Question

Let $a,\;b,\;c \in \mathbb{R}$be such that ${a^2} + {b^2} + {c^2} = 1$. If $a\;\cos \theta = b\;\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right) = c\;\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)$, where $\theta = \dfrac{\pi }{9},$ then the angle between the vectors $a\widehat i + b\widehat j + c\widehat k$ and $b\widehat i + c\widehat j + a\widehat k$ is:
(A) $\dfrac{\pi }{2}$
(B) $\dfrac{{2\pi }}{3}$
(C) $\dfrac{\pi }{9}$
(D) $0$

Answer 1

Hint: In order to find out the angle between two vectors we will use the dot product method. First, we will find the dot product of two vectors $\overrightarrow u $and $\overrightarrow v $. Then, we will find the magnitude of $\overrightarrow u $and $\overrightarrow v $. Next, we will use the cosine formula used in finding the angle between two vectors. Substitute the dot product and magnitude obtained above to get the final answer.

Formula Used:
$\cos \theta = \dfrac{{\overrightarrow u .\overrightarrow v }}{{\left| {\overrightarrow u } \right|\left| {\overrightarrow v } \right|}}$
$\left[ {\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)} \right]$
$\left[ {\cos \left( {\theta + \pi } \right) = - \cos \theta } \right]$

Complete step by step Solution:
We are given that,
$\overrightarrow u = a\widehat i + b\widehat j + c\widehat k$ ………………equation $(1)$
$\overrightarrow v = b\widehat i + c\widehat j + a\widehat k$ ………………equation $(2)$
Let us find the dot product of $\overrightarrow u $ and $\overrightarrow v $
$\overrightarrow u .\overrightarrow v = (ab) + (bc) + (ca)$ ………………equation $(3)$
$[\because $Multiply the $\widehat i,\widehat j,\widehat k$coefficients of both $\overrightarrow u $ and $\overrightarrow v $ separately$]$
Now, we will find the magnitude of both vectors.
$\left| {\overrightarrow u } \right| = \sqrt {{a^2} + {b^2} + {c^2}} $ ………………equation $(4)$
$\left| {\overrightarrow v } \right| = \sqrt {{a^2} + {b^2} + {c^2}} $ ………………equation $(5)$
Using the cosine formula to find the angle between two vectors using the dot product method,
$\cos \theta = \dfrac{{\overrightarrow u .\overrightarrow v }}{{\left| {\overrightarrow u } \right|\left| {\overrightarrow v } \right|}}$
Substituting the equations $(3)$, $(4)$ and $(5)$ in the above formula
\[\cos \theta = \dfrac{{(ab) + (bc) + (ca)}}{{\left( {\sqrt {{a^2} + {b^2} + {c^2}} } \right)\left( {\sqrt {{a^2} + {b^2} + {c^2}} } \right)}}\]
\[\cos \theta = \dfrac{{(ab) + (bc) + (ca)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}}\]
We are given that \[{a^2} + {b^2} + {c^2} = 1,\] substituting it in above equation
\[\cos \theta = \dfrac{{(ab) + (bc) + (ca)}}{1}\]
Multiply the numerator and denominator by $abc,$
\[\cos \theta = abc\left( {\dfrac{{ab}}{{abc}} + \dfrac{{bc}}{{abc}} + \dfrac{{ca}}{{abc}}} \right)\]
Simplifying it,
\[\cos \theta = abc\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)\] ………………equation $(6)$
Given,
$a\;\cos \theta = b\;\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right) = c\;\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right) = \lambda $(say)
We will use this equation to find the value of $a,b,c$
$a = \dfrac{{\cos \theta }}{\lambda }$
$b = \dfrac{{\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right)}}{\lambda }$
$c = \dfrac{{\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)}}{\lambda }$
Substituting the value of $a,b,c$ in equation $(6)$
$\cos \theta = abc\left[ {\dfrac{{\cos \theta }}{\lambda } + \dfrac{{\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right)}}{\lambda } + \dfrac{{\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)}}{\lambda }} \right]$
Applying trigonometric identity to simplify the equation,
$\cos \theta = \dfrac{{abc}}{\lambda }\left[ {\cos \theta + 2\cos \left( {\theta + \pi } \right)\cos \left( {\dfrac{\pi }{3}} \right)} \right]$ $\left[ {\because \cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)} \right]$
$\cos \theta = \dfrac{{abc}}{2}\left[ {\cos \theta + 2( - \cos \theta )\left( {\dfrac{1}{2}} \right)} \right]$ $\left[ {\because \cos \left( {\theta + \pi } \right) = - \cos \theta } \right],$ $\left[ {\because \cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}} \right]$
Solving it further,
$\cos \theta = \dfrac{{abc}}{2}\left[ {\cos \theta - \cos \theta } \right]$
\[\cos \theta = 0\]
Now, at last, finding $\theta $ to get the correct answer,
$\cos \theta = \cos \left( {\dfrac{\pi }{2}} \right)$ $\left[ {\because \cos \left( {\dfrac{\pi }{2}} \right) = 0} \right]$
On comparing the angles of L.H.S and R.H.S,
$\theta = \dfrac{\pi }{2}$

Hence, the correct option is A.

Note: Here, we have used the dot product using the cosine formula to find the angle between two vectors which is also known as the scalar product of the two vectors. Also, for any two vectors $a$ and $b$, if $a.b$is positive, then the angle lies between ${0^ \circ }$ and ${90^ \circ };$ if $a.b$ is negative, then the angle lies between ${90^ \circ }$ and ${180^ \circ }.$