In this question, we are given two vectors A, B and we need to find out angle between vectors (A + B) and (A – B).
Easiest way to find out angle between any two vectors is to use Dot Product
If a and b are two vectors then their dot product is
a.b = |a| |b| Cosθ
Where
a, b are two vectors
|a|, |b| are magnitudes of these vectors
θ is the angle between directions of vectors a, b
Replacing
a = (A + B)
b = (A – B)
in the dot product formula
(A + B).(A – B) = |A + B| |A – B| Cosθ
Simplifying Left Hand Side of this equation
A.(A – B) + B.(A – B) = |A + B| |A – B| Cosθ
A.A – A.B + B.A – B.B = |A + B| |A – B| Cosθ (Equation 1)
Dot Product of a vector with itself is square of it’s magnitude
If a is a vector then a.a = |a|2
Therefore
A.A = |A|2
B.B = |B|2
|A|2 – A.B + B.A – |B|2 = |A + B| |A – B| Cosθ
Moreover Dot Product of two vectors is also Commutative which means
a.b = b.a if a, b are vectors
Therefore A.B = B.A
|A|2 – A.B + A.B – |B|2 = |A + B| |A – B| Cosθ
|A|2 – |B|2 = |A + B| |A – B| Cosθ
Using algebraic formula a2 – b2 = (a – b)(a + b)
(|A| – |B|)(|A| + |B|) = |A + B| |A – B| Cosθ
⇒ Cosθ = 1
General Solution for Cosθ = 1 is θ = 2n𝛑 where n is integer
Thus if A and B are two vectors then possible values of angle between vectors (A + B) and (A – B) are 0, 2𝛑, 4𝛑, 6𝛑 and so on.
👇🏻 Key Concepts Used in This Question
1. Dot Product of two vectors a and b is defined as a.b = |a| |b| Cosθ
Where
a, b are two vectors
|a|, |b| are magnitudes of these vectors
θ is the angle between directions of vectors a, b
2. Dot Product of a vector with itself is square of it’s magnitude
This can be derived from formula of Dot Product
If a, b are two vectors
Then
Their Dot Product a.b = |a| |b| Cosθ
Where
|a|, |b| are magnitudes of these vectors
θ is the angle between directions of vectors a, b
Let’s put b = a in Dot Product formula
a.a = |a| |a| Cosθ
θ = 0 (Angle between vectors a and a is zero)
a.a = |a| |a| Cos0 = |a|2
a.a = |a|2
3. Dot Product of two vectors is Commutative
Which means a.b = b.a
If a, b are two vectors
This can also be proved from formula of Dot Product itself
a.b = |a| |b| Cosθ
b.a = |b| |a| Cosθ
⇒ a.b = b.a
Therefore Dot Product of two vectors is Commutative
4. In Algebra (a – b)(a + b) = a2 – b2
In order to prove this, just simplify Left Hand Side
(a – b)(a + b) = a(a + b) – b(a + b)
= a2 + ab – ba – b2
= a2 – b2
⇒ (a – b)(a + b) = a2 – b2
(Do note that here a, b are numbers not vectors)
5. General Solution for Cosθ = 1 is θ = 2n𝛑 where n is integer