Distance from a 3-Space Line to a Point

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Homework Statement

Find the distance between the line x=3t-1, y=2-t, z=t and the point (2,0,-5)

Homework Equations

The Attempt at a Solution

So I'm assuming I need to use the formula D=[ax+by+cz+d]/root(a^2+b^2+c^2) and I'm guessing I need to use (3,-1,1), extracted from the parametric line equation as the parallel vector, and possibly also the point (-1,2,0), extracted from the same. But not sure how to proceed??

 

[Dick]

Minimize the distance from (x,y,z) to (3,-1,1). The distance D just a function of t. Use calculus. Hint: you'll find it's nicer to minimize D^2.

 

[SrEstroncio]

There is a way to do this using vector projection, does anyone remember it?

 

[hbweb500]

I think you are on the right track. What I would do, instead of minimizing the distance, is to take advantage of the geometry.

Draw a vector \vec{p} from the origin to the point. Draw another vector from the origin to the parameterized vector function: \vec{r}(t) = (3t-1, 2-t, t).

Now, let \vec{d} be the vector from the point to the place on the line at time t. In other words, \vec{p} + \vec{d} = \vec{r}(t).

It might help to draw a picture. Once you do, you will notice that the minimum distance will occur when \vec{d} is perpindicular to \vec{r}(t) and hence also orthogonal to \vec{r}'(t).

So all you need to do is use what you know about the dot product and orthogonality to solve for when d and r'(t) are orthogonal, and you can find the distance from there.

 

[LCKurtz]

Call the point on the line P, the point off the line Q, and a direction vector for the line D. Draw the perpendicular from Q to the line hitting the line at S. Let

\vec V = PQ,\ \hat D = \frac {\vec D}{|\vec D|}\hbox{ and }\theta\hbox{ be the angle between }\vec V \hbox{ and }\hat D.

From the right triangle PQS you have formed you can see the distance from the point to the line is:

d = |\vec V| \sin\theta

But this is the same as the magnitude of the cross product:

d = |\vec V \times \hat D|

That is the most direct way to get the distance because it is easy to write down

\vec V \hbox{ and }\hat D

and take the cross product and its length.