Question

If the magnitude of the scalar product and the vector product of the two vectors are \[8\] and \[8\sqrt 3 \] respectively then the angle between them is …..
A. \[{60^ \circ }\]
B. \[{90^ \circ }\]
C. \[\dfrac{\pi }{4}\]
D. \[\dfrac{\pi }{6}\]

Answer 1

Hint:The given problem revolves around the concepts of ‘scalar and vector’ property, which are majorly differentiated as the only magnitude (such as - mass, speed, etc.) and the both magnitude as well as direction (such as – velocity, acceleration, etc.) respectively. As a result, dividing the both the conditions by substituting the respective magnitudes given, to obtain the desired angle.

Complete step by step answer:
Since, we have given that the magnitude and the vector product of the two vectors are \[8\] and \[8\sqrt 3 \] respectively.As a result, we know that the scalar product of any two vectors particularly is always the “dot product” (where, its product is in scalar quantity that is without any certain direction)! For instance, if ‘work’ is scalar and ‘force’ as well as ‘displacement’ is vector then mathematically it can be expressed as, \[W = \int
{\overrightarrow F .d\overrightarrow s } \] respectively.

Hence, scalar product of two vectors say, \[\overrightarrow A \] and \[\overrightarrow B \] will be,
\[\overrightarrow A .\overrightarrow B = |\overrightarrow A ||\overrightarrow B |\cos \theta \]
Substituting the given magnitude i.e. \[8\], we get
\[8 = |\overrightarrow A ||\overrightarrow B |\cos \theta \] … (i)
Where, \[\overrightarrow A ,\overrightarrow B \] are the magnitude of the two vectors,
\[\theta \] is the angle between these two vectors.

Similarly, the vector product of any two vectors in particular is always the “cross product” (also called a ‘vector product’)! For instance, the vector product of ‘force’ and ‘distance’ from the axis of rotation i.e. its direction then its vector quantity is expressed as, torque \[\tau = \overrightarrow F \times \overrightarrow r \] respectively.
Hence, vector product of these two vectors is,
 \[\overrightarrow A \times \overrightarrow B = |\overrightarrow A ||\overrightarrow B |\sin \theta \]
Substituting the given magnitude i.e. \[8\sqrt 3 \], we get
\[8\sqrt 3 = |\overrightarrow A ||\overrightarrow B |\sin \theta \] … (ii)
Where,
\[\overrightarrow A ,\overrightarrow B \] are the magnitude of the two vectors,
\[\theta \] is the angle between these two vectors.

Now, dividing (ii) and (i) so as to get the desired angle, we get
\[\dfrac{{8\sqrt 3 }}{8} = \dfrac{{|\overrightarrow A ||\overrightarrow B |\sin \theta }}{{|\overrightarrow A ||\overrightarrow B |\cos \theta }}\]
Solving it mathematically, we get
\[\dfrac{{8\sqrt 3 }}{8} = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \sqrt 3 \]
We know that, trigonometric ratio for \[\tan \theta \] is \[\dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\tan \theta = \sqrt 3 \]
\[\Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right)\]
As a result, substituting \[\tan {60^ \circ } = \sqrt 3 \], we get
\[\therefore \theta = {60^ \circ }\]

Hence, option A is correct.

Note: One must able to know the definition of SCALAR and VECTOR which states that scalar quantity has only magnitude (no direction at all) mathematically exists ‘dot product’ i.e. \[\bar a.\bar b\cos \theta \] whereas, the vector quantity contains both the magnitude as well as directions [mathematically resembles ‘cross product’ always, \[\overrightarrow a \overrightarrow { \times b} \sin \theta \]] that is both the parameters are totally different, so as to be sure of our final answer.