Question

Let $A,B,C$ be points with position vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}+2\hat{j}+\hat{k}$ and $3\hat{i}+\hat{j}+2\hat{k}$ respectively. Find the shortest distance between point $B$ and plane $OAC$.

Answer 1

Given points are,

$A(2,-1,1),B(1,2,1),C(3,1,2)$

Let the given points be

$x_1=0,y_1=0,z_1=0x_2=2,y_2=-1,z_2=1x_3=3,y_3=1,z_3=2$

Then ,

equation of plane passing through $O,A,C$ is

$\begin{array}{c}\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0\\ \\ \left|\begin{array}{ccc}x& y& z\\ 2& -1& 1\\ 3& 1& 2\end{array}\right|=0\\ \\ x\left(-2-1\right)-y\left(4-3\right)+z\left(2+3\right)=0\\ \\ 3x+y-5z=0\end{array}$

Now,

Shortest distance between point $B(1,2,1)$ and plane $OAC$ is

$d=\left|\frac{3\times 1+2\times 1-5\times 1}{\sqrt{3^2+1^2+(-5{)}^2}}\right|=0$

So, the point B is at no distance from the plane and hence it lies on the plane.