Question

f \[\overrightarrow a \] and \[\overrightarrow b \] are unit vectors, then angle between \[\overrightarrow a \] and \[\overrightarrow b \] for \[\sqrt 3 \overrightarrow a - \overrightarrow b \] to be a unit vector is
(a) \[{\rm{60}}^\circ \]
(b) \[{\rm{90}}^\circ \]
(c) \[{\rm{45}}^\circ \]
(d) \[{\rm{30}}^\circ \]

Answer 1

Hint:
Here, we will assume the angle between the vectors \[\overrightarrow a \] and \[\overrightarrow b \] to be \[\theta \], such that \[\sqrt 3 \overrightarrow a - \overrightarrow b \] is a unit vector. We will square the equation for the magnitude of the third vector. Then, we will use the formula for angle \[\theta \] between two vectors non-zero vectors \[u\] and \[v\] to simplify the expression and find the required angle between \[\overrightarrow a \] and \[\overrightarrow b \],

Formula used:
We will use the formula of the angle \[\theta \] between two vectors non-zero vectors \[u\] and \[v\] is given by \[\cos \theta = \dfrac{{u \cdot v}}{{\left| u \right| \times \left| v \right|}}\].

Complete step by step solution:
It is given that the vectors \[\overrightarrow a \] and \[\overrightarrow b \] are unit vectors.
Therefore, we get the magnitude of the two vectors as \[\left| {\overrightarrow a } \right| = 1\] and \[\left| {\overrightarrow b } \right| = 1\].
Let the angle between the vectors \[\overrightarrow a \] and \[\overrightarrow b \] be \[\theta \], such that \[\sqrt 3 \overrightarrow a - \overrightarrow b \] is a unit vector.
Therefore, the angle between the two vectors \[\overrightarrow a \] and \[\overrightarrow b \] is given by
\[ \Rightarrow \cos \theta = \dfrac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right|}}\]
Substituting \[\left| {\overrightarrow a } \right| = 1\] and \[\left| {\overrightarrow b } \right| = 1\] in the formula, we get
\[ \Rightarrow \cos \theta = \dfrac{{\overrightarrow a \cdot \overrightarrow b }}{{1 \times 1}}\]
Multiplying the terms in the denominator, we get
\[ \Rightarrow \cos \theta = \dfrac{{\overrightarrow a \cdot \overrightarrow b }}{1}\]
Therefore, we get
\[ \Rightarrow \cos \theta = \overrightarrow a \cdot \overrightarrow b \]
Now, the vector \[\sqrt 3 \overrightarrow a - \overrightarrow b \] is a unit vector.
Therefore, we get
\[\left| {\sqrt 3 \overrightarrow a - \overrightarrow b } \right| = 1\]
Taking the squares of both sides of the equation, we get
\[{\left( {\left| {\sqrt 3 \overrightarrow a - \overrightarrow b } \right|} \right)^2} = {1^2}\]
The square of the expression \[\left| {\sqrt 3 \overrightarrow a - \overrightarrow b } \right|\] can be written as \[{\left( {\sqrt 3 } \right)^2}{\left( {\left| {\overrightarrow a } \right|} \right)^2} + {\left( {\left| {\overrightarrow b } \right|} \right)^2} - 2\sqrt 3 \left( {\overrightarrow a \cdot \overrightarrow b } \right)\].
Therefore, we get
\[ \Rightarrow {\left( {\sqrt 3 } \right)^2}{\left( {\left| {\overrightarrow a } \right|} \right)^2} + {\left( {\left| {\overrightarrow b } \right|} \right)^2} - 2\sqrt 3 \left( {\overrightarrow a \cdot \overrightarrow b } \right) = 1\]
Substituting \[\left| {\overrightarrow a } \right| = 1\] and \[\left| {\overrightarrow b } \right| = 1\] in the equation, we get
\[ \Rightarrow {\left( {\sqrt 3 } \right)^2}{\left( 1 \right)^2} + {\left( 1 \right)^2} - 2\sqrt 3 \left( {\overrightarrow a \cdot \overrightarrow b } \right) = 1\]
Applying the exponents on the bases, we get
\[ \Rightarrow 3\left( 1 \right) + 1 - 2\sqrt 3 \left( {\overrightarrow a \cdot \overrightarrow b } \right) = 1\]
Multiplying the terms in the expression, we get
\[ \Rightarrow 3 + 1 - 2\sqrt 3 \left( {\overrightarrow a \cdot \overrightarrow b } \right) = 1\]
Simplifying the expression, we get
\[ \Rightarrow 3 - 2\sqrt 3 \left( {\overrightarrow a \cdot \overrightarrow b } \right) = 0\]
Substituting \[\overrightarrow a \cdot \overrightarrow b = \cos \theta \] in the equation, we get
\[ \Rightarrow 3 - 2\sqrt 3 \cos \theta = 0\]
Rewriting the equation by rearranging the terms, we get
\[ \Rightarrow 2\sqrt 3 \cos \theta = 3\]
Dividing both sides of the equation by \[2\sqrt 3 \], we get
\[ \Rightarrow \cos \theta = \dfrac{3}{{2\sqrt 3 }}\]
Therefore, we get
\[\begin{array}{l} \Rightarrow \cos \theta = \dfrac{{\sqrt 3 \times \sqrt 3 }}{{2\sqrt 3 }}\\ \Rightarrow \cos \theta = \dfrac{{\sqrt 3 }}{2}\end{array}\]
We know that the cosine of the angle measuring \[30^\circ \] is \[\dfrac{{\sqrt 3 }}{2}\].
Substituting \[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\] in the equation, we get
\[ \Rightarrow \cos \theta = \cos 30^\circ \]
Therefore, we get
\[ \Rightarrow \theta = 30^\circ \]
Therefore, we get the angle between the vectors \[\overrightarrow a \] and \[\overrightarrow b \], such that \[\sqrt 3 \overrightarrow a - \overrightarrow b \] is a unit vector as \[30^\circ \].

Thus, the correct option is option (d).

Note:
We used the term ‘unit vector’ and ‘magnitude’ in the solution. The magnitude of a vector is the length of the vector. The magnitude of a vector \[{\bf{v}} = \overline {x{\bf{i}}} + \overline {y{\bf{j}}} + \overline {z{\bf{k}}} \] is given by the formula \[\left| {\bf{v}} \right| = \sqrt {{x^2} + {y^2} + {z^2}} \]. A unit vector is a vector whose magnitude is equal to 1. If a vector \[{\bf{v}} = \overline {x{\bf{i}}} + \overline {y{\bf{j}}} + \overline {z{\bf{k}}} \] is a unit vector, then \[\left| {\bf{v}} \right| = 1\].