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?
Best Answer
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 $\sqrt{6}$.