3d plane in a cube

thedarktiger · 2014-02-15 18:30:08

Let ABCDEFGH be a cube of side length 5, as shown. Let P and Q be points on \overline{AB} and \overline{AE}, respectively, such that AP = 2 and AQ = 1. The plane through C, P, and Q intersects \overline{DH} at R. Find DR.
http://cache.artofproblemsolving.com/asyforum/1/1/8/118d23c1a4792612d4a97f539003b085e17242f2.png
agg how to do this

Bob · 2014-02-15 21:00:49

hi,

I can show you a method using 3D coordinate /vector geometry. I can short cut some of the theory but inevitably it'll be a steep learning curve.

The techniques will open the door to lots of neat stuff though.

Ready ?

Bob

thedarktiger · 2014-02-16 10:56:57

okay!
funfunfunfunfun...

Last edited by thedarktiger (2014-02-16 11:05:18)

Bob · 2014-02-16 19:47:52

hi thedarktiger,

OK, here we go.

Firstly, we need a coordinate system. I'll use your cube for this.

Make A the origin. It's coordinates will be (0,0,0)

Make AD the x axis. So D will be (5,0,0)

Make AB the y axis. So B will be (0,5,0)

If you imagine looking down from overhead the square ABCD will be in the usual x-y plane with C at (5,5,0)

Now to rise out of that plane.

Make AE the z axis. So E is (0,0,5)

F is (0,5,5)

P is (0,2,0)

So, your first exercise, to check you have understood so far.

What are the coordinates of Q, G and H ?

Lesson two after that.

Bob

thedarktiger · 2014-02-16 22:18:01

Q is (0,0,1).
G is (5,5,5).
H is (5,0,5).
Wow, you guys make great tutors.

Bob · 2014-02-17 01:14:14

Thanks.

Those are all correct!

You are probably familiar with y = mx + c as the equation of a line. If not, have a look here:

http://www.mathsisfun.com/equation_of_line.html

We can say that equation has one degree of freedom; you know the fixed numbers m and c, so if you choose an 'x', you can calculate the 'y' that goes with it. x is called the independent variable and y, the dependent variable.

But you could also switch the equation around to x = (y-c)/m and then choose y, and calculate x. Now y is the independent variable and x is dependent. Either way you can make one choice and that's what one degree of freedom means.

You can also write the equation like this: ax + by = d for some values a, b, and d.

For a plane I'd expect two degrees of freedom. I could choose to be somewhere along the line PC, but I could also, and independently, head out from there at right angles to PC, yet still remain in the plane. Or think of it like this: Choose an x and y, say, x = 3 and y = 4, in the x-y plane. That point won't be in the plane. But if I go vertically straight up from there, I'll hit the plane somewhere and that will fix z. And if I carry on up, I won't hit the plane again. So it would seem that z depends on the x and y I chose. That's the two degrees of freedom.

So I'm expecting an equation with this format:

ax + by + cz = d

There's about three lessons of vector theory to justify that, so I'll just skip that for now.

So how can we work out a, b c, and d ?

We know that P, Q, and C lie in the plane and that's just enough to work them out.

P lies in the plane => 0a + 2b + 0c = d => 2b = d

Q lies in the plane => 0a + 0b + 1c = d => c = d

C lies in the plane => 5a + 5b + 0c = d => 5a + 5d/2 = d => a = -3d/10

So I can get everything else in terms of d, but cannot get d as well.

But actually, that doesn't matter.

Say I let d = 1, then a = -3/10 b = 1/2 and c = 1

That makes the equation -3x/10 + y/2 + z = 1 ........(i)

Now let's say I let d = 2, then a = -6/10 b = 1 and c = 2

That makes the equation -6x/10 + y + 2z = 2

that's exactly what you'd get if you take equation (i) and double everything. So the two equations are really the same! So I might as well choose a 'd' that will make the working easier ... I think d = 10 will remove all the fractions and then I get

-3x + 5y + 10z = 10

Check it's right:

(0,2,0) gives -3 x 0 + 5 x 2 + 10 x 0 = 10 ... yes, that's ok.
(0,0,1) gives -3 x 0 + 5 x 0 + 10 x 1 = 10 ... yes, that's ok.
(5,5,0) gives -3 x 5 + 5 x 5 + 10 x 0 = -15 + 25 = 10 ... yes. that's ok.

