TITLE: Finding shortest distance from point to plane
QUESTION [8 upvotes]: I need you guys to check my homework question out if I'm wrong or not...
Given point $(1,4,1)$ in need to find the shortest distance between this and the plane $2x_1 - x_2 + x_3 = 5$.
So firstly, I found the normal $n =  
\left( \begin{array}{c}
-2\\ 
1\\
-1
\end{array} \right)$
Then transformed the plane to parametric form and found a point on the plane:
$ 
\left( \begin{array}{c}
5/2\\ 
0\\
0
\end{array} \right)$
Found the vector, $v$:
$ 
\left( \begin{array}{c}
-3/2\\ 
4\\
1
\end{array} \right)$
Then I found the distance (dot product n and v all over n), I get a distance of 1. However the answer is actually square root of 6. Any idea where I went wrong?

REPLY [10 votes]: Sometimes its better not to think in terms of formulae.
Initially, the point is at $(1,4,1)$ and wants to reach the plane in a straight line that is the shortest path, the direction it must travel is given by the normal of the plane (draw a picture and this is obvious), which is $(2,-1,1)$, so we need $(1,4,1)+t(2,-1,1)$ to be in the plane. So we solve for $t$, which comes to $1$. Now, the point has to move $1.(2,-1,1)$. And that's why the distance is $\sqrt{6}$.

REPLY [5 votes]: Another way of doing this problem. Just apply the formula given at this link. By that formula we have $$D = \frac{|2 \times 1 - 4 \times 1 + 1 -5|}{\sqrt{4+1+1}} = \frac{6}{\sqrt{6}} = \sqrt{6}$$