Question

If $\overline{a},\overline{b},\overline{c}$ are three vectors such that $|\overline{a}|=1,|b|=2,|c|=3$ and $\overline{a}\overline{b}=\overline{b}\overline{c}=\overline{c}\cdot \overline{a}=0$ . Then $\left[\begin{array}{ccc}\overline{a}& \overline{b}& \overrightarrow{c}\end{array}\right]=$

• A. $6$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $6$

We have given from the question

$\left|\overrightarrow{a}\right|=1\left|\overrightarrow{b}\right|=1\left|\overrightarrow{c}\right|=1$

So, $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$

Now, by using determinants method

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

$=\left|\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 9\end{array}\right|=1\left(4\times 9\right)=36$

$\Rightarrow {\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2=36$

$\therefore \left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=6$

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