Question

Find a unit vector perpendicular to the plane determined by the points $P(1,-1,2),Q(2,0,-1)$ and $R(0,2,1)$.

Answer 1

Consider the problem

Given points are

$P\left(1,-1,2\right),Q\left(2,0,-1\right)$ and $R\left(0,2,1\right)$

Let,

$x_1=1,y_1=-1,z_1=2$

$x_2=2,y_2=0,z_2=-1$

$x_3=0,y_3=2,z_3=1$

The equation of plane passing through $P,Q$ and $R$ is given by

$\begin{array}{c}\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0\\ \\ \left|\begin{array}{ccc}x-1& y+1& z-2\\ 1& 1& -3\\ -1& 3& -1\end{array}\right|=0\end{array}$

On expanding along $R_1$

$(x-1)(-1+9)-(y+1)(-1-3)+(z-2)(3+1)=0$

$2x+y+z-3=0$ ---- $\left(1\right)$

Direction ratios of the normal to the plane $1$ are $2,1,1$

Therefore,

Vector perpendicular to the plane

$\overrightarrow{P}=2\hat{i}+\hat{j}+\hat{k}$

$|\overrightarrow{P}|=\sqrt{4+1+1}=\sqrt{6}$

Hence unit vector perpendicular to the plane

$\frac{\overrightarrow{P}}{|\overrightarrow{P}|}=\frac{2}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$