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0° is answer?
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Please type the question directly as much as Possible.
Q: If A + B = C and |A| + |B| = |C| , then the angle between vector A and B is ?
Steps:
1) Take dot products with A, B and C vectors separately.
2) Add all three equations and use given condition to get angle between A and B as 0°.
This is bit Rigorous method, So There is one new method.
Method -2
Steps:
1) We know that,
A + B = C
and
|A| +|B| = |C|
where A, B and C are vectors.
=> |A + B | = |C|
=> |A+B|^2 = |C|^2
=> (A+B).(A+B) =( |A| + |B|)^2
=> |A|^2 + 2A. B + |B|^2 = |A|^2 + 2|A||B|+ |B|^2
=> 2A. B = 2|A||B|
=> 2|A||B| cos α = 2|A||B|
=> cos α = 1
=> α = 0°
Therefore, angle between A and B is 0°
Please type the question directly as much as Possible.
Q: If A + B = C and |A| + |B| = |C| , then the angle between vector A and B is ?
Steps:
1) Take dot products with A, B and C vectors separately.
2) Add all three equations and use given condition to get angle between A and B as 0°.
This is bit Rigorous method, So There is one new method.
Method -2
Steps:
1) We know that,
A + B = C
and
|A| +|B| = |C|
where A, B and C are vectors.
=> |A + B | = |C|
=> |A+B|^2 = |C|^2
=> (A+B).(A+B) =( |A| + |B|)^2
=> |A|^2 + 2A. B + |B|^2 = |A|^2 + 2|A||B|+ |B|^2
=> 2A. B = 2|A||B|
=> 2|A||B| cos α = 2|A||B|
=> cos α = 1
=> α = 0°
Therefore, angle between A and B is 0°
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