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Simple Vector Addition Question

  1. May 22, 2012 #1
    I'm not sure if this the correct area but Ill go ahead anyways.

    I have three points (A, B, and C) in 3D space that form a triangle. All three points have the (x,y,z) coordinates defined. Using the dot product I have solved for the angle θ (or in other notation the angle ABC) of the triangle.

    I want to extend the magnitude of vector AC by a known amount and re-define the coordinate points of C. A picture is always worth a thousand words, so one is attached.

    I do know that you can define the magnitudes of AB and BC and then simply use the law of cosines to calculate the new angle but I am trying to determine this alternative method but maybe that is why it won't work. Thanks for any help.

    *edit* just slight clarification that I know how to calculate the magnitude of vector AC and since I know the magnitude of the extension CC' they can simply be added together. What I can't determine is the coordinate points of C' such that the magnitude of AC' equals AC + CC'.
     

    Attached Files:

  2. jcsd
  3. May 22, 2012 #2

    HallsofIvy

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    Use "similar triangles". Let H be the length of line segment AC (in your picture A= (-10, 30, 30) and C= (-80, 40, 40) so AC has length [itex]\sqrt{(-10-(-80))^2+ (3- 40)^2+ (30- 40)^2}= \sqrt{5100}[/itex] or approximately 71.4 ). Let h be the length of CC' (in your picture h= 10). Then each coordinate of C' has distance from A equal to (h+ H)/h (in your 81.4/71.4 which is approximately 1.14). The x-distance from A to C is -80- (-10)= -70 so the x-distance from A to C' is -70(1.14)= -79.8 and so the x-coordinate of C' must be -10- 79.8= -89.8. Do the same with the other coordinates.
     
  4. May 22, 2012 #3
    This is perfect and exactly what I was looking for, thank you very much. Just to clarify, I believe you intended to say, "Then each coordinate of C' has distance from A equal to (h+ H)/H...". That part is really the heart of the matter and I am still trying to understand that graphically. The numbers obviously work out. Again thank you for the quick response. *edit* once i get home i will study up on similar triangles to see if it clears up my final confusion.
     
  5. May 23, 2012 #4
    HallsofIvy,

    I am still having some difficulty with this vector problem. I used your method to calculate the new coordinates of C' and using those coordinates to calculate the magnitude of the vector AC' it indeed does equal the magnitude of AC + CC'.

    The next part still doesn't seem to work for me. It is not shown in my original picture, but you can create a second triangle by drawing a line segment BC', hence where the similar triangles come in play. No matter which law i use, Law of Sine, Law of Cosines, or the Dot product I get a different answer for the angle created by CBC'.

    This is actually a real-world mechanical situation where the line segment AC extends and retracts causing something to pivot about the point B. My colleagues simply hold AB and BC fixed, then use the new magnitude of AC (AC') and use the law of cosines to calculate the new angle. When I use the new coordinates we defined to use the dot product or whatever, I get a different answer. Why can they hold those segments fixed when AC changes? wouldn't only AB stay the same but BC change? Thanks for anymore help you can provide.


    *edit* I think I figured out the problem. When deriving the new C' coordinates it is confusing whether you add or substract to get to the new coordiante number? Based on the system i defined it would seem x would increase in a negative direction and z would decrease and y would increase bu that is not the case. Although different C' coordinate points give you the same overall magnitude of AC' it does not calculate the angle properly with respect to the other segments. I guess i wasn't using the similar triangle rules properly. I think I got confused in your original response because you calculate a negative distance from A to C' but then subtract the magnitude of that number from the A point x coordinate to define the C' x coordinate and that tripped me up.
     
    Last edited: May 23, 2012
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