How many can you determine with ?

[Дьявол]

Homework Statement

1. How many planes are determined with 3 different lines?

2. How many planes can you determine using 1 line and 2 points which doesn't lay on the line?

Homework Equations

1 plane can be determined using 3 points

1 line can be created with min. 2 points

The Attempt at a Solution

1. If you ask me I would say that three different lines can create unlimited planes. In my book results it says 1 or three.

2. Again my answer would be unlimited. One line got unlimited points so I can create unlimited planes with the other 2 points. In my textbook results 1 or 2.

Please help! Thanks in advance.

 

[mutton]

1. Why unlimited planes? If you take any 2 nonparallel lines that don't intersect, they can't be "connected" to form a plane. Likewise, if you take any 3 nonparallel lines such that any 2 of them don't intersect, they don't determine a plane. So, at least 2 of the lines are parallel or at least 2 of the lines intersect. Go through each case, and you will get 1 or 3 planes as the answer.

2. When you connect the points of a line to a point not on the line, you get a bunch of lines which determine one plane, not unlimited planes. What happens if there's a second point? There are 2 cases, as suggested by the textbook.

 

[Дьявол]

Thanks for the reply.

Sorry for telling, but honestly I don't understand you. I made a picture. With 3 lines I can create as many as I want planes.

http://img387.imageshack.us/my.php?image=47519589zd3.jpg" the picture.

Thanks in advance.

 

[mutton]

You drew some lines and triangles on the same plane. The triangles are all different, but when you extend them, they "merge" into one plane on your computer screen. Remember that planes are flat surfaces that stretch into infinity.

The question is in 3-space. Use a ruler/stick to represent a line and a piece of paper to represent a plane, and move them around physically to get a feel for the question.

http://en.wikipedia.org/wiki/Line-plane_intersection

 

[Дьявол]

You drew some lines and triangles on the same plane. The triangles are all different, but when you extend them, they "merge" into one plane on your computer screen. Remember that planes are flat surfaces that stretch into infinity.

The question is in 3-space. Use a ruler/stick to represent a line and a piece of paper to represent a plane, and move them around physically to get a feel for the question.

http://en.wikipedia.org/wiki/Line-plane_intersection

mutton thanks for the reply. I understand what you are telling me. Again I take 3 random lines (different angles in the space), and again I get unlimited planes. Do you have any links or something that could help me explain and solve my problem?

Thanks in advance.

 

[Дьявол]

I was thinking, and now I understand what you are talking me.

I went through each case.

Are the answers correct on the picture:

http://img152.imageshack.us/img152/7596/43658451hq2.th.jpg

 

[mutton]

Cases II and III are essentially the same, and I think they both determine 1 plane only by two lines, even though the third line intersects that plane because the third line doesn't determine a plane with either of the other 2. It will help to know exactly what is meant by "determine a plane". I am just going by intuition, unfortunately.

Consider Case V: All 3 lines intersect at 1 point.

 

[Дьявол]

As you say my Case II and Case III aren't correct, but I don't understand why.

But in which case there are 3 planes determined?

Here is Case V:

http://img232.imageshack.us/img232/6991/69681614kr2.th.jpg

Actually, I am totally confused doing this things.

Thanks for the efforts.

Regards.

 

[HallsofIvy]

I am having extreme difficulty understanding exactly what the problem and question are!

If by "plane determined by lines", you mean a plane that contains all the lines, then with two lines in three dimensions, if they cross or are parallel, exactly one plane is determined. If they are skew, there is no plane that contains both lines.

With three lines, the only way there exist a plane containing all three lines is if
(1) They are all parallel.
(2) Two are parallel and the third intersect the other two.
(3) They all intersect each other.
Otherwise, there is no plane containing all three lines.

On the other hand, if by "plane determined by lines" you simply mean a plane that contains at least one of the lines, then the answer is infinitely many. Any one of the three lines is contained in infinitely many planes.

 

[vanesch]

You can even go further, and by "determine" only understand "are constructed based upon".

For instance, if you have two arbitrary, non-parallel, non-intersecting lines A and B, you can determine (if the space has an Euclidean metric) the point p on A closest to B, and have the plane containing p and B ; you can determine the plane orthogonal to the that plane and containing B. If q is the point on B closest to A, you can determine the plane through q orthogonal to the vector pq. You can determine the plane through p orthogonal to pq. You can determine the plane through q, orthogonal to the line A. You can determine the plane though p, orthogonal to the line A...

 

[Дьявол]

I am having extreme difficulty understanding exactly what the problem and question are!

If by "plane determined by lines", you mean a plane that contains all the lines, then with two lines in three dimensions, if they cross or are parallel, exactly one plane is determined. If they are skew, there is no plane that contains both lines.

With three lines, the only way there exist a plane containing all three lines is if
(1) They are all parallel.
(2) Two are parallel and the third intersect the other two.
(3) They all intersect each other.
Otherwise, there is no plane containing all three lines.

On the other hand, if by "plane determined by lines" you simply mean a plane that contains at least one of the lines, then the answer is infinitely many. Any one of the three lines is contained in infinitely many planes.

HallsofIvy sorry for not being explicit.

Even, when I put the question here, I didn't understand what was it for, but as I look at my textbook results I realized that I need to find plane containing all three lines.

Also I realized that I can find one plane, only by using two lines (no matter if they are parallel, intersecting, non-intersecting or non-parallel).

