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Direction cosines

  1. Sep 12, 2015 #1
    • Member warned about posting with no effort
    1. The problem statement, all variables and given/known data
    To find the direction cosine of a equation say (4x+5y+7z=13)

    2. Relevant equations

    Im not really sure what to do

    3. The attempt at a solution
    (I know this is really basic but i would be glad if someone helps me with this
     
  2. jcsd
  3. Sep 12, 2015 #2

    SammyS

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    Is that the complete question?

    What have you tried? Where are you stuck ?

    What is that the equation of?

    How can such an object have direction cosines?
     
  4. Sep 12, 2015 #3

    SteamKing

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    Looks like the equation for a plane. How do you usually find a normal to a plane?
     
  5. Sep 14, 2015 #4

    HallsofIvy

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    The problem appears to be that the OP really does NOT understand the basics of three dimensional lines. Rahul, as SteamKing said, the equation you give defines a plane in three dimensions, not a line. The "direction cosines" of a line in three dimensions are the cosines of the angles the line makes with lines parallel to the three coordinate axes. In addition, They are the dot products of a unit vector in the direction of the line with unit vectors in the directions of the three coordinate axes. In particular, if a line is given in parametric form, x= at+ b, y= ct+ d, z= et+ f, then the vector ai+ bj+ ck is in the direction of the line. Dividing by [itex]\sqrt{a^2+ b^2+ c^2}[/itex] is gives a unit vector in that direction. In other words, the three direction cosines are [itex]\frac{a}{\sqrt{a^2+ b^2+ c^2}}[/itex], [itex]\frac{b}{\sqrt{a^2+ b^2+ c^2}}[/itex], and [itex]\frac{c}{\sqrt{a^2+ b^2+ c^2}}[/itex].
     
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