# Find closest possible points between lines? (vectors)? Edit

1. May 28, 2015

### mathlabrat

1. The problem statement, all variables and given/known data

Find points P,Q which are closest possible with P lying on the line

x=8+1t
y=8+1t
z=7−3t

and Q lying on the line
x=231−6t
y=−10−17t
z=71−13t

3. Attempt at solution

Hi,
I am at loss as to how to do this.

I know that from the equations I can get point (8,8,7) and direction vector (1,1,-3) for the first line and similar for the second line. Then I found the normal to both lines as the shortest distance points should form a perpendicular.

I don't know how to proceed from there.

Any help will be appreciated

Last edited by a moderator: May 28, 2015
2. May 28, 2015

### ehild

Welcome to PF!

If you have found the normal of both lines (what did you get?) you can write the equation of the plane defined by this normal and the first line e. The second line f have to intersect that plane at point Q. The normal drawn from this point intersects the first line at P.

Last edited: May 29, 2015
3. May 29, 2015

### Ray Vickson

If line L1 is $x_1 = 8+t, y_1 = 8 + t, z_1 = 7 - 3t$ and line L2 is $x_2 = 231 - 6s, y_2 = -10 - 17 s, z_2 = 71 - 13 s$ we can let
$$D(s,t) = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$$
be the distance between points $(x_1,y_2,z_1)$ on L1 and $(x_2,y_2,z_2)$ on L2. We can minimize $D(s,t)$ using standard calculus methods. Even easier, we can solve the equivalent problem of minimizing $E(s,t) = D(s,t)^2$, which does not have a $\sqrt{\;\;}$ in it.