Question

if |a+b|=|a-b| calculate the angle between a and b​

Answer 1

Explanation:

Answer in the attachment

Answer 2

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I assume the question means |A+B|=|A-B| where A and B are non-zero vectors.

Squaring both sides,

|A+B|2=|A−B|2|A+B|2=|A−B|2

Since A.A=|A|2A.A=|A|2

|A+B|2=|A−B|2|A+B|2=|A−B|2

(A+B).(A+B)=(A−B).(A−B)(A+B).(A+B)=(A−B).(A−B)

A.A+A.B+B.A+B.B=A.A−A.B−B.A+B.BA.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|A|2+2A.B+|B|2=|A|2−2A.B+|B|2

4A.B=0

A.B=0

|A||B|cos(θ)=0

Since A and B are non-zero vectors, cos(θ)cos(θ)must be zero. It implies that A and B are perpendicular as in the interval [0,π][0,π] cos vanishes only at π/2π

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