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?
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.