Vectors in a tetrahedron?

Question

Hi there,

The tetrahedron OABC is such that OA=a, OB=b and OC=c. The point D is the midpoint of BC and E is the midpoint of OA. The line DE is a bimedian of the tetrahedron.

Find the position vector of G, the midpoint of DE (i.e. the vector OG) and show that G is also the midpoint of the other two bimedians of the tetrahedron.
This question has got me stumped. I did Vectors a long time ago at school but nothing about bimedians.
Any help would be warmly appreciated.
Thanks

Answer 1

I'm going to give you 2 methods.

Method 1: Pure geometry

* E is the midpoint of [OA]. That means that, vectorially, OE=1/2OA.

* D is the midpoint of [BC]. Therefore, from what you know from vector operations you will have OB+OC=2OD, so OD=1/2(OB+OC) ("parallelogram rule")

* Now G is the midpoint of [DE]. So, again, just as above, you should have the relation: OD+OE=2OG, so OG=1/2(OD+OE).

But you know how to expres OD and OE in terms of OA, OB and OC: OG=1/2(1/2OA+1/2(OB+OC))

Thence, OG=1/4(OA+OB+OC). From there on, it should be straitforward to show that G is indeed the midpoint of the other 2 bimedians. Define midpoints for AB, OC and AC, and simply repeat what we did above.

Method 2: Using a coordinate system

Define a coordinate system where O is the origin, OA is the x axis, OB is the y axis, OC is the z axis. Then, algebraically, we have:

OA=
OB=<0, b, 0>
OC=<0, 0, c>

E is the midpoint of OA, so OE=

D is the midpoint of BC. So OD=1/2(OC+OC)=<0, b/2, c/2>

Now G is the midpoint of DE, so again you have OG=1/2(OD+OE)

So OG=

So just use whichever method you prefer to finish the exercise.

Answer 2

Midpoint of line joining ends of two vectos (x, y) is (x+y)/2.
Remember it is vector addition.

OA = a
OB = b
OC = c

OD = (OB + OC)/2 = (b+c)/2 (midpoint of BC)
OE = (OO + OA)/2 = (0+a)/2 = a/2 (midpoint of OA)

OG = (OD + OE)/2 = [(b+c)/2 + a/2]/2 = (a+b+c)/4

For the proof that G is the midpoint of the other two bimedias:
Show that the other two bimedians give the same result.