# Distance from a 3 space line

## Homework Statement

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

## 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??

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Dick
Homework Helper
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.

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

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
Homework Helper
Gold Member
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.