Question

if |a+b| = |a-b| then show that angle between a and b is 90°​

Answer 1

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$\large\sf \red {|a + b| = |a - b| }$

Squaring both sides,

$\sf|A+B|^2=|A−B|^2$

$\sf{Since \: A.A=|A|^2}$

$|A+B|^2=|A−B|^2$

(A+B).(A+B)=(A−B).(A−B)

(A.A)+(A.B)+(B.A)+(B.B)=(A.A)−(A.B)−(B.A)+(B.B) Using distributive property)

$|A|^2+2A.B+|B|^2=|A|^2−2A.B+|B|^2$

4(A.B)=0

A.B=0

|A||B|cos(theta)=0

Since A and B are non-zero vectors, A and B must be perpendicular as cos (90°) = 0.

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