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Vector/Geometry problem

Hey guys, I've run into a bit of trouble with this question:

Show that for any tetrahedron, the lines joining the midpoints of opposite edges are concurrent.

Firstly, I defined vectors from some origin O to each other vertex A, B and C as a, b and c. Then I tried expressing each line joining the midpoints of opposite edges as a combination of these three vectors, but I think I've screwed up with this.

I tried getting 1/2(c-a) + 1/2b, 1/2(c-b) + 1/2a and 1/2(b-c) + 1/2a but I'm not sure where to go from here. Also I'm not sure if those expressions are correct.
Any help/hints/guidance is very much appreciated, thanks all!

Reply 1

DFranklin

18

Nichrome

Hey guys, I've run into a bit of trouble with this question:

Show that for any tetrahedron, the lines joining the midpoints of opposite edges are concurrent.

Firstly, I defined vectors from some origin O to each other vertex A, B and C as a, b and c. Then I tried expressing each line joining the midpoints of opposite edges as a combination of these three vectors, but I think I've screwed up with this.

I tried getting 1/2(c-a) + 1/2b, 1/2(c-b) + 1/2a and 1/2(b-c) + 1/2aIt would be helpful if you said what these vectors are actually supposed to represent.

If I were doing this, I'd do something like:

Find the midpoint of OA.
Find the midpoint of BC.
Write down the equation of the line joining these two points.
Find the midpoint of OB.
Find the midpoint of AC.
Write down the equation of the line joining these two points.
Find a point that lies on both lines. (Either by solving the equations or guessing).

Reply 2

Nichrome

OP

18

Right, I've attached an image to show what I meant by the vectors I defined!

I've got the equations of the two lines, and an equation of a third from midpoint of OC to midpoint of AB.

The equations I got are:

\frac{1}{2}\mathbf{a} + \frac{\lambda}{2}((\mathbf{c} - \mathbf{b}) - \mathbf{a})

\frac{1}{2}\mathbf{b} + \frac{\lambda}{2}((\mathbf{c} - \mathbf{a}) - \mathbf{b})

\frac{1}{2}\mathbf{c} + \frac{\lambda}{2}((\mathbf{b} - \mathbf{a}) - \mathbf{c}

I guess you have to equate, find the value of lambda and you're done? Is this the right method? Thanks alot for your help D.Franklin.

Tetrahedron.jpg9.8KB

Reply 3

Mathmoid

It won't be the same

\lambda

in each equation, use a different name for the coefficient in each equation or you'll probably end up very confused.

Reply 4

Nancarrow

4

Are you sure? Seems to me it would be the same value of lambda in each equation, after all we haven't picked out any distinguishing features of O, A, B and C.

I hope y'all get lambda = 1/2.

Reply 5

Nichrome

OP

18

No I guess he is right, since it is supposed to be for any tetrahedron, not just a regular one. I've used mu and nu instead now for the other two constants, but i still keep getting lost, like getting inconsistent equations. Very annoying.

Reply 6

ad absurdum

2

Nichrome - those equations aren't quite right, you've not calculated the midpoint of BC, CA, AB correctly. If you think of it, b-a isn't actually going to be anywhere near the midpoint of AB, can you think of something else that would be?

The correct answer is lambda=1/2 for all three equation, but as Mathmoid said it's not right to assume that.

Reply 7

Nancarrow

4

Hmm. I'm still sceptical. Call them l,m,n if you must (I'll figure out greek letters in a minute) but you will surely get l=m=n=1/2.

ETA: just to clarify, I'm not ignoring that the tetrahedron is a 'general' one. I'm saying those parameters are equal precisely because it is a general tetrahedron.

Suppose they were different. Then you would find a different point of intersection merely by permuting the labels of your starting vectors, which can't be right.

Reply 8

Nichrome

OP

18

So the mid point of AB isn't 1/2(b-a)?

I'm kind of confused, I'm sitting here with a diagram from my lecture notes which looks like this:

vectors.jpg5.6KB

Reply 9

Mathmoid

I really did mean that they're not the same lambda in each equation as they represent different things. I deliberately didn't say anything about the value of each lambda.

Reply 10

Mathmoid

Nichrome

So the mid point of AB isn't 1/2(b-a)?

I'm kind of confused, I'm sitting here with a diagram from my lecture notes which looks like this:

No, that's a vector in the direction of A to B, which has a length of half the distance between A and B. What you want is the position vector of the mid point of AB (which you can get by going to A first, and then travelling half the distance to B).

Reply 11

Nichrome

OP

18

Lol, what a tit I am, sorry guys. :redface:

Guess dumb moments happen sometimes.

Edit: So it would be a + 1/2(b-a) = 1/2(a+b) ?

Reply 12

Nancarrow

4

Mathmoid

I really did mean that they're not the same lambda in each equation as they represent different things. I deliberately didn't say anything about the value of each lambda.

Oh, OK then! :redface:

Reply 13

Nichrome

OP

18

I got it out, thanks guys for your help!

Vector/Geometry problem

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