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Direction Cosines

  1. Aug 26, 2005 #1
    We're trying to find the height of a mountain.
    First we have the measured altitude of 2 points at 3000 m above sea level.
    Then, these two points are 10,000 m apart on the same axis(x axis, z axis is pointed up)

    point A--> (0,0,0)
    point B-->(10000,0,0)

    We're trying to find the height of the mountain, the top of the mountain is point P.

    Direction cosines:

    Rap:
    cos theta x= .5179
    cos theta y= .6906
    cos theta z= .5048

    Rbp:
    cos theta x=-.3743
    cos theta y=.7486
    cos theta z=.5472

    Anybody? I've been stuck on this for a LONG time.

    edit: well let me show what i've done so far (isn't much)

    all i've written out is:

    rapx/rap=.5179
    rapy/rap=.6906
    etc

    but where do i go from here? I've drawn out triangles for each APx, APy, APz and same for BP, but none of this tells me anything or lets me continue. Nor do i know how the 10,000 plays a role in the equations other than |Rap| + |Rbp| =10,000m
     
    Last edited: Aug 26, 2005
  2. jcsd
  3. Aug 26, 2005 #2
    I'm not entirely sure I'm visualizing this correctly, or where the y axis come into this because you didn't mention the y axis. Furthermore i'm not sure what you mean by Rap and Rbp...but

    If you have two points on opposite sides of the mountain that are 10,000m apart, can you construct two right triangles?

    Let A equal the point at (0, 0, 0)
    Let B equal the point at (10,000, 0, 0)
    Let P equal the point at the top of the mountain
    Let Q equal a point on the line AB
    Let the line PQ intersect the line AB at a right angle (aka, drops straight down from P)

    You have two right triangles,
    PQA
    PQB

    They both have the side PQ in common, so this side is equal for both right triangles...
     
  4. Aug 26, 2005 #3
    i drew it out
     

    Attached Files:

  5. Aug 26, 2005 #4
    Are points A, B, and P all in the xz plane?
     
  6. Aug 26, 2005 #5
    A and B are on the x axis, and P is xyz
     
  7. Aug 26, 2005 #6
    anyone else? pleeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeease?
     
  8. Aug 28, 2005 #7
    bizzump ttt
     
  9. Aug 28, 2005 #8

    Fermat

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    Homework Helper

    You could draw it out in 3D.

    Draw out the x-, y-, and z-axes and plot the point P using the direction cosines.
    Since you have direction cosines for two separate points, A and B, the vectors from these points, AP and BP, will intersect at P.

    Once you have drawn out the 3D sketch, use trig to work out the height of P from the x-y plane.

    Alternatively, use the direction cosines to write out the point P as a vector, AP, and as a vector AP=AB + BP.
    Solve using vector algebra.
     
    Last edited: Aug 28, 2005
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