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 January 24th, 2012, 12:43 AM #1 Newbie   Joined: Jan 2012 Posts: 6 Thanks: 0 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?
 January 24th, 2012, 01:05 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: vector problem in three dimensions You are in the correct forum. How did you calculate the answers for questions a and b?
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"?

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.

 January 25th, 2012, 12:46 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: vector problem in three dimensions Can you post your working for part b?
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.

 January 25th, 2012, 02:41 AM #7 Newbie   Joined: Jan 2012 Posts: 6 Thanks: 0 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
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:

$\dfrac{3(i + 3j + 2k) + (5i - j - 2k)}{4}= 2i + 2j + k$

then you are assuming that F lies on BD not BC. That is why I think the problem might be a misprint.

 January 26th, 2012, 10:09 AM #9 Newbie   Joined: Jan 2012 Posts: 6 Thanks: 0 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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