Given 3 non parallel lines in space, is there always a 4th one that can be passed thru them all?

Question

Stopped at a light, I noticed 3 lines on various poles that , to my eye, crossed at one point. If I were to back away from my observation point, or move towards it, I would suspect that the projected image seen to me would still see the lines cross. (Please inform me if you think not). So, that line representing me backing away passes through all three lines. The question is, is there always such a 4th line?

Answer 1

Yes. There is always a fourth line that can pass through the three nonparallel lines. I'm sure this could be proven using equations, but I am using a purely conceptual argument.

Brief explanation:

If you can stand at a point on one line as see the other two lines cross anywhere in the distance, then you can draw a fourth line through all three. The only instance where two lines in 3-D space could appear to parallel is if you could approximate a perfect projection of that 3-D space into 2-D. In other words, they look parallel from one angle but not others. This can never happen since your vantage point is always a point. You will always be at an angle relative to that perfect projection.

Long explanation:

Recall first of all, that in two dimensions, two lines that are not parallel must intersect, and vice versa. Now imagine that you are able to place yourself within the three-dimensional world where these three lines exist and look at them from any vantage point you like. Remember, that when you look at something from a distance, you see a two-dimensional picture. We can also think about projecting the three-dimensional space onto two dimensions from various vantage points. For example, using cartesian coordinates (x, y, and z axes), we can imagine making a projection which would see a plane containing the Y and Z axes as a line. One way to think of this is as if you were looking at the cartesian world from a distance and you were sitting on the Z axis. You might see the X axis as a horizontal line, and the Y axis as a vertical line. You would not be able to tell that the plane in front of you was a plane because you would be IN the plane.

Here is the key concept that will be used to prove the point. If you are able to place yourself at any point on any line and see the other two lines intersect, then by definition, there is a line that will intersect all three, and it will pass through the intersection points you view in the other two lines, as well as the point at which you are located.

Using this rule try to imagine any configuration that would allow you to see the other two lines as not intersecting from EVERY point on the third line and repeating this test for all three lines. I will argue, this is impossible.

We are left considering only the case where from the vantage of a a point on one line, the other two lines appear parallel. Not strictly possible, but let’s leave that aside and consider projections first. We know that we can define a 2-D projection in which two nonparallel lines appear parallel. Let’s call these lines 1 and 2. Let’s also say that in a projection in which the plane containing both the Y and Z axes appears as a line, that lines 1 and 2 are parallel not only to each other, but also to the Y axis. For simplicity, let’s say that in this projection, line 1 is coincident with the Y axis, and line 2 intersects the X axis at a point we will call +1.

Let’s review. In our projection, we see two vertical lines, one of which coincides with the vertical Y axis, and the horizontal X axis.

Since we know these lines are not parallel in 3-D, then we know that any projection that does not see the Y-Z plane as a line will also not show lines 1 and 2 as parallel, but rather as intersecting. Since we defined our lines as both intersecting the X axis, then if we look at a projection along the X axis, orthogonal to our previous projection, and in which the X-Y plane appears as a line, then we will see lines 1 and 2 intersecting at the intersection of the Y and Z axes.

Now can any third line be drawn in such a way that from any point on that line the other two lines appear parallel. I think to some this answer is intuitively clear, but let’s try to flesh it out. First, is there any vantage point from which lines 1 and 2 look parallel? We could argue that from a distance on a third line we might approximate what it looks like to project the other two lines onto 2-D, and that as for the projection, we will see them as parallel only if we also see the Y-Z plane as parallel. Therefore also, if any part of line 3 extends far out of the Y-Z plane, then from those parts of that line, the Y-Z plane will not appear as a line, but as a plane, and lines 1 and 2 will not appear parallel.

Therefore, the only remaining possibility is a line that does not extend far beyond the Y-Z plane. The only ones that fit this criterion (since lines are infinitely long) are ones IN the Y-Z plane or in a plane parallel to it. If it were in the Y-Z plane, then it would by definition have to intersect in space with line 1, since they cannot be parallel. If they intersected, then we could easily draw a fouth line through that intersection point and any point on line 2. Therefore line 3 must be in a plane parallel to the Y-Z plane.

