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Finding Cosine of an Angle in 3D space

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data

    a) Find the cosine of the angles that the line r = [-3,2,5] + t [2,2,√2 ] makes with the coordinate axes.

    b) If a,b and c are the angles that the line makes with the x,y and z axis respectively, find the value of cos^2a + cos^2b + cos^2c.

    c) What is the magnitude of the vector [cosa,cosb,cosc]

    d) What can you say about the direction of the vector [cosa,cosb,cosc] and the direction of the vector line?


    2. Relevant equations

    I believe we have to use CosTheta = r . t / |r||t| but it doesn't seem right because this is a vector equation involving direction.
     
    Last edited by a moderator: Jan 29, 2012
  2. jcsd
  3. Jan 28, 2012 #2

    gneill

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    Actually, you'll want to use the dot product of a vector that is parallel to the line with unit vectors on your coordinate axes. So if you construct vector [itex] \vec{r} [/itex] to lie parallel to your line, then
    [tex] cosa = \frac{\vec{r} \cdot \hat{i}}{|\vec{r}||\hat{i}|} [/tex]
    and so on for the unit vectors [itex] \hat{i},\hat{j},\hat{k} [/itex] to yield cosa, cosb, cosc.
    Yes, there will be three angle cosines to find, one for each axes of the coordinate system.

    Your first step should be to find a vector r that is parallel to your line.
     
  4. Jan 28, 2012 #3
    How do I find a the line parallel to R?
     
  5. Jan 28, 2012 #4

    gneill

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    You have a line given in parametric form: [-3,2,5] + t [2,2,√2 ]. The [-3,2,5] bit is just an offset. You could shift the line to pass through the origin by subtracting this offset, and the result would obviously be a line that is parallel to the original. Thus the line r' = t [2,2,√2 ] is parallel to the original line. Plug in any value of t that you like (1 is a good choice!) to find a vector that lies in the line.
     
  6. Jan 28, 2012 #5

    SammyS

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    Hello hunt3rshadow. Welcome to PF !

    Usually you should post no more than two questions when opening a thread. These all look fairly closely related, so I suppose it's OK.

    I'll try to get you headed in the right direction, but it's up to you to actually solve the problem(s).

    For (a):
    The vector, r = [-3,2,5] + t [2,2,√2 ], is a position vector having its tail at the origin, and its head it the point (x, y, z) = ( -3 + 2t, 2 + 2t, 5 + (√2)t ), i.e.
    x = -3 + 2t,
    y = 2 + 2t,
    z = 5 + (√2)t .​
    As t goes from -∞ to +∞, the head of vector r(t) traces a straight line. The vector [-3,2,5], has nothing to do with the direction of this line. It only specifies the location of a point, (-3,2,5), on the line that corresponds to t = 0. The direction is given by the vector that t multiplies, namely, [2,2,(√2) ]. Note that this is also equal to dr/dt.
    [itex]\displaystyle\frac{d\textbf{r}}{dt}=\ \left<2,2,\sqrt{2}\ \right>[/itex]​

    You have to find the cosine of the angle this vector makes with respect to (w.r.t) the x-axis, ... then the cosine of the angle this vector makes w.r.t. the y-axis,... then the cosine of the angle this vector makes w.r.t. the z-axis.

    The x-axis is in the direction of [1, 0, 0].
    This is where you use that equation you have:

    [itex]\displaystyle\cos(\theta)=\frac{ \left\langle 2,\,2,\,\sqrt{2} \right\rangle\cdot \left\langle 1,\,0,\,0 \right\rangle}{\left| \left\langle 2,\,2,\,\sqrt{2} \right\rangle \right|\ \left| \left\langle 1,\,0,\,0 \right\rangle \right|}[/itex]

    Do similarly for the other two axes.
     
  7. Jan 28, 2012 #6
    Oh my gosh guys I really owe you! You two made it seem so easy :cry:
     
    Last edited: Jan 29, 2012
  8. Jan 28, 2012 #7

    SammyS

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    A vector parallel to the line is the vector < 2, 2, (√2) > . That's basically the dr/dt vector I discussed.
     
  9. Jan 28, 2012 #8
    Ahh I see, thanks.
     
    Last edited: Jan 29, 2012
  10. Jan 28, 2012 #9

    gneill

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    Just do as the formula states and add the sum of the squares of the cosines of the angles; you don't need to find the angles themselves, just use the cosines. The result should be particularly satisfying :wink:.
     
  11. Jan 28, 2012 #10
    so much more simple then I imagined! Thanks
     
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