MCQ
If $\vec a = \hat i + \hat j + \hat k,\,\,\vec b = \hat i - \hat j + \hat k,\,\,\vec c = \hat i + 2\hat j + \hat k$ ,  then the value of $\left| {\begin{array}{*{20}{c}}
  {\vec a.\vec a}&{\vec a.\vec b}&{\vec a.\vec c} \\ 
  {\vec b.\vec a}&{\vec b.\vec b}&{\vec b.\vec c} \\ 
  {\vec c.\vec a}&{\vec c.\vec b}&{\vec c.\vec c} 
\end{array}} \right|$ is
  • A
    $2$
  • B
    $4$
  • $16$
  • D
    $64$

Answer

Correct option: C.
$16$
c
$[\bar{a} \bar{b} \bar{c}]^{2}=\left|\begin{array}{lll}{\bar{a} \cdot \bar{a}} & {\bar{a} \cdot \bar{b}} & {\bar{a} \cdot \bar{c}} \\ {\bar{b} \cdot \bar{a}} & {\bar{b} \cdot \bar{b}} & {\bar{b} \cdot \bar{c}} \\ {\bar{c} \cdot \bar{a}} & {\bar{c} \cdot \bar{b}} & {\bar{c} \cdot \bar{c}}\end{array}\right|$

and $[\bar{a} \bar{b} \bar{c}]=\left|\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {-1} & {1} \\ {1} & {2} & {-1}\end{array}\right|=4$

$\Rightarrow[\bar{a} \bar{b} \bar{c}]^{2}=16$

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