If the magnitude of the cross product of two vectors is $\sqrt 3 $times to the magnitude of their scalar product, the angle between the two vectors will be
A) $\pi $.
B) $\dfrac{\pi }{2}$.
C) $\dfrac{\pi }{3}$.
D) $\dfrac{\pi }{6}$.

Hint
Recall the concept of dot product of two vectors and cross product of two vectors. The dot product between two vectors a and b is given by. The cross product between two vectors a and b is given by and then use the given condition and put the values to solve this question.

Complete step by step answer
The dot product between two vectors a and b is given by.
It is given in the question that the magnitude of the cross product of two vectors is $\sqrt 3 $times to the magnitude of their scalar product. Therefore,
Using the above mentioned formulas, it can be simplified as,
On further solving we have,
$\Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \sqrt 3 $
$\Rightarrow \tan \theta = \sqrt 3 $
$\Rightarrow \theta = {60^{^ \circ }} $
Which can be represented as $\dfrac{\pi }{3}$radians.
Therefore, the angle between the given vectors is $\theta = {60^{^ \circ }}$and so the correct option is option (C).

Note
i) Cross product of two vectors a and b is perpendicular to both the vectors a as well as b and is represented by $A \times B$.
ii) Cross product is anticommutative while dot product is commutative.