If |vector A - vector B| = |vector A|=|Vector B|,the angle between vector a and b is
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when vector A = vector B then the angle between them is 60°
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I assume the question means |A+B|=|A-B| whereA 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=04A.B=0
A.B=0A.B=0
|A||B|cos(theta)=0|A||B|cos(theta)=0
Since A and B are non-zero vectors, A and B must be perpendicular as cos (90 degrees) = 0
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=04A.B=0
A.B=0A.B=0
|A||B|cos(theta)=0|A||B|cos(theta)=0
Since A and B are non-zero vectors, A and B must be perpendicular as cos (90 degrees) = 0
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