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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 midpoint 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: 7,923 Thanks: 1123 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  
Senior Member Joined: Feb 2010 Posts: 704 Thanks: 138  Re: vector problem in three dimensions Quote:
 
January 24th, 2012, 11:13 PM  #4  
Newbie Joined: Jan 2012 Posts: 6 Thanks: 0  Re: vector problem in three dimensions Quote:
Quote:
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: 7,923 Thanks: 1123 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  
Senior Member Joined: Feb 2010 Posts: 704 Thanks: 138  Re: vector problem in three dimensions Quote:
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  
Senior Member Joined: Feb 2010 Posts: 704 Thanks: 138  Re: vector problem in three dimensions Quote:
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.  
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 + (31)j + (21)k = 2i + 2j + k Now I’m aware that my error lies in assuming the vector BF equals onethird 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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TheÂ point FÂ onÂ BC suchÂ thatÂ BF/FDÂ =Â 1/3 Use your answersÂ toÂ drawÂ aÂ sketch showingÂ theÂ relative positionsof A,Â B,,find the point f on bc such that bf/ fd=1/3
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