6. Product of Two Vectors and Scalar (or dot) product of two vectors

Product of Two Vectors and Scalar (or dot) product of two vectors:

1. scalar product of vectors or dot product
2. vector product of vectors or cross product

Dot Product of Vectors:

The scalar or dot product of two non-zero vectors a⃗  and b⃗ , denoted by a⃗ .b⃗  is

a⃗ .b⃗  = |a⃗ | |b⃗ | cosθ

where θ is the angle between a⃗  and b⃗  and 0 ≤ θ ≤ π as shown in the figure below.

It is important to note that if either a⃗  = 0⃗  or b⃗  = 0⃗ , then θ is not defined, and in this case

a⃗ .b⃗  = 0

 We can express the scalar product as:

where || and || represent the magnitude of the vectors  and  while cos θ denotes the cosine of the angle between both the vectors and .  indicate the dot product of the two vectors.

Projection of Vectors:

BP is known to be the projection of a vector a on vector b in the direction of vector b given by |a| cos θ.

Similarly, the projection of vector b on a vector a in the direction of the vector a is given by |b| cos θ.

The projection of vector    on a vector  

Question: Find the angle between two vectors a⃗  and b⃗  with magnitudes √3 and 2 ,
respectively having a⃗ .b⃗  = √6.

Solution: By definition of the scalar or dot product of vectors, we know that

a⃗ .b⃗  = |a⃗ | |b⃗ | cosθ

where θ is the angle between a⃗  and b⃗ . In this question, we have,

|a⃗ | = √3, |b⃗ | = 2, and a⃗ .b⃗  = √6

Replacing these values in the formula, we get             

a⃗ .b⃗  = |a⃗ | |b⃗ | cosθ
⟺ √6 = √3 x 2 x cosθ
⟺ √3 x √2= √3x √2 x √2x cosθ

Cancelling the common terms on both sides, we get

1 = √2 x cosθ
Or, cosθ = √12
Therefore, θ = π/4
Hence, the angle between the vectors a⃗  and b⃗  is π/4.

Dot Product Properties of Vector:

• Property 1: Dot product of two vectors is commutative i.e. a.b = b.a = ab cos θ.
• Property 2: If a.b = 0 then it can be clearly seen that either b or a is zero or cos θ = 0 

•   q = π/2

• Property 3: Also we know that using scalar product of vectors (pa).(qb)=(pb).(qa)=pq (a.b)
• Property 4: The dot product of a vector to itself is the magnitude squared of the vector i.e. a.a = a.a  cos 0 = a²
• Property 5: The dot product follows the distributive law also i.e. a.(b + c) = a.b + a.c
• Property 6: In terms of orthogonal coordinates for mutually perpendicular vectors it is seen that
• 

Example : Let there be two vectors [6, 2, -1] and [5, -8, 2]. Find the dot product of the vectors.

Solution:

Given vectors: [6, 2, -1] and [5, -8, 2] be a and b respectively.

a.b = (6)(5) + (2)(-8) + (-1)(2)

a.b = 30 – 16 – 2

a.b = 12

Example 2: Let there be two vectors |a|=4 and |b|=2 and θ = 60°. Find their dot product.

Solution:

a.b = |a||b|cos θ

a.b = 4.2 cos 60°

a.b = 4.2 × (1/2)

a.b = 4

Example :

Find the angle between two vectors

And

Solution:

Given:

And

The formula to find the angle between two vectors is given by:

Hence,

= 1-1-1

= -1

Therefore, 

Now, substituting the value in the formula, we get

Cos θ = -⅓

Hence, the angle between two vectors is θ = cos^-1(-⅓).

Question:  Ifare such thatis perpendicular to, then find the value of λ.

Solution:

We know that the