Solve Vector Angle Problem: 120 Degrees & Resultant Vector C

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The discussion revolves around determining the relationship between the magnitudes of the resultant vector C and the difference of two vectors A and B, given that the angle between them is 120 degrees. The user initially believes that the magnitude of C must be less than |A-B|, but the text suggests it is greater. Through calculations, they derive the expressions for |C| and |A-B|, concluding that |A-B| is indeed greater than |C| unless one of the vectors is zero. Other participants agree with the user's reasoning, leading to confusion about the correct answer. The consensus leans towards the conclusion that the answer should be iii), indicating that |C| is greater than |A-B| under the given conditions.
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This seems to be a simple problem in vectors, but for some reason, I can't solve it correctly. Hope you guys can help me out here! :D

Q: Let the angle between two non-zero vectors A and B be 120 degrees, and their resultant be vector C. Then, which of the following is true?

i) |C| must be equal to |A-B|

ii) |C| must be less than |A-B|

iii) |C| must be greater than |A-B|

iv) |C| may be equal to |A-B|

Here's what I attempted at an answer:

|C| = SQRT(A^2 + B^2 + 2 AB cos120 ) [Magnitude of resultant of two vectors A and B]

= SQRT(A^2 + B^2 -AB) [cos120 = -1/2]--------eq(1)

Now, |A-B|^2 = A^2 + B^2 - 2(A.B)

= A^2 + B^2 - 2ABcos120
= A^2 + B^2 + AB

Therefore,
|A-B| = SQRT(A^2 + B^2 + AB) -------------eq(2)

Now, it's clear that the RHS of equation (2) is greater than equation (1), with equality occurring only if A or B is a Zero vector.

So, I think the right option ought to be (ii), but the text says the answer is (iii)! It would be really helpful if someone could point out my mistake. Thanks in advance! :D

Also, please pardon my crude writing style. I'm using the mobile version of the site, and my phone doesn't seem to support symbols.
 
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I think you've done it all correctly. I would have thought the answer should be ii), I don't know why the text says it should be iii) ...
 
jobsism said:
This seems to be a simple problem in vectors, but for some reason, I can't solve it correctly. Hope you guys can help me out here! :D

Q: Let the angle between two non-zero vectors A and B be 120 degrees, and their resultant be vector C. Then, which of the following is true?

i) |C| must be equal to |A-B|

ii) |C| must be less than |A-B|

iii) |C| must be greater than |A-B|

iv) |C| may be equal to |A-B|

Here's what I attempted at an answer:

|C| = SQRT(A^2 + B^2 + 2 AB cos120 ) [Magnitude of resultant of two vectors A and B]

= SQRT(A^2 + B^2 -AB) [cos120 = -1/2]--------eq(1)

Now, |A-B|^2 = A^2 + B^2 - 2(A.B)

= A^2 + B^2 - 2ABcos120
= A^2 + B^2 + AB

Therefore,
|A-B| = SQRT(A^2 + B^2 + AB) -------------eq(2)

Now, it's clear that the RHS of equation (2) is greater than equation (1), with equality occurring only if A or B is a Zero vector.

So, I think the right option ought to be (ii), but the text says the answer is (iii)! It would be really helpful if someone could point out my mistake. Thanks in advance! :D

Also, please pardon my crude writing style. I'm using the mobile version of the site, and my phone doesn't seem to support symbols.

Just wondering if those options should have been

i) |C| must be equal to |A|-|B|

ii) |C| must be less than |A|-|B|

iii) |C| must be greater than |A|-|B|

iv) |C| may be equal to |A|-|B|
 
Thank you for your replies, BruceW and PeterO!

Glad to know that I wasn't wrong after all... :)

I think the options indeed ought to be as you say, PeterO. The answer then ought to be iii) going by a similar proof, right?
 
If the options are as peter has mentioned, then I don't think the answer is iii)
 
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