Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three vectors such that $\overline{a}.\overline{a}=\overline{b}.\overline{b}=\overline{c}.\overline{c}=1;\overline{a}.\overline{b}=\frac{1}{2};\overline{b}.\overline{c}=\frac{1}{\sqrt{2}};\overline{c}.\overline{a}=\frac{\sqrt{3}}{2}then\left[\overline{a}\overline{b}\overline{c}\right]$ is

• A. $\frac{\sqrt{3}-1}{2\sqrt{2}}$
  
• B. $\frac{\sqrt{3}+1}{2\sqrt{2}}$
  
• C. $\frac{\sqrt{\sqrt{6}+2}}{2}$
  
• D. $\frac{\sqrt{\sqrt{6}-2}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{\sqrt{6}-2}}{2}$

${\left[\overline{a}\overline{b}\overline{c}\right]}^2=\left|\begin{array}{ccc}\overline{a}.\overline{a}& \overline{a}.\overline{b}& \overline{a}.\overline{c}\\ \overline{b}.\overline{a}& \overline{b}.\overline{b}& \overline{b}.\overline{c}\\ \overline{c}.\overline{a}& \overline{c}.\overline{b}& \overline{c}.\overline{c}\end{array}\right|$
=$\left|\begin{array}{ccc}1& \frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{1}{2}& 1& \frac{1}{\sqrt{2}}\\ \frac{\sqrt{3}}{2}& \frac{1}{\sqrt{2}}& 1\end{array}\right|$
$=1(1-\frac{1}{2})-\frac{1}{2}(\frac{1}{2}-\frac{\sqrt{3}}{2\sqrt{2}})+\frac{\sqrt{3}}{2}(\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2})=\frac{1}{2}-\frac{1}{4}+\frac{\sqrt{3}}{4\sqrt{2}}+\frac{\sqrt{3}}{4\sqrt{2}}-\frac{3}{4}=\frac{2\sqrt{3}}{4\sqrt{2}}-\frac{1}{2}=\frac{2\sqrt{3}-2\sqrt{2}}{4\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{2}}$