A unit vector perpendicular to the plane determined by the points P(1,-1,2), Q

Question -

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

Options:

A .

\frac{2^i +^j +^k}{\sqrt{6}}

B .

\frac{2^i +^j +^k}{3}

C .

\frac{2^i -^j -^k}{\sqrt{3}}

D .

\frac{2^i -^j -^k}{3}

Answer: Option A
:
A

- - \to O P =^i -^j + 2^k, - - \to O Q = 2^i -^k, - - \to O R = 2^j +^k \Rightarrow - - \to P Q = - - \to O Q - - - \to O P =^i +^j - 3^k, - - \to P R = - - \to O R - - - \to O P = -^i + 3^j -^k

- - \to P Q \times - - \to P R = ∣ ∣∣∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 1 & - 3 - 1 & 3 & - 1 \end{matrix} ∣ ∣∣∣ ∣ = 8^i + 4^j + 4^k; | - - \to P Q \times - - \to P R | = \sqrt{64 + 16 + 16} = \sqrt{96} = 4 \sqrt{6}

Required unit vectors =

\pm \frac{8^i + 4^j + 4^k}{4 \sqrt{6}} = \pm \frac{2^i +^j +^k}{\sqrt{6}}

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