We have two responses for you

Robin,

If the lines are coplanar, they either intersect (in a single point), or are the same line (colinear) or are parallel (no intersection). If you know they intersect (perhaps from the context of the question), you can immediately look for the single point.

At this point, all the equations must work so:

(x-5)/4=(y-7)/4 AND
(y-7)/4=-(z+3)/5 AND
(x-8)/7=y-4 AND
y-4=(z-5)/3

You have three unknowns, but you have four equations here, so you can solve this the same way as you would with three variables and two equations: use the substitution and/or the elimination method.

I would:

1. Identify the two equations that don't have z in them.
   
   
2. Solve each for x.
   
   
3. Make the two expressions that equal x, equal each other. Solve for y.
   
   
4. Check this triple (x, y, z) in both lines to make sure I didn't make an arithmetic mistake.

Cheers,
Stephen La Rocque.

Hi Robin,

Look at the East wall of the room you are in. The line where the East wall meets the ceiling (line 1) and the line where the East wall meets the floor (line 2) are parallel lines and they are coplanar. The East wall is the plane that contains them both.

Now look at the vertical line where the North wall meets the East wall (line 3). Line 1 and line 3 meet at the upper North East corner of the room and again these lines are coplanar. The East wall is the plane that contains them both.

Finally look at the line where the North wall meets the floor (line 4). There is no point that is on lines 1 and 4 and these lines are not coplanar.

The room you are in is special because the lines I have identified are at right angles to each other. You can however imagine a room where the corners are not square and the ceiling and floor are not parallel and the situation is similar. Two lines in 3-space are either parallel (and are hence coplanar), meet at a point (and are hence coplanar) or are not parallel and do not meet (and are hence not coplanar).

I hope this helps,
Penny