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The Question

$\angle$ between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$. If $|a|=3$ and $|b|=5$, find $|a+b|$ and $|a-b|$. Hint use law of cosines

Before I used law of cosines, I did something like below, but didn't get the right asnwer:

Is the method wrong or something?

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up vote 0 down vote accepted

You have done exactly correct up to $|a+b|^2=(\frac{3\sqrt{3}}{2})^2+(5+\frac{3}{2})^2$. Then you done the mistake as $(5+\frac{3}{2})^2=(\frac{13}{2})^2=\frac{169}{4}$ and you have written $16$.

BTW, your question has given hint to use cosine law, and the label says without cosine.

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I know the hint is there. But before I used Law of Cos, I used usual trig. and got the wrong answer. I was wondering why – Jiew Meng Nov 11 '11 at 1:54
@jiewmeng: I have retag the question to to vector analysis as I think it is not suitable in vector space tag. – Tapu Nov 11 '11 at 2:00
@jiewmeng: Forgot to mention, I like your approach of doing everything explicitly. Probably it was a test..? – Tapu Nov 11 '11 at 2:05
Actually revision for exam :) What do you mean by explicitly? Maybe you mean most people skip steps? – Jiew Meng Nov 11 '11 at 6:21
@jiewmeng: I meant I like drawing the figure, labels, no step jump, all calculation in rough in the right side (like you did). So, looking at your answer sheet, I remembered my school days. I did exactly like you (of course those silly mistakes, which people now say "typo"). Alas, I lost those good habits later which has lead me to making typos very often! – Tapu Nov 11 '11 at 16:38

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