Question

Find the correct option
If the projection of the vector \[\overrightarrow a \] on the vector \[\overrightarrow b \] is \[\left| {\overrightarrow a \times \overrightarrow b } \right|\] and \[3\overrightarrow b = \widehat i + \widehat j + \widehat k\] , then, the angle between \[\overrightarrow a \] and \[\overrightarrow b \] is equal to
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{2}\]
C. \[\dfrac{\pi }{4}\]
D. \[\dfrac{\pi }{6}\]

Answer 1

Hint: By using the projection formula, the projection of one vector on another vector is found and then an equation is established with the given data to calculate the angle between the two vectors and the correct option is chosen.

Formula used: The following formulas are used to solve this problem:
If \[\overrightarrow a \] and \[\overrightarrow b \] are two vectors and \[\theta \] is the angle between them, then,
1. The projection of the vector \[\overrightarrow a \] on the vector \[\overrightarrow b \] is equal to \[\dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow b } \right|}}\] .
2. The dot product of the two vectors \[\overrightarrow a \] and \[\overrightarrow b \] represented as \[\overrightarrow a \overrightarrow {.b} \] is equal to \[\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \times \cos \theta \] .
3. The magnitude of cross product of the two vectors \[\overrightarrow a \] and \[\overrightarrow b \] represented as \[\left| {\overrightarrow a \times \overrightarrow b } \right|\] is equal to \[\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \times \sin \theta \] .
4. If \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\] , then, the magnitude of the vector \[\overrightarrow a \] represented as \[\left| {\overrightarrow a } \right|\] is equal to \[\sqrt {{a_1}^2 + {a_2}^2 + {a_3}^2} \] .

Complete step-by-step solution:
We have been given that \[\overrightarrow a \] and \[\overrightarrow b \] are two vectors, the projection of the vector \[\overrightarrow a \] on the vector \[\overrightarrow b \] is \[\left| {\overrightarrow a \times \overrightarrow b } \right|\] and \[3\overrightarrow b = \widehat i + \widehat j + \widehat k\] .
Let, \[\theta \] is the angle between the two vectors and \[p\] is the magnitude of the projection of the vector \[\overrightarrow a \] on the vector \[\overrightarrow b \]
Then, \[p = \dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow b } \right|}}\] [Using the formula]
But, it is given that the above projection is equal to \[\left| {\overrightarrow a \times \overrightarrow b } \right|\]
Establishing the equation, we have
\[\dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow b } \right|}} = \left| {\overrightarrow a \times \overrightarrow b } \right|\]
\[ \Rightarrow \dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow b } \right|}} = \left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \times \sin \theta \] [Since \[\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \times \sin \theta \] ]
\[ \Rightarrow \dfrac{{\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \times \cos \theta }}{{\left| {\overrightarrow b } \right|}} = \left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \times \sin \theta \] [Since \[\overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \times \cos \theta \] ]
\[ \Rightarrow \dfrac{{\cos \theta }}{{\left| {\overrightarrow b } \right|}} = \sin \theta \]
Further solving the above
\[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{1}{{\left| {\overrightarrow b } \right|}}\]
\[ \Rightarrow \tan \theta = \dfrac{1}{{\left| {\overrightarrow b } \right|}}\] ………………………………equation (1)
Now, we will calculate the magnitude of the vector \[\overrightarrow b \] as follows:
Given, \[3\overrightarrow b = \widehat i + \widehat j + \widehat k\]
\[ \Rightarrow \overrightarrow b = \dfrac{1}{{\sqrt 3 }}\widehat i + \dfrac{1}{{\sqrt 3 }}\widehat j + \dfrac{1}{{\sqrt 3 }}\widehat k\]
\[ \Rightarrow \left| {\overrightarrow b } \right| = \sqrt {{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}} \] [Using formula \[\left| {\overrightarrow a } \right| = \sqrt {{a_1}^2 + {a_2}^2 + {a_3}^2} \], if \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\] ]
\[ \Rightarrow \left| {\overrightarrow b } \right| = \sqrt {\dfrac{1}{9} + \dfrac{1}{9} + \dfrac{1}{9}} \]
Further, simplifying the above
\[\left| {\overrightarrow b } \right| = \sqrt {\dfrac{3}{9}} \]
\[ \Rightarrow \left| {\overrightarrow b } \right| = \dfrac{1}{{\sqrt 3 }}\]
Putting the value of \[\left| {\overrightarrow b } \right|\] in the equation (1)
\[\tan \theta = \dfrac{1}{{\dfrac{1}{{\sqrt 3 }}}}\]
\[ \Rightarrow \tan \theta = \sqrt 3 \]
\[ \Rightarrow \tan \theta = \tan \dfrac{\pi }{3}\]
\[ \Rightarrow \theta = \dfrac{\pi }{3}\]
Thus, the angle between \[\overrightarrow a \] and \[\overrightarrow b \] is equal to \[\dfrac{\pi }{3}\] .

Hence, option A. is the correct answer.

Note: The length of the projection of one vector on the other vector is obtained by dividing their dot product by the magnitude of the other vector. The cross product of two vectors is also a vector, while their dot product is a scalar quantity.