A) \[\frac{\sqrt{3}-1}{2\sqrt{2}}\]
B) \[\frac{\sqrt{3}+1}{2\sqrt{2}}\]
C) \[\frac{\sqrt{\sqrt{6}+2}}{2}\]
D) \[\frac{\sqrt{\sqrt{6}-2}}{2}\]
Correct Answer: D
Solution :
\[\because \,\,\left[ \vec{a}\,\vec{b}\,\vec{c} \right]=\left| \begin{matrix} \vec{a}.\vec{a} & \vec{a}.\vec{b} & \vec{a}.\vec{c} \\ \vec{b}.\vec{a} & \vec{b}.\vec{b} & \vec{b}.\vec{c} \\ \vec{c}.\vec{a} & \vec{c}.\vec{b} & \vec{c}.\vec{c} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{1}{2} & 1 & \frac{1}{\sqrt{2}} \\ \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{2}} & 1 \\ \end{matrix} \right|\] \[=\left( 1-\frac{1}{2} \right)-\frac{1}{2}\left( \frac{1}{2}-\frac{\sqrt{3}}{2\sqrt{2}} \right)+\frac{\sqrt{3}}{2}\left( \frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2} \right)\] \[=\frac{\sqrt{6}-2}{4}\] \[\therefore \,\,[\vec{a}\,\vec{b}\,\vec{c}]=\sqrt{\frac{\sqrt{6}-2}{4}}\].
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