Angle Between a & b When |a| = |b| = |a-b| or |a+b|

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SUMMARY

The discussion centers on the conditions under which the angle between two vectors a and b is determined by their magnitudes. It establishes that if |a| = |b| = |a-b|, then the angle between a and b is 0 degrees, indicating that a and b are identical vectors. Conversely, when |a| = |b| = |a+b|, the angle remains 0 degrees as well, since it implies that a and b are equal in magnitude and direction. However, the assertion that |a| = |b| implies a = b is incorrect, as demonstrated by the example of the vectors (1,0) and (0,1).

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Homework Statement



If a and b are vectors such that|a|=|b|=|a-b|, what is then the angle between a and b? What changes for |a|=|b|=|a+b|


Homework Equations





The Attempt at a Solution



|a|=|b| ⇔ a=b ∴ a=b ⇔|a-b|=0 ...identity of discernibles

angle between them is then 0.

|a|=|b|=|a+b| ⇔ |a+b| = |2a| or |2b| = |2||a| = 2|a|.

The magnitude of a or b changes but angle between a and b is 0 since a=b.

Is this correct logic?
 
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No, it's not. |a|=|b| doesn't imply a=b. For example, |(1,0)| = |(0,1)|.
 

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