Given that A = B and AperpendicularB. What is the angle
between (A+B) and (A-B) ?
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Answer:
Squaring both sides,
|A+B|2=|A−B|2
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
4A.B=0
A.B=0
|A||B|cos(θ)=0
Explanation:
Squaring both sides,
|A+B|2=|A−B|2
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
4A.B=0
A.B=0
|A||B|cos(θ)=0
Since A and B are non-zero vectors, cos(θ) must be zero. It implies that A and B are perpendicular as in the interval [0,π] cos vanishes only at π/2
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