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January 24th, 2012, 12:43 AM   #1
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vector problem in three dimensions

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

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?
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January 24th, 2012, 01:05 AM   #2
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Re: vector problem in three dimensions

You are in the correct forum.

How did you calculate the answers for questions a and b?
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January 24th, 2012, 03:24 AM   #3
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Re: vector problem in three dimensions

Quote:
Originally Posted by furor celtica
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
Is there a misprint here? Should this be "The point F on BD such that BF/FD = 1/3"?
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January 24th, 2012, 11:13 PM   #4
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Re: vector problem in three dimensions

Quote:
Originally Posted by mrtwhs
Quote:
Originally Posted by furor celtica
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
Is there a misprint here? Should this be "The point F on BD such that BF/FD = 1/3"?
no that's not a misprint, thats from the book. maybe its a typo, but if it was i don't think they would ask me to find the relative position of D later on.

Quote:
Originally Posted by greg1313
You are in the correct forum.

How did you calculate the answers for questions a and b?
Is that important? I can post all my work if it is, but I didn't because it was pretty straightforward stuff that I had no problem with. It doesn't seem like my difficulties with the last question are related; my problem is pretty specific, namely how E and F can be the same point and yet A, B and C are not collinear.

Thanks for your time btw.
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January 25th, 2012, 12:46 AM   #5
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Re: vector problem in three dimensions

Can you post your working for part b?
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January 25th, 2012, 02:34 AM   #6
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Re: vector problem in three dimensions

Quote:
Originally Posted by furor celtica

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

I had no difficulty solving a. and b.: both E and F have position vectors (2i, 2j, k)
(2i,2j,k) is not the answer to part b. The point F is on BC and you said that was correct (not a typo). The equation of the line BC is

x = 1 + 3t
y = 3
z = 2 + 2t

Clearly (2,2,1) is not on that line. Use (1+3t,3,2+3t) as F. Calculate the distances from F to B and F to D. Set the ratio equal to 1/3 and solve for t. Then plug it into the equation for BC to find the point. I think you get a quadratic equation but it is ugly.
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January 25th, 2012, 02:41 AM   #7
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Re: vector problem in three dimensions

hmm i'll post my work then. however the answer given in the textbook IS 2i + 2j + k and I remember coming to that exact conclusion very convincingly. That could hardly be a coincidence, but I'll be back with my work anyway. Also, what is 't' in your equation of BC?
Once again, thanks to both of you for helping out
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January 25th, 2012, 05:23 AM   #8
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Re: vector problem in three dimensions

Quote:
Originally Posted by furor celtica
hmm i'll post my work then. however the answer given in the textbook IS 2i + 2j + k and I remember coming to that exact conclusion very convincingly. That could hardly be a coincidence, but I'll be back with my work anyway. Also, what is 't' in your equation of BC?
Once again, thanks to both of you for helping out
t is a parameter in the parametric form for the equation of the line BC.

If you got point F in part b from:



then you are assuming that F lies on BD not BC. That is why I think the problem might be a misprint.
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January 26th, 2012, 10:09 AM   #9
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Re: vector problem in three dimensions

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, as abovementioned but which I have only just understood. So I'm starting to think this must just be a typo.
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