6. MULTIPLICATION OF TWO VECTORS
Two types of multiplication
6.1 Scalar (or dot) product of two vectors.
6.2 Vector (or cross) product of two vectors.
6.1 Scalar (or dot) product of two vectors :
The scalar or dot product of two vectors a and b denoted by a. b (read as a dot b) its defined as
a . b = |a| |b| cos θ = ab cos θ where a = |a| ; b = |b| and θ is the angle between a and b
6.1.1 Remarks :
(1) If a = 0 or b = 0 we define a . b = 0 (as θ is meaningless)
(2) The dot product of two vectors is a scalar quantity
(3) If θ = 0 (i.e. a and b are like vectors) a . b = ab (as cos 0 = 1)
(4) If θ = π (i.e. a and b are unlike vectors) a . b = – ab (as cos π = –1)
(5) Condition for two vectors to be perpendicular;
If a and b are perpendicular , the angle between them is π/2 and we obtain a. b = ab cos π/2 = 0
Conversely, if a. b = 0 i.e. if ab cos θ = 0 then either a = 0 or b = 0 or cos θ = 0 it follows that either (or both) of the vectors is a zero or else they are perpendicular.
(6) Note that a.b > 0 if 0 ≤ θ < π/2 i.e. if angle between vectors in acute. a.b = 0 if θ = π/2 i.e. if a and b are perpendicular. a.b < 0 if π/2 < θ ≤ π if angle between vectors is obtuse.
6.1.2 Geometrical interpretation of scalar product
Let OA and OB represent vectors a and b respectively. Then

a = |a| = |OA| = OA
b = |b| = |OB| = OB
Let M, N be the feet of the perpendicularsfrom A, B on OB, OA respectively.
Then magnitude of projection of a on b
= OM = OA cos θ (∴ cos θ = OM/OA in Δ OMA) = a cos θ
∴ a.b = ab cos ? = b(a cos ?) = b. (projection of a on b)
Similarly magnitude of projection of b on a.
= ON = OB cos θ(∴ cos θ = ON/OB in ΔONB) = b cos θ
a . b = ab cos θ = a (b cos θ) = a. projection of b on a. Thus a. b can be defined as the product of the modulus of one vector and the projection of the other vector upon it.
6.1.3 Remarks :
(1) Given two vectors a and b, the projection of one vector on another can be found by using the formula.
projection of a on b = a cos θ = a cos θ
and projection of b on a = b cos θ
(ii) Angle between two vectors :
Angle between two vectors a and b can be found by using
a . b = ab cos θ
⇒ cos θ =
(iii) Square of a vector
The scalar product of a vector a with itself is called the square of the vector a, and is written asa. a or (a)²
(a)² = a.a = aa cos 0 = a.a.1 = a² = |a|²
The magnitude of a vector can be found by using
|a|= √ a.a (∴ |a|² = a . a)

6.1.4 Squares and scalar products of
Since,and are unit vectors along the co-ordinate axes i.e. along three mutually perpendicular lines, we have :
. = 1.1 cos 0 = 1 similarly . = 1 ; . = 1
Also, . = 1.1 cos 90º = 1.1.0 = 0, similarly . = 0; . = 0; . = 0; . = 0.
6.1.5 Properties of scalar (or dot product)
(1) The scalar product is commutativei.e. a. b = b.a
(2) If a, b are any vectors and m is any real number (scalar) then(ma) . b = m(a. b) = a . (m b)
Corollary 1 : a (–b) = –(a . b) = (–a) . (b)
Corollary 2 : (–a) . (–b) = a. b
6.1.6 Scalar product of two vectors in terms of their rectangular components.
Let a = a₁ + a₂ + a₃ and b = b₁ + b₂ + b₃ then a.b = (a₁ + a₂ + a₃ ) . (b₁ + b₂ + b₃ ) = a₁b₁+ a₂b₂+ a₃b₃
Corollary 1 : a. b = 0
If a₁b₁+ a₂b₂+ a₃b₃ = 0. Thus if a ⊥ b or if a₁ b₁+ a₂ b₂+ a₃ b₃ = 0
Corollary 2 : a || b (collinear)
If a₁/b₁= a₂/b₂= a₃/b₃
6.1.7 Angle between two vectors
cos θ = a.b / ab
Let a = and b =
Then a. b = a₁b₁+ a₂b₂+ a₃b₃

6.1.8 Examples of dot product :
(i) Work (W) is the dot product of force (F) and displacement (r), i.e.W = F . r
(ii) Power (P) is the dot product of force (F) and velocity (v), i.e.P = F . v
(iii) Electric flux (f) is the dot product of intensity of electric field (E) and normal area A i.e.f = E . A

6.2 Cross product or vector product of two vectors
The vector product or cross product of two vectors is defined as a vector having a magnitude equal to the product of the magnitudes of two vectors with the sine of angle between them and direction perpendicular to the plane containing the two vectors in accordance with right hand screw rule.
Thus if & are two vectors, then their vector product written as × is a vector defined by = × = AB Sin θ
Where A and B are the magnitudes of and respectively and θ is the smaller angle between the two. Where is the unit vector whose direction is perpendicular to the plane containing the two vectors, in accordance with right hand screw rule.
6.2.1 Right hand screw rule :
(a) A right hand screw whose axis is perpendicular to the plane framed by and is rotated from to through the smaller angle between them, then the direction of advancement of the screw gives the direction of × i.e.

(b) Place the vector and tail to tail (this defines a plane). Now place stretched fingers and thumb of right hand perpendicular to the plane of and such that the fingers are along the vectors .If the fingers are now closed through smaller angle so as to go towards . The thumb gives the direction of × i.e. .

(c) Cross product non-commutative :
× ≠ ×
(d) Follows distributive law :
× ( + ) = × + ×
(e) Does not follows associative law :
× ( × ) ≠ ( × ) ×
(f) × = ; × = ; × = and × = 0; × = 0, × = 0
(g),
× = () × ()
= (a_yb_z – a_zb_y) + (a_zb_x – a_xb_z) + (a_xb_y – a_yb_x)
6.2.2 Properties of vector (or cross product)
(i) If a and b are any vectors, and m is any real number (positive or negative) then (m a) × b = m (a × b) = a × (m b)
(ii) The vector product is distributive w.r.t. additiona × (b + c) = a × b + a × c
(iii) Vector product of two vectors in terms of their rectangular componentsa = , b =
a × b =
a × b = (a₂b₃ – a₃b₂) – (a₁b₃ – a₃b₁) + (a₁b₂ – a₂b₁)
6.2.3 Unit vector perpendicular to two given vectors
Let be a unit vector perpendicular to two (non-zero) vectors a, b and positive for right handed rotation from a to b and θ be the angle between the vectors a, b then
a × b = ab sin θ
|a × b| = ab sin θ
Thus we get a × b /|a × b|