# Vector angle problem

1. Feb 16, 2012

### jobsism

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 occuring 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.

2. Feb 16, 2012

### BruceW

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) ...

3. Feb 16, 2012

### PeterO

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|

4. Feb 17, 2012

### jobsism

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?

5. Feb 17, 2012

### BruceW

If the options are as peter has mentioned, then I don't think the answer is iii)