Results 1 to 5 of 5

Math Help - Find value of c; Plane perpendicular to the line

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    4

    Find value of c; Plane perpendicular to the line

    Question:

    Find the value of c so the plane perpendicular to the line x =3+t y=4t z=-4-t that contains the point (1,5,6) also contains the point (c,3c,5c).

    Attempt at solution:

    First I found the vector of the line:

    Point P when t =0 (3,0,-4)
    Point Q when t=1 (4,4,-5)

    Vector PQ <4-3,4-0,-5-(-4)> --> <1,4,-1>

    So to get the plane through the first point I just used the vector and the point to get => 1(x-1)+4(y-5)-1(z-6) = 0

    Then I used the same vector and the second point to get =>

    1(x-c)+4(y-3c)-1(z-5c) = 0

    This is where I got stuck. I set the two equations equal to each other:


    1(x-1)+4(y-5)-1(z-6) =1(x-c)+4(y-3c)-1(z-5c)


    ...but in order to solve for c I need to know what x, y, and z are supposed to be. I would substitute the parametric equations for x, y, and z, but that would only be if I was solving for t, right?

    Any suggestions? Thanks

    ----------------

    **Update: I realized I could cancel out all of the x,y, and z terms to give me c =15/8 can anyone tell me if I took the right approach to solving this problem? Thanks

    I meant to put 15/8
    Last edited by CocoLily; November 14th 2009 at 01:37 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by CocoLily View Post
    Question:

    Find the value of c so the plane perpendicular to the line x =3+t y=4t z=-4-t that contains the point (1,5,6) also contains the point (c,3c,5c).
    The equation of that plane is (x-1)+4(y-5)-(z-6)=0. WHY?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    4
    Quote Originally Posted by Plato View Post
    The equation of that plane is (x-1)+4(y-5)-(z-6)=0. WHY?

    Equation of plane a(x-x0)+b(y-y0)+c(z-z0)=0 when

    vector = <a,b,c> = <1,4,-1> and point = (x0,y0,z0) = (1,5,6)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by CocoLily View Post
    Equation of plane a(x-x0)+b(y-y0)+c(z-z0)=0 when

    vector = <a,b,c> = <1,4,-1> and point = (x0,y0,z0) = (1,5,6)
    So solve for c.
    (c-1)+4(3c-5)-(5c-6)=0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    4
    Quote Originally Posted by Plato View Post
    So solve for c.
    (c-1)+4(3c-5)-(5c-6)=0.

    Ah okay cool, I think I did it right. Your approach is a lot easier though, haha!
    Last edited by CocoLily; November 14th 2009 at 01:38 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Point that is perpendicular to a line of a plane?
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 4th 2011, 06:55 AM
  2. Replies: 5
    Last Post: March 22nd 2011, 06:50 AM
  3. Vector equation of line perpendicular to plane
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 4th 2010, 03:20 AM
  4. Vector Algebra (line perpendicular to plane)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2009, 02:30 AM
  5. Replies: 3
    Last Post: September 21st 2008, 07:00 PM

Search Tags


/mathhelpforum @mathhelpforum