Question 1192278: 1.Let a,b,c,d,e be five points, no three of which are collinear. How many lines contain two of these five points?
2.If no four of the five points are coplanar, how many planes contain three of the five points?
Hint: (for 1 and 2, list all the lines and the planes as sets)
3. Suppose you have n points, no three of which are collinear. How many lines contain two of these n points?
4.If no four of the n points are coplanar, how many planes contain three of the n points?
Hint: (for 3 and 4, generalize in a form of a formula)
5.Prove theorem 1.1.4. The steps in the proof are already given: you just have to supply the reasons for each step.
Theorem 1.1.4. If two lines intersect, then their union lies in exactly one plane.
Proof: Let and be two intersecting lines.
a. A ∩ B is a point p.
b. B contains a point q ≠ p.
c. There is a plane E, containing A and q.
d. E contains A ∪ B.
e. No other plane contains A ∪ B.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I'll do the first two problems to get you started. In the future, please only post one problem at a time.
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Problem 1
We have n = 5 points and we need r = 2 of them to form a straight line.
The order in which we pick the points doesn't matter. Something like line ab is the same as line ba.
Use the nCr combination formula to get
n C r = (n!)/(r!(n-r)!)
5 C 2 = (5!)/(2!*(5-2)!)
5 C 2 = (5!)/(2!*3!)
5 C 2 = (5*4*3!)/(2!*3!)
5 C 2 = (5*4)/(2!)
5 C 2 = (5*4)/(2*1)
5 C 2 = (20)/(2)
5 C 2 = 10
There are 10 lines total.
Optionally you can look in Pascal's Triangle.
Locate the row that has 1,5,10,10,5,1. Then note that 10 is in the third slot corresponding to 5C2.
We start at slot zero which may be a bit strange at first.
Here are all of them listed- ab
- ac
- ad
- ae
- bc
- bd
- be
- cd
- ce
- de
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Problem 2
Once again we have n = 5 unique points to pick from.
This time we're selecting r = 3 points to form a plane.
We'll assume the conditions from problem 1 still hold that no three points are collinear; otherwise, we wouldn't be able to form a unique plane through the 3 points.
Use the nCr formula to get
n C r = (n!)/(r!(n-r)!)
5 C 3 = (5!)/(3!*(5-3)!)
5 C 3 = (5!)/(3!*2!)
5 C 3 = (5*4*3!)/(3!*2!)
5 C 3 = (5*4)/(2!)
5 C 3 = (5*4)/(2*1)
5 C 3 = (20)/(2)
5 C 3 = 10
Or you could refer to Pascal's Triangle again if you prefer. This time you're looking at the fourth slot instead of the third slot. Stay in the same row of course.
It's not a coincidence that we end up with the same result as the previous problem, even though r is now 1 unit larger.
Why is this? Well when selecting the two points to form a line, whatever three points we didn't pick will be left to form their own group and form that particular plane.
For example, if we picked 'a' and b in problem 1 to form line ab, then we'll have c,d,e left over to form plane cde.
This is why Pascal's Triangle exhibits such mirror symmetry.
Here are all ten planes- abc
- abd
- abe
- acd
- ace
- ade
- bcd
- bce
- bde
- cde
The order doesn't matter.
For example, plane abc is the same as plane cba.
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