If a be a constant vector and b be a vector of constant magnitude such that $∣\vec {b}∣$=$\dfrac{a}{2}$​ and $∣\vec {a}∣ \neq 0$. then maximum value of angle between $\vec a$ and $\vec a + \vec b$ is:
A) $\dfrac{{5\pi }}{6}$
B) $\dfrac{{5\pi }}{3}$
C) $\dfrac{{2\pi }}{3}$
D) $\dfrac{\pi }{6}$

Hint: A vector is represented by a line with an arrowhead. The point where the arrow starts is called the origin of the vector and the point where the arrow ends is called the tip. A vector remains unchanged when it is displaced parallel to itself.

Complete step by step solution:
Consider the vector b makes an angle $\theta $ with the negative side of vector a. Then the resultant component in the x-direction is given by
 $(a + b)_x$ $ = a - b\cos \theta $
 $ \Rightarrow $ $(a + b)_x$ $ = a - \dfrac{a}{2}\cos \theta $ $\left( {\because b = \dfrac{a}{2}} \right)$
$ \Rightarrow $ $(a + b)_x $ = $\dfrac{a}{2}(2 - \cos \theta )$
 Similarly, the resultant vector of the y component is given by
 $(a + b)_y $ = $b\sin \theta $
$ \Rightarrow $ $(a + b)_y $ = $\dfrac{a}{2}\sin \theta $\[\]$\left( {\because b = \dfrac{a}{2}} \right)$
We know that according to the parallelogram law of vector addition, the angle between the resultant with the first vector is given by,
 $\tan \alpha = \left( {\dfrac{{\dfrac{a}{2}\sin \theta }}{{\dfrac{a}{2}\left( {2 + \cos \theta } \right)}}} \right) = \dfrac{{\sin \theta }}{{(2 + \cos \theta )}}$
If α has to be maximum, then $\dfrac{{d\alpha }}{{d\theta }}$ must be equal to zero, thus we have $\dfrac{d}{{d\theta }}\left( {\tan \alpha } \right) = \dfrac{d}{{d\theta }}\left( {\dfrac{{\sin \theta }}{{2 + \cos \theta }}} \right)$
 \[ \Rightarrow se{c^2}\alpha \left( {\dfrac{{d\alpha }}{{d\theta }}} \right) = \dfrac{{\sin \theta \left( { - \sin \theta } \right) - \left( {2 + \cos \theta } \right)\left( {\cos \theta } \right)}}{{{{(2 + cos\theta )}^2}}}\]
$
\Rightarrow - si{n^2}\theta - 2\cos \theta - co{s^2}\theta = 0 \\
\Rightarrow 2\cos \theta = - 1 \\
\Rightarrow \theta = \dfrac{{2\pi }}{3} \\
$
 Thus for the above value $\theta $, the angle between the resultant and vector a is maximum.
 $\therefore $The maximum value of the angle \[\alpha \]is,
 $\tan \alpha = \dfrac{{\sin \theta }}{{(2 + \cos \theta )}}$
 Substituting the value of $\theta $, we get
 \[ \Rightarrow \tan \alpha = \dfrac{{\sin \left( {\dfrac{{2\pi }}{3}} \right)}}{{(2 + \cos \left( {\dfrac{{2\pi }}{3}} \right))}}\] \[(\because \theta = \dfrac{{2\pi }}{3})\]
 $ \Rightarrow \tan \alpha = \dfrac{{\;\sqrt {\dfrac{3}{2}} \;}}{{(2 - \dfrac{1}{2})}}$
 $ \Rightarrow \tan \alpha = \dfrac{1}{{\sqrt 3 }}$
 \[ \Rightarrow \alpha = ta{n^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]
\[ \Rightarrow \alpha = \dfrac{\pi }{6}\]
$\therefore $ The maximum value of the angle, \[\alpha = \dfrac{\pi }{6}\]\[\;\;\]

Hence the correct option is D.

Note: 1) When a vector is multiplied by a scalar quantity then we get a new vector in the same direction but having a new magnitude. The vector product is distributive and the addition is commutative. The vector has a definite direction. Vectors obey the law of parallelograms of addition.
 2) Zero vector is the one that has zero magnitudes and also no specific direction.
 3) If two vectors have equal magnitude and are in the same direction, then they are called equal vectors.