If |a+b|=|a-b|,show that a and b are perpendicular vectors

Given: |a+b|=|a-b|

To Prove: a and b are perpendicular vectors

 

So, if you square both the sides, then you will get 

 

(|a+b|)^2 = (|a-b|)^2

 

Since, any |a|^2 can be written as the dot product of a with itself i.e. a.a , so here also (|a+b|)^2 can be represented as (a+b).(a+b). Using the same logic 

 

(a+b).(a+b) = (a-b).(a-b)

On expanding, we will get

a.a +b.b +2a.b = a.a +b.b -2a.b

4a.b=0

4|a|*|b|*cos(theta) = 0 

Since, a and b vectors are non-zero vectors, hence

cos(theta) = 0

and hence, theta = 90

 

So, the angle between a and b vectors is 90. 

 

Thank you