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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?

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I've changed it to 5/2 instead of -1/2. – meiryo Jun 10 '11 at 17:11
+1 for showing your working. – Chris Taylor Jun 10 '11 at 17:24
Normalize your normal. – Ben Jackson Jun 10 '11 at 23:04
up vote 8 down vote accepted

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 $\sqrt{6}$.

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

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