Question

If $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1$ and $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=\cos \theta$, then the maximum value of $\theta$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is C $\frac{2\pi }{3}$

${[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]}^2$
$=\left|\begin{array}{ccc}\overrightarrow{a}.\overrightarrow{a}& \overrightarrow{a}.\overrightarrow{b}& \overrightarrow{a}.\overrightarrow{c}\\ \overrightarrow{b}.\overrightarrow{a}& \overrightarrow{b}.\overrightarrow{b}& \overrightarrow{b}.\overrightarrow{c}\\ \overrightarrow{c}.\overrightarrow{a}& \overrightarrow{c}.\overrightarrow{b}& \overrightarrow{c}.\overrightarrow{c}\end{array}\right|$
$=\left[\begin{array}{ccc}1& \cos \theta & \cos \theta \\ \cos \theta & 1& \cos \theta \\ \cos \theta & \cos \theta & 1\end{array}\right]$
$=1-{\cos }^2\theta -\cos \theta (\cos \theta -{\cos }^2\theta )+\cos \theta ({\cos }^2\theta -\cos \theta )$
$=1-3{\cos }^2\theta +2{\cos }^3\theta$
$=(1-\cos \theta )(1+\cos \theta -2{\cos }^2\theta )$ on simplification
$={(1-\cos \theta )}^2(1+2\cos \theta )\ge 0$
$\Rightarrow 1+2\cos \theta \ge 0$
$\Rightarrow \cos \theta \ge \frac{-1}{2},\theta \in [0,\frac{2\pi }{3}]$