Given three unit vector \[a, b, c\] ; no two of which are collinear and satisfies \[a \times \left( {b \times c} \right) = \dfrac{1}{2} \times b\]. The angle between \[a\] and \[b\] is equal to
A) \[\dfrac{\pi }{3}\]
B) \[\dfrac{\pi }{4}\]
C) \[\dfrac{\pi }{2}\]
D) None of these

Hint:
Here, we will use the given condition with the vector triple product and we will simplify the given equation. Then by using the condition of collinearity and scalar product of vectors, we will find the angle between the given vectors.

Formula Used:
We will use the following formula:
1) Vector Triple Product: \[a \times \left( {b \times c} \right) = \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c\]
2) Scalar Product of Vectors: \[a \cdot b = ab\cos \theta \]
3) Trigonometric Ratio: \[{\cos ^{ - 1}}0 = \dfrac{\pi }{2}\]

Complete step by step solution:
We are given three unit vectors \[a,b,c\] .
It is given that no two points of which are collinear and satisfy \[a \times \left( {b \times c} \right) = \dfrac{1}{2} \times b\].
Now, by using the vector triple product formula \[a \times \left( {b \times c} \right) = \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c\], we get
\[ \Rightarrow \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c = \dfrac{1}{2} \times b\]
Rewriting the equation, we get
\[ \Rightarrow \left( {a \cdot c} \right)b - \dfrac{1}{2} \times b = \left( {a \cdot b} \right)c\]
Now, by taking out the common factors, we get
\[ \Rightarrow \left( {\left( {a \cdot c} \right) - \dfrac{1}{2}} \right)b = \left( {a \cdot b} \right)c\].
Since the given vectors are collinear, thus the possibilities are \[a \cdot c = \dfrac{1}{2}a\] and \[a \cdot b = 0\].
Now, by using the scalar product of vectors \[a \cdot b = ab\cos \theta \], we get
\[ab\cos \theta = 0\]
Now, by rewriting the equation, we get
\[ \Rightarrow \cos \theta = 0\]
 \[ \Rightarrow \theta = {\cos ^{ - 1}}0\]
We know that Trigonometric Ratio: \[{\cos ^{ - 1}}0 = \dfrac{\pi }{2}\]
\[ \Rightarrow \theta = \dfrac{\pi }{2}\]
Therefore, the angle between two vectors \[a\] and \[b\] is \[\dfrac{\pi }{2}\].

Thus, option (C) is the correct answer.

Note:
We know that a vector is an object which has both a magnitude and a direction. We should know that the resultant of two vectors is equal to either of them if and only if the two vectors are of same magnitude and the vectors are inclined to each other at an angle. The conditions for three vectors to be collinear are:
1) Two vectors \[\overrightarrow a \] and \[\overrightarrow b \] are collinear if there exists a number \[\eta \] such that \[\overrightarrow a = \eta \cdot \overrightarrow b \]
2) Two vectors are collinear if relations of their coordinates are equal only if one of the components of the vector is zero.
3) Two vectors are collinear if their cross-product is equal to the zero vector.