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

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 $\left| {\bf{v}} \right| = 1$.