 1

k.k = 1.1 cos0º = 1

Thus, i.i = j.j = k.k

(v) Dot product of perpendicular vectors:- The angle between two perpendicular vectors is 90º

i.e., θ = 90º or cos 90º = 0

If two vectors vector(A and B) are perpendicular, then

vector(A.B) = AB cos 90º = 0

Thus, two vectors are perpendicular if their dot product is zero.

Now i.j = 1 x 1 x cos 90º = 0

j.k = 1 x 1 x cos 90º = 0

k.i = 1 x 1 x cos 90º = 0

Thus, i.j = j.k = k.i = 0

(vi) Dot product of two anti-parallel vectors:- The angle between two anti-parallel vectors is 180º. i.e., θ = 180º or cos 180º = -1

If vectors vector(A and B) are anti-parallel vectors, then

vector(A.B) = AB cos 180º = -AB

(vii) Dot-product of two vectors in terms of their components:- The vectors vector(A and B) in terms of their components are written as:

vector A = A_xi + A_yj + A_zk and

vector B = B_xi + B_yj + B_zk

Hence, vector(A.B) = (A_xi + A_yj + A_zk).(B_xi + B_yj + B_zk)

= A_xB_x(i.i) + A_xB_y(i.j) + A_xB_z(j.k) + A_yB_x(j.i) + A_yB_y(j.j) + A_yB_z(j.k) + A_zB_x(k.i) + A_zB_y(k.j) + A_zB_z(k.k)

We know, i.i = j.j = k.k = 1

and, i.j = j.k = k.i = 0

Thus vector(A.B) = A_xB_x + A_yB_y + A_zB_z