Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
Since you are saying this is homework, could you please add the [homework] tag? –  Arturo Magidin Jun 10 '11 at 17:06
    
I've changed it to 5/2 instead of -1/2. –  meiryo Jun 10 '11 at 17:11
3  
+1 for showing your working. –  Chris Taylor Jun 10 '11 at 17:24
1  
Normalize your normal. –  Ben Jackson Jun 10 '11 at 23:04

2 Answers 2

up vote 7 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 $\sqrt{6}$.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.