there r two vectors A and B such that A+B=A-B then find the angle between A and B
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Answered by
26
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=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
Anonymous:
but ans is pie/2
Answered by
1
Answer:
Angle, ∅= 120°
Explanation:
Given;-
R = A + B
and, R = A = B ______(1)
Now, we know that when R = A = B;-
∵ R = 2A cos∅/2
∴ R = 2A cos 120°/2 [Putting ∅= 120 through trial and error method]
R = 2A cos 60°
R = 2A × 1/2 [since, cos 60°= 1/2]
R = A ______(2)
From, (1) and (2), we obtain;-
R = A = B [at ∅ = 120°]
Hope it helps ;-))
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