Now we have all three lines in planes that are parallel to each other, and our original projection (which sees the Y-Z plane as a line) now sees three parallel lines. Based on our earlier argument, though, all three lines are seen as intersecting in any projection that does not see the Y-Z plane as a line.

Using one of the lines in one of the outer two planes, ask yourself if you can look back at the other two lines and ever see them as parallel. The answer is no, because the only time you can ever see them as parallel is in a perfect projection which sees all three planes as lines. Since you cannot be “in” all three planes, you will necessarily see the other two as planes, and not as lines. Therefore, you cannot possibly see the other two lines as parallel. This also means there is point on this line from which you cannot see an intersection of the other two lines and there will actually be an infinite number of ways to draw line 4 through the other three.

Answer 2

There is always a fourth line - in fact, there are an infinite number of ways to choose a fourth line that intersects all three skew lines. (Two straight lines are said to be "skew" if they are neither parallel nor intersecting. It doesn't mean they're curved.)

Suppose we call the three lines L, M, and N. Pick any point on line M, and call it B. Next, construct the plane P that contains L and passes through B. (In three-dimensional Euclidean geometry, there is always one and only one plane that contains a given line and passes through a given point not on that line.)

Now, find the intersection of plane P and line N. (Every plane has a point of intersection with every line, unless the line and the plane are parallel. I'll save for later the possibility that P and N are parallel.) Say that P and N intersect in point C.

Finally, construct the line through B and C, and call it J. Line J is the fourth line that you were looking for. J intersects N at C; it intersects M at B; and both lines J and L lie in plane P, so they must intersect - unless they're parallel.

Well, then, what if they're parallel? In that case, choose a different point B' on line M. Construct the plane P' that passes through B' and contains L. Let the intersection of P' and N be called C'. Draw line J' through B' and C'.

You can repeat this procedure for any number of different points on M: B'', B''', B'''', and so on. It's possible that one of the planes P, P', P'', etc. is parallel to line N, but it's not possible that more than one is. If two different planes P and P', both containing the line L, are both parallel to N, then L is parallel to N, and we've assumed from the very start that L and N are not parallel. It's also possible that one of the lines J, J', J'', etc. is parallel to line L, but no more than one.

So we've found an infinite number of lines J, J', J'', J''', etc. (except that one of them may not exist if one of the planes P, P', P'', P''', etc. is parallel to N), and each of those J lines intersects with all three of the lines L, M, and N (except that one of them might not intersect with L because they're parallel).

You're right in thinking that if you backed away from, or moved toward, the apparent crossing of the three lines, they would still seem to cross - as long as you were moving along the direct line of sight. As you got closer, you might have to stand on a stepladder. You could also move left or right and still see them crossing, but you might have to back up at the same time, or move forward - it's hard to predict exactly how you would have to move.

Answer 3

Assuming that the three lines exist on the same plane then yes, there will always be a line that could pass through all 3. Since lines extend into infinity, any two lines that are not parallel (share the same slope) will at one point in space intersect. Any line that has a different slope from all three previous lines will intersect at some point in space.

In response to Kathleen if you skew them or twist them, they cease to be lines and your are changing the question altogether

Answer 4

No, there does not have to be a 4th line where all 3 cross. There is easily the possibility that you could have several lines that are neither parallel or intersecting, and this would be fairly easy to see in 3-D space.

Answer 5

slope is comparable to the adaptation interior the y variables divided via the adjustments interior the x-variables, in many cases spoke of as upward push/run. So, the slope of the line that runs by pts G & H is (-3 - 4) / (2-(-a million)) or -7/3. Any line it fairly is parallel to a line working by G & H would have the comparable slope.

Answer 6

No, you could have a bunch of skew lines that are neither parallel nor intersecting. They are all contained in different planes. Imagine 4 parallel lines and then "twist" them (but don't "tilt" them) so that they are all going in different directions but not intersecting.

Answer 7

Just thinking about the geometry of it, my instincts tell me the answer is yes. I don't really have time to set down and derive a tight proof of that, but maybe someone else does. Good question, I don't think I've ever seen any Theorems that say much about that.

Doug