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Intersection of line and plane

  1. Aug 14, 2011 #1
    Intersection of line and plane!!!!!!!!!!!!!

    1. The problem statement, all variables and given/known data

    Intersection of line and plane!!!!!!!!!!!!!
    Okay i to find the common points of line and plane

    Question r=i+j+A(2k-j) and r . (i+j) = 4

    2. Relevant equations

    I heard that it is easier to use the vector equation in the form r . n = p

    3. The attempt at a solution

    So if i do that to above

    You get

    1 0 1

    1 + A -1 . 1 =4

    0 2 0

    The above doesn't sound right i wont get an x, y, z value i defo need help on this or can anyone start my off

    kind regards

    And also how do you wrap the vectors in brackets on this forum lol
     
  2. jcsd
  3. Aug 14, 2011 #2

    HallsofIvy

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    Re: Intersection of line and plane!!!!!!!!!!!!!

    It's not clear to me what the first and third lines are but it looks like the second line is almost the result of putting r=i+j+A(2k-j)= i+ (1- A)jk+ 2Ak into the equation r . (i+j) = 4.
    Rather than what you have, you should get 1+ (1- A)= 2- A= 4.

    Solve That for A, then put that value into the equation of the line.


     
  4. Aug 14, 2011 #3
    Re: Intersection of line and plane!!!!!!!!!!!!!

    Sorry about the way its laid out. your right thats what i have done

    i put r=i+j+A(2k-j) in to r . (i+j) = 4

    which gives me

    i+j+A(2k-j) . (i+j) = 4

    I tried writing this in vectorial format but dont khow how to get the brackets to wrap i,j,k on this website. If you can help with that i would be gratefull

    Anyway to continue on with question we break them down in components. Correct me if im wrong

    1+A(0) . 1=4
    1+A(-1) . 1=4
    0+A(2) . 0=4

    Which is where i got previously. Which is probably wrong

    Now to continue with what you did im trying to work out how you derived the above.

    I broke the vector r=i+j+A(2k-j) into its components ready to put the A values in

    so you get

    1+A(0)=X This gives x=1
    1+A(-1)=Y
    0+A(2)=Z

    Just need to know how you derived A whcih i quotes above. If you did put put r=i+j+A(2k-j) in to r . (i+j) = 4 To get the equation i+j+A(2k-j) . (i+j) = 4. And from this you managed to get A. Can you please explain to me step by step please

    Thanks for quick reply and Kinds regards
     
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