1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

No idea how to find the equation of a plane

  1. Aug 12, 2008 #1
    would this be ok for this question because i thought we had to find normals and things and every site i go on about plane equations is different and i need to know this for an exam tomorrow. find the equation of a plane that passes through the points (2,-2,1)(4,-1,6)(3,-3,-2)

    (1)2a-2b+c=d
    (2)4a-b+6c=d
    (3)3a-3b-2c=d

    then re arrange somehow to get values in terms of d for a b and c and then what? infact i have no clue what im talking about i have no idea how to find the equation of a plane
     
  2. jcsd
  3. Aug 12, 2008 #2
    Re: confused

    scrap that lets use the cross product methord using PQ=(0,1,-1) and PR=(-2,1,0) giving a normal vector of (1,2,2) after this it tells me

    "(Q - P) x (R - P) = (1, 2, 2) = normal vector A and the equation is A . X = d for some d

    Using the method in the example above, we can find d = A . P = 5. Thus the equation is A . X = 5, which is the same as one of the equations in the earlier example."

    what the heck does this mean? i get everything up to finding the normal vector then it just all sounds forign
     
  4. Aug 12, 2008 #3
    Re: confused

    hang on so does this mean that A.P=D and then we just stick the normal vector in as (A,B,C) using the formula Ax+By+Cz=D giving us an equation of x + 2y + 2Z=5 also if this is the case can someone check my one below that im about to do
     
  5. Aug 12, 2008 #4
    Re: confused

    1) find two direction vectors
    e.g. (4,-1,6) - (2,-2,1)

    2) find normal by taking cross product those two vectors

    3) same problem as you asked before
     
  6. Aug 12, 2008 #5
    Re: confused

    find eqn of plane through points (2,-2,1)(4,-1,6)(3,-3,-2)

    normal is cross product of (b-a)(2,1,5)X(b-c)(1,2,8)=-2,11,3 so then we have N.A=-15 and we get -2x+11y+3z=-15??
     
  7. Aug 12, 2008 #6
    Re: confused

    i know its the same as before but i still dont get it from before lol
     
  8. Aug 12, 2008 #7
    Re: confused

    and ye your right they did just assume i knew this even though ive never studdied discrete maths
     
  9. Aug 12, 2008 #8
    Re: confused

    >> p1 = [2,-2,1];
    >> p2 = [4,-1,6];
    >> p3 = [3,-3,-2]; %pos vectors
    >> d1 = p2-p1;
    >> d2=p3-p2; %direction vectors
    >> n = cross(d1,d2) %finding normal
    n =

    2 11 -3 %output
    >> dot(n,p1)

    ans =

    -21

    >> dot(n,p2)

    ans =

    -21

    >> dot(n,p3)

    ans =

    -21


    =============
    so everything is right but your d (above is matlab code ... - makes life easier - didn't have casio near which also do dots and cross)
     
  10. Aug 12, 2008 #9
    Re: confused

    isnt it 2,-11,-3 and then when we get that do we do N.P1=D=23 so then the equation is 2x-11y-3z=23?
     
  11. Aug 12, 2008 #10
    Re: confused

    You must have done something wrong.

    When we find n then

    n dot any pos vector = should give equal value ..

    >> n=[2,-11,-3];
    >> dot(n,p1)

    ans =

    23

    >> dot(n,p2)

    ans =

    1

    >> dot(n,p3)

    ans =

    45
     
  12. Aug 12, 2008 #11
    Re: confused

    ok ye just checked over it math error, apart from that tho is the rest of my working right subbing in to get the Ax+By+Cz=D?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: No idea how to find the equation of a plane
Loading...