The position vectors of the points A, B, C are respectively $(1, 1, 1); (1, - 1, 2); (0, 2, - 1)$ . Find a unit vector parallel to the plane determined by ABC and perpendicular to the vector $(1, 0, 1)$

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Solution

Equation of plane is
$∣ ∣ ∣ ∣ \begin{matrix} x - 1 & y - 1 & z - 1 0 & - 2 & 1 - 1 & 1 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow 3 (x - 1) + y - 1 - 2 (z - 1) = 0$
$\Rightarrow 3 x + y - 2 z - 2 = 0$
let $\to a = a_{1}^i + a_{2}^j + a_{3}^k$
$3 a_{1} + a_{2} - 2 a_{3} = 0$
$a_{1} + 0. a_{2} + a_{3} = 0$
$\frac{a_{1}}{1} = \frac{- a_{2}}{3 + 2} = \frac{a_{3}}{- 1}$
$\Rightarrow \frac{a_{1}}{1} = \frac{a_{2}}{- 5} = \frac{a_{3}}{- 1} = K$
$a_{1} = K, a_{2} = - 5 K, a_{3} = - K$
$a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 1$
$\Rightarrow 27 K^{2} = 1$
$\Rightarrow K = \pm \frac{1}{3 \sqrt{3}}$
$∴ \to a = \frac{1}{3 \sqrt{3}}^i \frac{- 5}{3 \sqrt{3}}^j \frac{- 1}{3 \sqrt{3}}^k$
or
$= - \frac{1}{3 \sqrt{3}}^i + \frac{5}{3 \sqrt{3}}^j + \frac{1}{3 \sqrt{3}}^k$

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