Question

Find unit vector perpendicular to the plane passing through the points $(1,2,3),(2,-1,1)$ and $(1,2,-4).$

Answer 1

Let the points be $A(1,2,3),B(2,-1,1),C(1,2,-4)$

Let,

$x_1=1,y_1=2,z_1=3$

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

$x_3=1,y_3=2,z_3=-4$

Equation of plane passing through $A,B$ and $C$ is

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

$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 1& -3& -2\\ 0& 0& -7\end{array}\right|=0$

Expanding along $R_3$

$\Rightarrow -7\left[-3\left(x-1\right)-\left(y-2\right)\right]=0$

$\Rightarrow 3x+y-5=0$

Directional ratios of normal to the plane are $3,1,0$

Therefore, let a vector perpendicular to the plane is

$\overrightarrow{P}=3\hat{i}+\hat{j}+0\hat{k}$

$|\overrightarrow{P}|=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2+{\left(0\right)}^2}=\sqrt{10}$

Hence a unit vector perpendicular to the plane $=\frac{\overrightarrow{P}}{|\overrightarrow{P}|}=\frac{3\hat{i}+\hat{j}+0\hat{k}}{\sqrt{10}}$

$=\frac{3}{\sqrt{10}}\hat{i}+\frac{1}{\sqrt{10}}\hat{j}$