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Calculus III Problem: find smallest angle

  1. Sep 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the smallest angle between 3x-2y-4z = 5 and {x = 1 + 2t, y = 3 + 4t, z = 5 + 6t}.


    2. Relevant equations
    3x-2y-4z=5 is a plane, the other 3 equations constitute a line.


    3. The attempt at a solution
    My understanding is that the angle between a plane and a line can be found by using the direction vector of the line and the normal vector of the plane, and applying the dot product formula,
    n . v = |n||v|cos Θ.
    Doing this, I get that theta is the inverse cosine of -29 / (square root of 29 * square root of 56), which comes out to 130.18 degrees. Since the smallest angle would be the acute angle, I subtracted this from 180 and got 49.82 degrees.

    Professor says this is the wrong answer. Can anyone help steer me in the right direction?
     
  2. jcsd
  3. Sep 17, 2014 #2

    Simon Bridge

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    You should look to the details of your working - could be as simple as an arithmetic mistake.
    Note: if n is the normal vector to the plane, and v is the vector pointing along the line, then arccos(n.v/|n||v|) gives the angle between n and v, not between v and the plane.
     
  4. Sep 18, 2014 #3
    Thanks, Simon. I have checked the arithmetic of n . v and also |n| and |v|, and those are correct. The inverse cosine of course I have to rely on my calculator for. :)

    I think what you're saying is that since the angle that results here is the angle between n and v, that the angle I found by subtracting from 180 is still the angle between n and v, just on the other "side" of the plane, and that this further needs to be subtracted from 90 to find the actual angle -- is that right??
     
  5. Sep 18, 2014 #4

    Simon Bridge

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    Well done - you should be able to verify this by a sketch.
     
  6. Sep 18, 2014 #5
    Great! Thank you!
     
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