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The angle between a line and a plane

  1. Nov 27, 2014 #1
    1. The problem statement, all variables and given/known data
    In Exercises 45-46, show that the plane and line with the given equations intersect, and then find the acute angle of intersection between them.
    45. The plane given by x + y + 2z = 0 and the line given by
    x = 2 + t
    y = 1 - 2t
    z = 3 + t


    Verbatim from Poole - Linear Algebra: A Modern Introduction.
    2. Relevant equations
    [tex]\cos{\theta}=\frac{\vec{u}\cdot\vec{v}}{||u||\cdot{}||v||}[/tex]

    3. The attempt at a solution
    The normal vector of the plane is n = [1, 1, 2].
    The direction vector of the line is d = [1, -2, 1].
    The acute angle between the line and the plane should be the complement of
    [tex]\cos^{-1}{\frac{\vec{n}\cdot\vec{d}}{||n||\cdot{}||d||}}[/tex]
    [tex]\frac{\vec{n}\cdot\vec{d}}{||n||\cdot{}||d||}=\frac{1}{6}[/tex]
    The complement is, approximately, 9.59406822686046 degrees.

    The book says it is close to 80.4 degrees. (Not the complement, but theta on my calculations.)

    What am I doing wrong?
     
  2. jcsd
  3. Nov 27, 2014 #2

    ehild

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    Your solution is correct.
     
  4. Nov 27, 2014 #3
    I won't ask for confirmation.

    It seems that many reviewers also found errors in this book. I know that this is not the best place to ask, but could you point me a better textbook on the subject? I don't think I am willing to put up with more hundreds of pages and a all the mistakes that come with them.
     
  5. Nov 27, 2014 #4

    LCKurtz

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    Just to add a comment, since ##\hat n\cdot \hat d## came out positive, you know that both of those vectors point to the same side of the plane. That is why you take the complement of the ##\cos^{-1}##. Had the dot product come out negative, what would you have done?
     
  6. Nov 27, 2014 #5
    I would subtract [itex]\pi{}/2[/itex] from [itex]\cos^{-1}[/itex]. (Based on what I am "rendering" in my brain.)

    BTW, would you like to suggest a book?
     
  7. Dec 2, 2014 #6
    I did not misunderstand the term acute angle. But the calculation gives me the cosine of the angle between the vector and the normal of the plane, right? So if I want the acute angle between the vector and the plane I should get the complement of that angle.
     
  8. Dec 2, 2014 #7

    ehild

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    @Mark,
    The angle of a plane and a straight line is equal to the angle the line encloses with its projection onto the plane. See: http://www.netcomuk.co.uk/~jenolive/vect14.html
    cos-1(1/6) is the angle between the normal of the plane and the straight line. The angle between the plane and the line is 90°-cos-1(1/6), that is sin-1(1/6)
     
  9. Dec 2, 2014 #8
    Could you comment the answer I gave LCKurtz, he never came back.
     
  10. Dec 2, 2014 #9

    Mark44

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    You are right, and I apologize for misunderstanding what you're trying to do. I mistakenly thought you were trying to find the angle between the plane's normal and the line rather than the angle between the plane itself and the line.

    In answer to your question about other textbooks, here are a few that I have.
    • Linear Algebra and Differential Equations, Charles G. Cullen
    • An Introduction to Linear Algebra with Applications, Steven Roman
    • Linear Algebra, P. G. Kumpel and J.A. Thorpe
    • Elementary Linear Algebra, 5th Ed., Howard Anton

    I've had these since about the mid-90s, so they might be out in newer editions.
     
  11. Dec 2, 2014 #10

    Mark44

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    Yes, I misunderstood what was being asked here.
     
  12. Dec 2, 2014 #11
    Thanks, and no problem about that. But it was somewhat settled and I was only waiting for LCKurtz to comment my on my answer. Also thanks for the books, I guess the first one is a bit out of what I can grasp but will look into the others.
     
  13. Dec 2, 2014 #12

    ehild

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    It looks correct.
     
  14. Dec 2, 2014 #13
    Make up your mind, I am writing a "proof" here. I was distracted but I still think I got it right.
     
  15. Dec 2, 2014 #14

    Mark44

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    Using the original two vectors, you had ##cos(\theta) = 1/6##, from which we get ##\theta \approx 80.4°##. If ##cos(\theta) = -1/6##, you get ##\theta \approx 99.6°##. Does it make sense to subtract 90° here if you want the angle between the two vectors (not the angle between the line direction and the plane)?
     
  16. Dec 2, 2014 #15
    Imagine the "normalized" normal being [0, 1, 0]. Let the vector v be [a, b, c] where are b is negative. The dot product is b, therefore negative. The cos^-1 gives me the angle (obtuse) between the normal and v. Let [itex]\alpha[/itex] be this angle. Let [itex]\beta + \gamma = \alpha[/itex] and Let [itex]\beta[/itex] be the angle between the normal and the plane (90 degrees) and [itex]\gamma[/itex] the angle between the direction vector and the plane. I get rid of beta and all that remains is the angle between the plane and the vector.

    But I want the angle between the line and the plane (the angle of intersection), right?
     
  17. Dec 2, 2014 #16

    Mark44

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    Yes, in that case subtracting 90° gives you the angle between the plane and the line.

    In your example with a normal of <0, 1, 0>, the plane in question is parallel to the x-z plane. A simpler example in two-d might be helpful. Instead of <0, 1, 0> let's look at the vector <1, 0>, a vector that is normal to the y-axis. For the line direction, let's use <-1, 1>.

    Using the formula in your first post, which still applies here, we get ##cos(\theta) = -1/2## so ##\theta## = 120°. The acute angle that <-1, 1> makes with the normal is 180° - 120° = 60°. The angle that <-1, 1> makes with the y-axis (standing in for the plane in your example) is 120° - 90° = 30°.

    I hope everything is clear now.
     
  18. Dec 2, 2014 #17
    Yes it is (and somewhat was) clear. I just wanted to know that my answer to Kurtz was correct, and it was. I have read so many examples with xyz that I forgot to use just xy in my "demonstration".
     
  19. Dec 2, 2014 #18

    LCKurtz

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    Sorry I didn't get back to you mafagafo. I had to go to Chicago last week so was away for a while. But, yes, your answer to my question was correct.
     
  20. Dec 2, 2014 #19
    Thanks and no problem, it is nice that you commented another possible case.
     
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