# Probably quite simple but i'm stuck

1. May 5, 2005

### .....

I'm stuck on this question... if someone could give me a hand that'd be great..

What is the minimum distance between the point (3,-2,4) and the line defined by

x = 1 + t
y = 4 - 3t
z = -2 + 2t

My approach was to find a point on the line by letting t = 1

I got (2, 1, 0)

then I found a vector between this point and the one given:

v = i -3j + 4k

Here's a picture to clarify:

Then the distance would be the scalar projection of this vector onto the normal vecor of the line

But i'm not sure how to find the normal to the line.

Could someone show me how?

And if there's a simpler approach could you show me that too?

Thanks.

2. May 5, 2005

### dextercioby

Geometrically,can u see what is the minimum distance...?Come on,it's something i learnt when i was 12...

Daniel.

3. May 5, 2005

### whozum

A single line in R^3 doesn't have a single normal vector, since it lies on an infinite number of planes.

The way I would try this to find two points in cartesian coordinates and find the line equation in those coordinates. Then using the distance formula and some max/min calculus, minimize the distance function.

4. May 5, 2005

### JonF

here is what i would try

$$L = \sqrt{(x_0-x_1)^2 +(y_0-y_1)^2 (z_0-z_1)^2$$
$$L = \sqrt{(1+t-3)^2 +(4-3t+2)^2 (-2+2t-4)^2}$$
$$= \sqrt{(t-2)^2 +(-3t+6)^2 (2t-6)^2}$$
$$= |t-2|\sqrt{36t^2-216t+325}$$

Then take the derivative in respect to t. Find the critical points, and then pick the minimum one.

5. May 5, 2005

### mansi

why not do something simpler? all you need to do is to find the point Q on the line where the normal vector passing through (3,-2,4) cuts it...that point looks like (1+t,4-3t,-2+2t) for some t. that is the t you need to find...and to do that use the fact that PQ is normal to the given line...

6. May 5, 2005

### .....

^^that works, i got it.

thanks a lot mansi.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook