Vectors \[\vec{a},\vec{b}\] and \[\vec{c}\] are such that \[\vec{a}+\vec{b}+\vec{c}=0\] and \[\left| {\vec{a}} \right|=3\], $\left| {\vec{b}} \right|=5$ and $\left| {\vec{c}} \right|=7$. Find the angle between $\vec{a}$ and $\vec{b}$.

Hint: In order to solve this question, we have to manipulate the equation \[\vec{a}+\vec{b}+\vec{c}=0\] to get the angle between $\vec{a}$ and $\vec{b}$. First shift the \[\vec{c}\] on the RHS side and then square both sides. Squaring of a vector means taking dot product with itself i.e. ${{\vec{m}}^{2}}=\vec{m}.\vec{m}$ (where $\vec{m}$ is any arbitrary vector). So, after squaring we will get the term $\vec{a}.\vec{b}$which according to dot products of vectors can also be written as $\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta $ where $\theta $ indicates the angle between $\vec{a}$ and $\vec{b}$. Hence, solve the equation and get the value of $\theta $.

Complete step by step answer:
It is given that \[\vec{a}+\vec{b}+\vec{c}=0\] and \[\left| {\vec{a}} \right|=3\], $\left| {\vec{b}} \right|=5$ and $\left| {\vec{c}} \right|=7$.
So, let us consider \[\vec{a}+\vec{b}+\vec{c}=0\]
Shifting \[\vec{c}\] to the RHS, we get
$\Rightarrow \vec{a}+\vec{b}=-\vec{c}$
Squaring both side, we get
\[\Rightarrow {{\left( \vec{a}+\vec{b} \right)}^{2}}={{\left( -\vec{c} \right)}^{2}}\]
\[\Rightarrow \left( \vec{a}+\vec{b} \right).\left( \vec{a}+\vec{b} \right)=\left( -\vec{c} \right).\left( -\vec{c} \right)\]
Expanding the dot product, we get
$\Rightarrow {{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+2\vec{a}.\vec{b}={{\left| {\vec{c}} \right|}^{2}}$
Now, we know that \[\left| {\vec{a}} \right|=3\], $\left| {\vec{b}} \right|=5$ and $\left| {\vec{c}} \right|=7$, substituting these values in the above equations, we get
$\Rightarrow {{3}^{2}}+{{5}^{2}}+2\vec{a}.\vec{b}={{7}^{2}}$
$\Rightarrow 2\vec{a}.\vec{b}=49-25-9=15$
As from dot product of vectors we know that $\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta $ where $\theta $ indicates the angle between $\vec{a}$ and $\vec{b}$, Substituting it in the above equation we get
\[\Rightarrow 2\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta =15\]
$\Rightarrow 2\times 3\times 5\times \cos \theta =15$
$\Rightarrow \cos \theta =\dfrac{15}{30}=\dfrac{1}{2}=\cos \dfrac{\pi }{3}$
$\Rightarrow \theta =\dfrac{\pi }{3}$
So, the value of $\theta $ is equal to $\dfrac{\pi }{3}$.
Hence, the angle between $\vec{a}$ and $\vec{b}$is $\dfrac{\pi }{3}$ i.e. ${{60}^{\circ }}$.

Note:
This a typical question of vectors where we have to manipulate the given equation in order to get the desired value. Students should take a note of two things here, first one is squaring a vector means taking dot product of that particular vector with itself. The second thing is shifting which vector to the RHS as in our case we need the angle between $\vec{a}$ and $\vec{b}$ which is possible only when we get the term $\vec{a}.\vec{b}$ and hence we shifted \[\vec{c}\] to the RHS. After taking note of these things we can easily get the answer. This procedure can be followed whenever we are faced with such questions.