MCQ
Let $\vec a,\vec b,\vec c$ are unit vectors and $\vec a.\vec b = \vec b.\vec c = \vec c.\vec a = \cos \theta $ then maximum value of $\theta $ is where $\theta  \in \left[ {0,\pi } \right]$
  • A
    $\frac{\pi }{3}$
  • $\frac{2\pi }{3}$
  • C
    $\frac{5\pi }{6}$
  • D
    $\frac{\pi }{6}$

Answer

Correct option: B.
$\frac{2\pi }{3}$
b
${\left[ {\begin{array}{*{20}{c}}
{\overrightarrow {\rm{a}} }&{\overrightarrow {\rm{b}} }&{\overrightarrow {\rm{c}} }
\end{array}} \right]^2} = \left| {\begin{array}{*{20}{l}}
{\widehat {\rm{a}} \cdot \widehat a}&{\widehat a \cdot \widehat b}&{\widehat a \cdot \widehat c}\\
{\widehat b \cdot \widehat a}&{\widehat b \cdot \widehat b}&{\widehat {\rm{b}} \cdot \widehat c}\\
{\widehat c \cdot \widehat a}&{\widehat c \cdot \widehat b}&{\widehat c \cdot \widehat c}
\end{array}} \right|$

${\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right]^2} = \left| {\begin{array}{*{20}{c}}
1&{\cos \theta }&{\cos \theta }\\
{\cos \theta }&1&{\cos \theta }\\
{\cos \theta }&{\cos \theta }&1
\end{array}} \right|$

${\left[ {\begin{array}{*{20}{l}}
{\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c }
\end{array}} \right]^2} = {(1 - \cos \theta )^2}(1 + 2\cos \theta ) \ge 0$

${\cos \theta \geq-\frac{1}{2}} $

${\theta=\frac{2 \pi}{3}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free