So now I can work out R.

It is directly above D so it has coordinates (5, 0, z)

plug into the equation

-3 x 5 + 5 x 0 + 10z = 10 => -15 + 10z = 10 => 10z = 25 => z = 2.5

So DR = 2.5

Bob

ps. That makes CR parallel to PQ. I did wonder if that was true as it looks like it on the diagram. But I wasn't sure. Now I am. Maybe it's because PQ lies in a vertical plane and so does CR. hhmmmm. Is there a general result here?

Last edited by Bob (2014-02-17 01:35:52)

thedarktiger · 2014-02-18 00:21:40

Thank you! I think I got it. Thats a cool solution!

championmathgirl · 2015-06-22 09:04:08

Quick question, if plane through C, P, and Q intersect AD at S, and the same plane through C, P, and Q intersects DH at R, which two dimensional figures do they make?

Bob · 2015-06-22 19:37:35

hi championmathgirl

Did you follow how I got the equation for the plane?

-3x + 5y + 10z = 10

To find S, on AD (the x axis):

Points on the x axis have y = z = 0

Substitute into the equation:

-3x = 10, so S is (-10/3, 0 ,0)

As the line RQ is in the xz plane, it will, if extended, cut the x axis at S.

So RSD is a right angled triangle, with RD = 2.5 and SD = 5 + 10/3.

Bob

dazzle1230 · 2016-06-18 11:25:26

Could we do this without a 3D coordinate plane?
I got a hint to solve this problem:
Let the plane through C, P, and Q intersect \overline{AD} at S. What can you say about the diagram? Which two-dimensional figures can you work with?

Bob · 2016-06-18 19:55:11

hi dazzle1230

Oh yes; championmathgirl gave that hint in post 8. I hadn't explored it because I had my solution, but let's do so now.

S lies in the plane of ABCD and in the plane of PQRC so CP extended must go through S

Triangles SAP and SDC are similar. Triangles SAQ and SDR are similar.

From the first pair SA/SD = PA/DC = 2/5

So SA/(SA+5) = 2/5 => 5xSA = 2xSA + 10 => SA = 10/3

From the second pair SA/SD = AQ/RD

=> RD = 1 x (5+10/3)/(10/3) = (15+10)/10 = 25/10 = 2.5

Thanks championmathgirl

Bob

dazzle1230 · 2016-06-19 02:15:01

Thank you so much bob!

pineapple12 · 2016-06-21 06:26:52

How do we know triangles SAQ and SDR are similar? (I hope I have my diagram correct) So, is SR a straight line and if it is, how do we know that? To be clear, point S is created when we extend CP and AD right??? Thank you in advance!

dazzle1230 · 2016-06-21 07:15:16

pineapple12 wrote:

How do we know triangles SAQ and SDR are similar? (I hope I have my diagram correct) So, is SR a straight line and if it is, how do we know that? To be clear, point S is created when we extend CP and AD right??? Thank you in advance!

i) Prove the triangles are similar using AA similarity.
ii) We don't need to prove that RS is straight. We can just say that we extended QR and DA and the point of intersection is S.
iii) Yes, point S is created when we extend CP and AD.

pineapple12 · 2016-06-21 07:33:29

Oh yeah... ok THANKS SO MUCH!!!

rileywkong · 2016-06-21 16:14:31

An alternate solution would be to consider triangles QAP and RDC.

(AoPS community please don't hate on me )

pineapple12 · 2016-06-23 01:04:18

@rileywkong Haha, thanks. I actually ended up using those triangles instead because the solution was shorter and easier in my opinion.

Wangotango · 2016-08-20 02:26:42

Where does the plane intersect at s? Doesn't the plane go above line AD so it has to intersect above?

