Thread: Distance from a point to a line in r^3

1. Distance from a point to a line in r^3

Hey guys. I am curious if I am doing this correctly. For some reason I am doubting my math here.

Question
Find the distance from the point (-3,0,1) to the line $\frac{x-2}{2} = \frac{y+2}{-2} = \frac{z}{-1}$

Solution
I know how to solve the problem when the line is in the form Ax + By + Cz + D, but I am unsure if I put it in that form correctly.

I first took the x and y and put both them equal to zero. So:
$-x - y - 3 = 0$

**Now, this is where I am unsure of myself. Can I set $-x - y - 3$ equal to $\frac {z}{-1}$?

If so, I get the equation: $x + y -z + 3 = 0$. Which I can run through the formula I have, and I get the distance of:
$\frac{-1\sqrt{3}}{3}$

Hopefully someone can clear this up for me! There is no example similar to this in my text. There is with something equal to something else, but not equal to two other things... if you know what I mean. Hard to explain.

Here $P(-3,0,1),~Q(2,-2,0)~\&~d=2i-2j-k$

3. Originally Posted by Plato

Here $P(-3,0,1),~Q(2,-2,0)~\&~d=2i-2j-k$
The answer I am looking for is whether I am doing the arithmetic to get to the form Ax + By + Cz + D correctly. Thank you for helping though!

4. Originally Posted by Kakariki
I know how to solve the problem when the line is in the form Ax + By + Cz + D, but I am unsure if I put it in that form correctly.
I deliberately did not address that statement.
Why? Because it is totally wrong. A line never has that form.
There are three forms for a line.

Vector $(a,b,c)+t(d_1,d_2,d_3)$.

Symmetric $\frac{x-a}{d_1}=\frac{y-b}{d_2} =\frac{z-c}{d_3};~~d_i\ne0$

Parametric $\ell (t) = \left\{ \begin{gathered}
x = a + td_1 \hfill \\
y = b + td_2 \hfill \\
z = c + td_3 \hfill \\
\end{gathered} \right.$