A unit vector perpendicular to the plane determined by the points… - Vidyadip

MCQ

A unit vector perpendicular to the plane determined by the points $ (1, -1, 2), (2, 0, -1) $ and $ (0, 2, 1) $ is

✓
$ \pm \,\frac{1}{{\sqrt 6 }}\,(2i + j + k)$
B
$\frac{1}{{\sqrt 6 }}\,(i + 2j + k)$
C
$\frac{1}{{\sqrt 6 }}\,(i + j + k)$
D
$\frac{1}{{\sqrt 6 }}\,(2i - j - k)$

✓

Answer

Correct option: A.

$ \pm \,\frac{1}{{\sqrt 6 }}\,(2i + j + k)$

a
(a) $a = i + j - 3k,$ $b = - 2i + 2j + 2k$

$a \times b = \left| {\begin{array}{*{20}{c}}i&j&k\\1&1&{ - 3}\\{ - 2}&2&2\end{array}} \right| = 8i + 4j + 4k$

Hence unit vector $ = \pm \frac{{2i + j + k}}{{\sqrt 6 }}.$

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