Question

If $\overline{a}\cdot \overline{a}=\overline{b}\cdot \overline{b}=\overline{c}\cdot \overline{c}=1,\overline{a}\cdot \overline{b}=\frac{1}{2},\overline{b}\cdot \overline{c}=\frac{1}{\sqrt{2}}\cdot \overrightarrow{c}\cdot \overline{a}=\frac{\sqrt{3}}{2}$ then, $\left[\begin{array}{ccc}\overline{a}& \overline{b}& \overline{c}\end{array}\right]$ 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 $\frac{\sqrt{\sqrt{6}-2}}{2}$

We have

$\overrightarrow{a}.\overrightarrow{a}=\overrightarrow{b}.\overrightarrow{b}=\overrightarrow{c}.\overrightarrow{c}=1$

$\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}\overrightarrow{b}.\overrightarrow{c}=\frac{1}{\sqrt{3}}\overrightarrow{c}.\overrightarrow{a}=\frac{\sqrt{3}}{2}$

${\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2=\left|\begin{array}{ccc}\overrightarrow{a}.\overrightarrow{a}& \overrightarrow{a}.\overrightarrow{b}& \overrightarrow{a}.\overrightarrow{c}\\ \overrightarrow{b}.\overrightarrow{a}& \overrightarrow{b}.\overrightarrow{b}& \overrightarrow{b}.\overrightarrow{c}\\ \overrightarrow{c}.\overrightarrow{a}& \overrightarrow{c}.\overrightarrow{b}& \overrightarrow{c}.\overrightarrow{c}\end{array}\right|$

$=\left|\begin{array}{ccc}1& \frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{1}{2}& 1& \frac{1}{\sqrt{2}}\\ \frac{\sqrt{3}}{2}& \frac{1}{\sqrt{2}}& 1\end{array}\right|$

$=1\left(1-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{\sqrt{3}}{2\sqrt{2}}\right)+\frac{\sqrt{3}}{2}\left(\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2}\right)$

$=-\frac{1}{2}+\frac{\sqrt{3}}{2\sqrt{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{2}}$

$=\frac{\sqrt{6}-2}{4}$

$\therefore \left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\frac{\sqrt{\sqrt{6}-2}}{2}$

Hence, the option $D$ is the correct answer.