In my textbook there are 3 questions "How many planes contain 3,4,5 lines?". The results are:
-3 lines (1 or 3 planes)
-4 lines (1,4 or 6 planes)
-5 lines (1,5,8 or 10 planes)

By using these results I realized that I can determine the maximum number of planes, by using the fact that two of them can determine one plane. For example:
-3 lines ... $$C_2^3=\frac{3!}{1!*2!}=3$$
-4 lines ... $$C_2^4=\frac{4!}{2!*2!}=6$$
-5 lines ... $$C_2^5=\frac{5!}{3!*2!}=10$$

$$C_n^k=\frac{n!}{k!*(n-k)!}$$ - combination.

Could you possibly tell me what is "skew"?

Do you refer to this:

If you refer to the lines of the picture I think that I still can create a plane using those ones in 3d space, by taking one point from the first line, and taking two points from the second one, since there must be at least 3 points to determine the plane.

Is this true:

(1) Three lines parallel can determine only 1 plane

(2) Two are parallel and the third intersect the other two can determine 3 plane

(3) If they all intersect, they can determine only 1 plane

vanesch thanks for the efforts, but I could not understand you. Could you possibly give me some graph or picture?

Thanks all for the replies.

Regards.

 

[Дьявол]

Can somebody please confirm or deny something? Did I get the thing right?

Thanks in advance.

Regards.

 

[mutton]

As you say my Case II and Case III aren't correct, but I don't understand why.

But in which case there are 3 planes determined?

Here is Case V:

http://img232.imageshack.us/img232/6991/69681614kr2.th.jpg

This is actually only a subcase, so let's split into Cases V and VI.

Case V, as you drew, has 3 intersecting lines all on one plane. If you wrote the slopes of the 3 lines in vector form, one of the vectors would be a linear combination of the other two.

Case VI: What if the 3 intersecting lines do not lie on one plane? The 3 corresponding vectors must be linearly independent. For a particular example, consider the x-axis, the y-axis, and the z-axis. These are 3 lines that intersect at the origin.

The x-axis and the y-axis determine the xy-plane.
The x-axis and the z-axis determine the xz-plane.
The y-axis and the z-axis determine the yz-plane.

This is how you get 3 planes. You can tilt/slant the axes and get different lines that also generate 3 planes, as long as the 3 lines don't lie on one plane.

 

[mutton]

HallsofIvy sorry for not being explicit.

Even, when I put the question here, I didn't understand what was it for, but as I look at my textbook results I realized that I need to find plane containing all three lines.

I think you actually want to find planes containing at least 2 of the 3 lines, and all my comments above are based on this. Otherwise, your answers don't make sense.

(1) Three lines parallel can determine only 1 plane

No. If the 3 lines do not lie on one plane, then 3 planes are determined. For example, consider these lines.

Define line A by (t, 0, 0) for real t. (Alternatively, define A by y = z = 0.)
Define line B by (t, 1, 0) for real t. (Alternatively, define B by y = 1, z = 0.)
Define line C by (t, 0, 1) for real t. (Alternatively, define C by y = 0, z = 1.)

These 3 lines are parallel because they are all perpendicular to the yz-plane.

The xy-plane (z = 0) contains A and B.
The xz-plane (y = 0) contains A and C.
The plane y + z = 1 contains B and C.

Thus, the lines A, B, C determine 3 planes.

(2) Two are parallel and the third intersect the other two can determine 3 plane

No. The 2 parallel lines determine 1 plane. Let's call that plane P. If the 3rd non-parallel plane doesn't intersect either of the other 2 lines, it doesn't matter at all. If you think intersecting P is relevant, then an infinite number of planes would be determined, not just 3. This is because there's no reason to just pick some arbitrary pair of planes that contain the 3rd line and P.

(3) If they all intersect, they can determine only 1 plane

No, see my previous post.

By using these results I realized that I can determine the maximum number of planes, by using the fact that two of them can determine one plane. For example:
-3 lines ... $$C_2^3=\frac{3!}{1!*2!}=3$$
-4 lines ... $$C_2^4=\frac{4!}{2!*2!}=6$$
-5 lines ... $$C_2^5=\frac{5!}{3!*2!}=10$$

Correct.

Could you possibly tell me what is "skew"?

http://en.wikipedia.org/wiki/Skew_lines

 

[Дьявол]

No. The 2 parallel lines determine 1 plane. Let's call that plane P. If the 3rd non-parallel plane doesn't intersect either of the other 2 lines, it doesn't matter at all. If you think intersecting P is relevant, then an infinite number of planes would be determined, not just 3. This is because there's no reason to just pick some arbitrary pair of planes that contain the 3rd line and P.

I agree that two parallel lines does create 1 plane, and the 3rd intersect line doesn't matter at all, but if the 3rd line is skew or parallel then the 3 lines determine 3 planes, as you said in your post.

Thank you for the help. You are completely right. I agree with you. But there are too many cases

For ex. If take 4 lines, I know that I can make maximum C₄²=4!/(2!*2!)=6 planes.

Now I need to determine the other ones.

I case: all intersecting in one point (lie on 1 plane)
sub case I, they are all intersecting but do not lay on same plane (determine 6 planes)
II case: all parallel (lie on 1 plane)
sub case I, do not lay lay on 6 planes.
III case: 3 intersecting, 1 parallel
sub case I, three intersecting lay on same plane,1 parallel (determine 6 planes)
sub case II, three intersecting lines don't lay on same plane, 1 parallel (determine 4 or 6 planes)
...
...
I think there are 1,4 or 6 planes. Is there any straight way for determining this stuff?

Thanks in advance.