If the angle between vectors $\vec a$ and $\vec b$ is $\dfrac{\pi }{3}$ then the angle between vectors $2\vec a$ and $ - 3\vec b$ will be ?

Hint:In physics, there are quantities which have magnitude but not direction such as temperature and mass, these are known as scalar quantities whereas some quantities have magnitude and also need a direction for their representation such are known as vectors for example velocity and acceleration are few examples of vector quantities.

Formula used:
For any given two vectors $\vec A$ and $\vec B$ whose dot product is calculated as,
$\vec A.\vec B = \left| A \right|\left| B \right|\cos \theta $
Where, $\left| A \right|(and)\left| B \right|$ are the magnitudes of vector $A$ and vector $B$ and $\theta $ denotes the angle between these two vectors.

Complete step by step answer:
According to the question we have given that, for vector $\vec a$ and $\vec b$ the angle $\theta = \dfrac{\pi }{3}$ ,let us calculate the dot product of these two vectors using the formula $\vec A.\vec B = \left| A \right|\left| B \right|\cos \theta $ so we have,
$\vec a.\vec b = \left| a \right|\left| b \right|\cos \dfrac{\pi }{3}$
Since, $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ so
$\vec a.\vec b = \dfrac{1}{2}\left| a \right|\left| b \right| \to (i)$

Now, let us calculate the dot product of vectors $2\vec a$ and $ - 3\vec b$ and assume the angle between them is $\phi $ so using the formula
$\vec A.\vec B = \left| A \right|\left| B \right|\cos \theta $ We have,
$\Rightarrow (2\vec a).( - 3\vec b) = \left| {2a} \right|\left| { - 3b} \right|\cos \phi $
$\Rightarrow ( - 6)\vec a.\vec b = 6\left| a \right|\left| b \right|\cos \phi $
$ \Rightarrow \vec a.\vec b = - \left| a \right|\left| b \right|\cos \phi \to (ii)$

From equation $i(and)ii$ on comparing we get,
$\dfrac{1}{2}\left| a \right|\left| b \right| = - \left| a \right|\left| b \right|\cos \phi $
$\Rightarrow \cos \phi = - \dfrac{1}{2}$
And as we know,
$\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}$
$ \therefore \phi = \dfrac{{2\pi }}{3}$

Hence, the angle between the vectors $2\vec a$ and $ - 3\vec b$ will be $\dfrac{{2\pi }}{3}$.

Note: It should be remembered that while solving such questions, the basic laws of vector algebra are used such as $({k_1}\vec A).({k_2}\vec B) = {k_1}{k_2}(\vec A.\vec B)$ where ${k_1}(and){k_2}$ are the scalar numbers multiplied to the vectors A and B, and when a scalar is multiplied by a vector the, resultant quantity is always a vector quantity. In physical terms, the dot product between two vectors is the area of the parallelogram formed whose two adjacent sides are represented by two vectors.