Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The position vectors of the points are $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$, $\vec{b}=2\hat{i}-\hat{k}$, and $\vec{c}=2\hat{j}+\hat{k}$.

The vectors in the plane are:
$\overrightarrow{AB}=\vec{b}-\vec{a}=(2-1)\hat{i}+(0-(-1))\hat{j}+(-1-2)\hat{k}=\hat{i}+\hat{j}-3\hat{k}$
$\overrightarrow{AC}=\vec{c}-\vec{a}=(0-1)\hat{i}+(2-(-1))\hat{j}+(1-2)\hat{k}=-\hat{i}+3\hat{j}-\hat{k}$

A vector perpendicular to the plane is given by the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$:
$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -3\\ -1& 3& -1\end{array}\right|$

$=\hat{i}(-1-(-9))-\hat{j}(-1-3)+\hat{k}(3-(-1))$
$=8\hat{i}+4\hat{j}+4\hat{k}$

Let $\vec{n}=8\hat{i}+4\hat{j}+4\hat{k}=4(2\hat{i}+\hat{j}+\hat{k})$.

The magnitude of this vector is:
$|\vec{n}|=4\sqrt{2^2+1^2+1^2}=4\sqrt{4+1+1}=4\sqrt{6}$

The unit vectors perpendicular to the plane are:
$\hat{n}=\pm \frac{\vec{n}}{|\vec{n}|}=\pm \frac{4(2\hat{i}+\hat{j}+\hat{k})}{4\sqrt{6}}=\pm \left(\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$

Answer: $\pm \left(\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$