MCQ
If $A\,( - 1,\,\,2,\,\,3),\,\,B\,(1,\,\,1,\,\,1)$ and $C\,(2,\,\, - 1,\,\,3)$ are points on a plane. A unit normal vector to the plane  $ABC$ is
  • $ \pm \,\left( {\frac{{2i + 2j + k}}{3}} \right)$
  • B
    $ \pm \,\left( {\frac{{2i - 2j + k}}{3}} \right)$
  • C
    $ \pm \,\left( {\frac{{2i - 2j - k}}{3}} \right)$
  • D
    $ - \,\left( {\frac{{2i + 2j + k}}{3}} \right)$

Answer

Correct option: A.
$ \pm \,\left( {\frac{{2i + 2j + k}}{3}} \right)$
a
(a) $\overrightarrow {AB} = 2i - j - 2k,$ $\overrightarrow {AC} = 3i - 3j + 0k$

$\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}i&j&k\\2&{ - 1}&{ - 2}\\3&{ - 3}&0\end{array}} \right| = ( - 6i - 6j - 3k)$

Hence unit vector $ = \pm \left( {\frac{{2i + 2j + k}}{3}} \right)\,.$

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