In this question, we’re given that |A + B| = |A – B| and A, B are vectors

What we’re trying to find out is angle between vectors A and B

So let’s start with what’s given |A + B| = |A – B|

Squaring both sides of this equation**|A + B| ^{2} = |A – B|^{2}** (Equation 1)

Dot Product of a vector with itself is square of it’s magnitude

This means if a is a vector

Then

|a|

^{2}= a.a

Using this formula

(Equation 1) can be written as

(A + B).(A + B) = (A – B).(A – B)

Simplifying

A.(A + B) + B.(A + B) = A.(A – B) – B.(A – B)

A.A + A.B + B.A + B.B = A.A – A.B – B.A + B.B

Cancelling out values on both sides

**A.B + B.A = – A.B – B.A**(Equation 2)

Dot Product of vectors is Commutative

Which means

If a, b are two vectors

Then a.b = b.a

Using this formula (Equation 2) can be written as

A.B + A.B = – A.B – A.B

2 A.B = – 2 A.B

2 A.B + 2 A.B = 0

4 A.B = 0

A.B = 0

|A| |B| Cosθ = 0

⇒ Cosθ = 0

General Solution for Cosθ = 0 is θ = (2n + 1)𝛑/2 where n is some Integer

Therefore if |A + B| = |A – B| and A, B are two vectors then possible values of angles between vectors A and B are 𝛑/2, 3𝛑/2, 5𝛑/2 and so on.

👇🏻 Key Concepts Used in This Question**1. 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}**2. 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**3. General Solution for Cosθ = 1 is θ = 2n𝛑 where n is integer**