# Vector problem in three dimensions

1. Jan 25, 2012

### furor celtica

1. The problem statement, all variables and given/known data
First of all, sorry if this isn't the right place to post this.

Four points A, B,C and D have coordinates (0, 1, -2), (1, 3, 2), (4, 3, 4) and (5, -1, -2) respectively. Find the position vectors of
a. The mid-point E of AC
b. The point F on BC such that BF/FD = 1/3
Use your answers to draw a sketch showing the relative positions of A, B, C and D

2. Relevant equations

3. The attempt at a solution
I had no difficulty solving a. and b.: both E and F have position vectors (2i, 2j, k)
This confuses as with respect to the last exercise, though. If E and F have the same position vectors, they should be the same point. Since E lies on AC and F on BC, shouldn’t A, B and C lie on the same line? However, vector AC = 4i + 2j + 6k and vector BC = 3i + 2k: they are not multiples, so A, B and C are not collinear.
I’m new to the study of vectors in three dimensions so there’s probably a noob mistake I’m making somewhere. What am I missing?

2. Jan 25, 2012

### stripes

Take a look again at your midpoint for part (a). When finding the midpoint between A and C, we have ((4-0)/2, (3-1)/2, (4+2)/2). So the midpoint E doesn't have the coordinates you mentioned.

Part (b), try finding the point F again: Let F = (a, b, c). We have 1/3 = BF/FD, which means that (1/3)FD = BF. Recall that FD = (5, -1, -2)-(a, b, c), and BF = (a, b, c)-(1, 3, 2).

Then you will have a system of equations, and you will be able to find the coordinates of F.

3. Jan 25, 2012

### vela

Staff Emeritus
You should write either (2, 2, 1) or $2\hat{i}+2\hat{j}+\hat{k}$. Don't mix the notations.

I agree that E is (2, 2, 1), but I get a much messier answer for F. How did you find F?

4. Jan 25, 2012

### vela

Staff Emeritus
This isn't correct.

This isn't correct either. BF and FD denote lengths.

5. Jan 25, 2012

### stripes

My apologies. I should try not to answer questions when I'm sleepy.

6. Jan 26, 2012

### furor celtica

Alright so here’s my work on b. :
We have 3BF = FD, f = BF + b
And since we know b, all we need is BF to find f position vector of F
BF + FD = BD => 4BF = d – b = 4i – 4j – 4k => BF = i – j – k
=> f = (1+1)i + (3-1)j + (2-1)k = 2i + 2j + k

Now I’m aware that my error lies in assuming the vector BF equals one-third of vector FD, as this proportion only applies to the lengths of the vectors and not the vectors themselves. It WOULD mean the same thing, however, if F lies on BD and not BC: so I’m starting to believe that problem is simply a typo in the textbook, in which case my working would be valid.

7. Jan 26, 2012

### vela

Staff Emeritus
So do you still have the question about sketching the four points?

8. Jan 27, 2012

### furor celtica

yes! i'm not sure how to sketch a three-dimensional vector at all, especially if the points are not collinear.