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Locating a point

  1. Sep 5, 2012 #1
    Given information:
    |vector AB| = 3*|vector AC|

    angle betweem AB and AC is 30 deg

    The coodinate points are given in the paint picture

    Is it possible to locate the point C???
     

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  2. jcsd
  3. Sep 5, 2012 #2

    gabbagabbahey

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    Hint: take a look at the inner product between AB and AC.
     
  4. Sep 5, 2012 #3
    the inner product would be

    9(squrt(3)/2) = 3*(x-4) + 3(y-5) + 3(z-6)

    x + y + z = 3*squrt(3)/2 + 11
     
  5. Sep 5, 2012 #4

    gabbagabbahey

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    That's what you get when you compute the inner product using components, what does the geometric form of the inner product equation ([itex]\vec{AB} \cdot \vec{AC} = |\vec{AB}||\vec{AC}|\cos\theta[/itex]) tell you?

    Edit: I guess you have used the geometric form of the inner product to get the [itex]9\frac{\sqrt{3}}{2}[/itex] on the LHS of the equation. In components, shouldn't you have [itex]\vec{AC}=(x-1)\vec{i} + (y-2)\vec{j} + (z-3)\vec{k}[/itex]? You have 3 unknowns (x,y & z), so you will need at least 2 more equations to solve for the vector uniquely.

    Hint: What does the fact that |vector AB| = 3*|vector AC| tell you?

    Hint 2: In your picture, it looks like AB, AC, and BC are intended to be coplanar, what does that tell you about their cross-products?
     
    Last edited: Sep 5, 2012
  6. Sep 5, 2012 #5
    its the x component of BC times the magnitude of AB
     
  7. Sep 5, 2012 #6

    gabbagabbahey

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    What is? :confused:

    (I've edited my above post, so you may want to re-read it)
     
  8. Sep 5, 2012 #7
    AC = (x-1)i + (y-2)j + (z-3)k

    Inner product of AB and BC = |AB||BC|Cos(∏/6) = (√27)(√27/3)(√3/2) =
    =9√3/2 = 3(x-4) + 3(y-5) + 3(z-6) ... x + y + z = 1/3(9√3/2 + 45)

    Eq 1: AC = (x-1)i + (y-2)j + (z-3)k
    Eq 2: x + y + z = 1/3(9√3/2 + 45)
    Eq 3:

    Sorry I am unfamilliar with coplanar points... this is a problem i made up so if you could tell me about the formula you are referring to about coplanar points I can look it up in my calc book.
     
  9. Sep 5, 2012 #8
    Wow I am sorry.... The angle between AB and BC is 30 deg.. please look at the picture and I ment to say the magnitude of vector AB is 3 time the magnitude of vecot BC
     
  10. Sep 5, 2012 #9

    gabbagabbahey

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    In your picture you have the angle between AB and BC as 30°, while in your problem statement you have 30° being the angle between AB and AC and |AB|=3|AC|. Did you mean for your problem statement to say that the angle between AB and BC is 30° and |AB|=3|BC| instead (which is what it looks like in your picture)?

    Eq 1 doesn't help you solve for x,y & z. What does the fact that |AB|=3|BC| (which is what I assume you meant in your problem statement) tell you in terms of components?

    If 3 or more vectors are coplanar, the cross-product between any two pairs of non-parallel vectors will point in the direction of the normal to the plane. This doesn't actually help you in this case though.
     
  11. Sep 5, 2012 #10
    Would the two other equation be

    vector Projection of BC onto AB

    and (Look at picture to see point Vector PC)

    PC = BC - (vector Projection of BC onto AB)
     

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  12. Sep 5, 2012 #11

    gabbagabbahey

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    You are looking for equations that involve only x,y,z and numbers. An equation like [itex]\text{Proj}_{\vec{AB}}(\vec{BC})=f(x,y,z)[/itex] won't help you since the LHS is not an expression involoving only x,y,z and scalar numbers.

    Again, how do you find the magnitude of a vector using components? What does that tell you about the the condition |AB|=3|BC|?
     
  13. Sep 5, 2012 #12
    Ok to find the magnitude of a vector using components you square root the square of its components

    so if the components of AB are x,y and z then I believe that the components of BC would be 9x,9y, and 9z because there must have been a factor of 9 for each component to pull a 3 out infront of |BC|
     
  14. Sep 5, 2012 #13

    gabbagabbahey

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    So, if AB=(3,3,3), BC=(x-4,y-5,z-6) and |AB|=3|BC|, what do you have?
     
  15. Sep 5, 2012 #14
    Bc= < 13/3,16/3,19/3>
     
  16. Sep 5, 2012 #15
    So im guessing I could have solved this another way by finding three equations that delt with scalars rather than three equations that delt with vectors. Is that correct?
     
  17. Sep 5, 2012 #16

    gabbagabbahey

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    No.

    |AB| = √(32+32+32 = 3√3

    and

    |BC| = √[(x-4)2 + (y-5)2 + (z-6)2]

    So |AB|=3* |BC| just tells you that 3√3=3√[(x-4)2 + (y-5)2 + (z-6)2] or, equivalently (x-4)2 + (y-5)2 + (z-6)2=3.

    What kind of surface does this describe? What kind of surface is described by you other equation, x+y+z=(3/2)√3 +15? What does their intersection look like?

    No.
     
  18. Sep 5, 2012 #17
    The equation x+y+z=(3/2)√3 +15 describes the surface whose coordinates of point (x,y,z) add up to (3/2)√3 +15

    (x-4)2 + (y-5)2 + (z-6)2=3 defines all points whose distance from (4,5,6) to (x,y,z) is squrt(3).

    The intersection of these two would be some type of circle im guessing.. is any of this correct?
     
  19. Sep 5, 2012 #18
    Thanks for your help, Im obviously doing a problem over my head because I've only been in this calc class for 2 weeks.

    For one im not sure what the intersection of those two equations will tell me because dont I need a third equation and find the intersection of those three equations to be relevent to finding the point im looking for.
    And given the two equations above I wouldnt know what the intersection looks like because I dont know what the equation x+y+z=(3/2)√3 +15 looks like.. I havent learned any of this yet... im guessing it is a line of some sort.. Anyways thank you for your help and I will take what you gave me and ask me teacher
     
  20. Sep 5, 2012 #19

    gabbagabbahey

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    Sure, but notice that this surface is a plane.

    Yes, but notice that this surface is spherical.

    Yes, their intersection is a circle. You can find the equation of that circle by solving the two simultaneous equations for x,y and z in terms of a single free parameter.

    Now, go back and look at your drawing again. Notice that if you rotate BC about an axis coincident with AB, the angle between the two vectors is still 30 degrees, and the magnitude of AB is still 3 times the magnitude of BC. Therefor, you should expect that any point C that lies on the circle you get by rotating BC 360° about AB, will satisfy your problem statement. Thus the solution is not a unique point, but any point that lies on that circle.
     
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