Bob · 2016-08-20 03:52:57

hi Wangotango

Define S to be where the plane CPQ intersects the x axis (AD).

S lies in the plane of ABCD and in the plane of PQRC so CP extended must go through S

This method (from championshipgirl) is much simpler than my original method. But it does show that many methods may all lead to the same result and shows the unity of geometry, vector algebra and coordinate geometry. Neat!

Bob

kk63353 · 2016-10-15 23:36:33

bob bundy wrote:

SA = 10/3
From the second pair SA/SD = AQ/RD
=> RD = 1 x (5+10/3)/(10/3) = (15+10)/10 = 25/10 = 2.5

Can't we just say SA/SD=2/5=AQ/RD, so since AQ=1, RD=5/2?

Bob · 2016-10-16 00:32:33

hi kk63353

Welcome to the forum.

I think you can say that, so well done. I do not claim that my methods are the quickest, neatest, or even correct.

That's why the forum has other members!!!

Bob

!nval!d_us3rnam3 · 2017-03-27 10:08:02

kk63353 wrote:

bob bundy wrote:
SA = 10/3
From the second pair SA/SD = AQ/RD
=> RD = 1 x (5+10/3)/(10/3) = (15+10)/10 = 25/10 = 2.5
Can't we just say SA/SD=2/5=AQ/RD, so since AQ=1, RD=5/2?

I was thinking the exact same way. Why not use the slope of the line, and the height it gains as it crawls along the edge of the cube, rather than a big long solution that I partially didn't understand? This is my reasoning:

PA = 2, and AQ = 1, so line PQ has a slope of 1/2. Therefore, when the line QR, which is just a continuation of PQ, goes 5 units laterally, it climbs 2.5 units. Therefore, DR = 2.5.

Q.E.D.

Last edited by !nval!d_us3rnam3 (2017-03-27 10:19:29)

Bob · 2018-10-01 01:28:18

hi n0tjyxgsgs

Welcome to the forum.

This problem came up ages ago so I've mostly forgotten it. Some other posters provided quicker solutions than mine; that's fine; I never said mine was the only one nor that it was quick. The OP was happy to be taught some new maths along the way which I was happy to do.

PQRC is a sloping plane. So QR isn't a continuation of PQ. Not sure what you mean here.

Bob

Math Is Fun Forum

#1 2014-02-15 18:30:08

3d plane in a cube

#2 2014-02-15 21:00:49

Re: 3d plane in a cube

#3 2014-02-16 10:56:57

Re: 3d plane in a cube

#4 2014-02-16 19:47:52

Re: 3d plane in a cube

#5 2014-02-16 22:18:01

Re: 3d plane in a cube

#6 2014-02-17 01:14:14

Re: 3d plane in a cube

#7 2014-02-18 00:21:40

Re: 3d plane in a cube

#8 2015-06-22 09:04:08

Re: 3d plane in a cube

#9 2015-06-22 19:37:35

Re: 3d plane in a cube

#10 2016-06-18 11:25:26

Re: 3d plane in a cube

#11 2016-06-18 19:55:11

Re: 3d plane in a cube

#12 2016-06-19 02:15:01

Re: 3d plane in a cube

#13 2016-06-21 06:26:52

Re: 3d plane in a cube

#14 2016-06-21 07:15:16

Re: 3d plane in a cube

#15 2016-06-21 07:33:29

Re: 3d plane in a cube

#16 2016-06-21 16:14:31

Re: 3d plane in a cube

#17 2016-06-23 01:04:18

Re: 3d plane in a cube

#18 2016-08-20 02:26:42

Re: 3d plane in a cube

#19 2016-08-20 03:52:57

Re: 3d plane in a cube

#20 2016-10-15 23:36:33

Re: 3d plane in a cube

#21 2016-10-16 00:32:33

Re: 3d plane in a cube

#22 2017-03-27 10:08:02

Re: 3d plane in a cube

#23 2018-10-01 01:28:18

Re: 3d plane in a cube

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