find the angle between A(vector) and B(vector) given c=b/2
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Given that
Thus, the angle between b and a is 30 degrees.
^_^
Thus, the angle between b and a is 30 degrees.
^_^
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Fazyl:
but the answer is 150°
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yuThank you for A2A.
This is a question which I loved while answering. Thank you for giving me an opportunity to answer this one.
We have,
a+b=c ……………………(1)
Also,
|a|+|b|=|c| …………(2)
Square the equation, (1), we get,
(a+b).(a+b)=c.c
Implies,
|a|2+|b|2|+2(a.b)=|c|2 ……….(3)
From (2), we can get,
|a|2+|b|2|+2|a||b|=|c|2 ………(4)
Since the RHS’s of (3) and (4) are same, we can equate them, we get,
|a|2+|b|2|+2(a.b)=|a|2+|b|2|+2|a||b|
Implies,
a.b=|a||b|
Dot Product of 2 vectors is the product of absolute values of the vectors with the cosine of angle between them. So,
|a||b|cosx=|a||b|
Here, x is the angle between a and b.
So,
cosx=1
Take cosine inverse or arccos on both sides, we get,
x=cos−1(1)
Implies,
x=0
So, the angle between the 2 given vectors is 0, which means they’re parallel.
I hope my answer was helpful. If you’ve any questions regarding Mathematics, Physics or Computers, ask me.
Good day my math learning friend!!!
Once again, thank you for A2A.
This is a question which I loved while answering. Thank you for giving me an opportunity to answer this one.
We have,
a+b=c ……………………(1)
Also,
|a|+|b|=|c| …………(2)
Square the equation, (1), we get,
(a+b).(a+b)=c.c
Implies,
|a|2+|b|2|+2(a.b)=|c|2 ……….(3)
From (2), we can get,
|a|2+|b|2|+2|a||b|=|c|2 ………(4)
Since the RHS’s of (3) and (4) are same, we can equate them, we get,
|a|2+|b|2|+2(a.b)=|a|2+|b|2|+2|a||b|
Implies,
a.b=|a||b|
Dot Product of 2 vectors is the product of absolute values of the vectors with the cosine of angle between them. So,
|a||b|cosx=|a||b|
Here, x is the angle between a and b.
So,
cosx=1
Take cosine inverse or arccos on both sides, we get,
x=cos−1(1)
Implies,
x=0
So, the angle between the 2 given vectors is 0, which means they’re parallel.
I hope my answer was helpful. If you’ve any questions regarding Mathematics, Physics or Computers, ask me.
Good day my math learning friend!!!
Once again, thank you for A2A.
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