Let a, b c $\in$ R be such that a² + b² + c² = 1. If
$$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$$,
where ${\theta = {\pi \over 9}}$, 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 :

Let, $\overrightarrow {{a_1}} = a\widehat i + b\widehat j + c\widehat k$<br><br>and $\overrightarrow {{a_2}} = b\widehat i + c\widehat j + a\widehat k$<br><br>We know, Angle between two vectors<br><br>$$\cos \alpha = {{\overrightarrow {{a_1}} \,.\,\overrightarrow {{a_2}} } \over {|\overrightarrow {{a_1}} \,|.|\,\overrightarrow {{a_2}} |}}$$<br><br>$$ = {{ab + bc + ac} \over {\sqrt {{a^2} + {b^2} + {c^2}} .\sqrt {{a^2} + {b^2} + {c^2}} }}$$<br><br>$= {{ab + bc + ac} \over {({a^2} + {b^2} + {c^2})}}$<br><br>Given, ${a^2} + {b^2} + {c^2} = 1$<br><br>$\therefore$ $\cos \alpha = ab + bc + ac$<br><br>$= ab\left( {{1 \over a} + {1 \over b} + {1 \over c}} \right)$ .....(1)<br><br>Given, $$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right) = \lambda $$ (Assume)<br><br>$\therefore$ ${1 \over a} = {{\cos \theta } \over \lambda }$<br><br>$${1 \over b} = {{\cos \left( {\theta + {{2\pi } \over 3}} \right)} \over \lambda }$$<br><br>$${1 \over c} = {{\cos \left( {\theta + {{4\pi } \over 3}} \right)} \over \lambda }$$<br><br>$\therefore$ $${1 \over a} + {1 \over b} + {1 \over c} = {1 \over \lambda }\left[ {\cos \theta + \cos \left( {\theta + {{2\pi } \over 3}} \right) + \cos \left( {\theta + {{4\pi } \over 3}} \right)} \right]$$<br><br>$$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos \left( {{{2\theta + 2\pi } \over 2}} \right)\cos \left( {{{{{2\pi } \over 3}} \over 2}} \right)} \right]$$<br><br>$$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos (\theta + \pi )\cos \left( {{\pi \over 3}} \right)} \right]$$<br><br>$$ = {1 \over \lambda }\left[ {\cos \theta + 2( - \cos \theta ) \times {1 \over 2}} \right]$$<br><br>$= {1 \over \lambda } \times 0$<br><br>$= 0$<br><br>Putting value of $\lambda$ in equation (1),<br><br>cos$\alpha$ = ab(0) = 0<br><br>$\Rightarrow$ $\alpha = {\pi \over 2}$