Results 1 to 4 of 4

Math Help - Equation of the Plane

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    221

    Equation of the Plane

    I have to find the eq. of the plane which contains the following lines:

    x = 1 - t
    y = 2 + 3t
    z = 2t

    and

    x = 1 - s
    y = 5
    z = 4 - 2s

    So we have the vectors <-1, 3, 2> and <-1, 0, -2>

    I found the cross product which is <-6, -4, 3> which is my normal vector.

    Now what?

    I know a_1(x - x_1) + a_2(x - x_2) + a_3(x - x_3) = 0; I have my a_1, ..., a_3, but how do I get the points? Apparently the point is (1, 2, 0), but I don't see how to get that.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, Ideasman!

    You may kick yourself . . .


    I have to find the eq. of the plane which contains the following lines:

    . . \begin{Bmatrix}x \:= \:1 - t \\<br />
y \:= \:2 + 3t \\ z = 2t\end{Bmatrix} .and . \begin{Bmatrix}x \:= \:1 - s \\ y \:= \:5 \\ z \:= \:4 - 2s\end{Bmatrix}

    So we have the vectors \langle-1, 3, 2\rangle and \langle-1, 0, -2\rangle . . . Yes!

    I found the cross product which is \langle-6, -4, 3\rangle which is my normal vector. . . . Right!

    Now what?

    I know a_1(x - x_1) + a_2(x - x_2) + a_3(x - x_3) \:= \:0
    I have my a_1,\,a_2,\,a_3, but how do I get the points?
    Apparently the point is (1, 2, 0), but I don't see how to get that.

    Use any point on either line.

    For example, in the first line, let t =0 . . . got it?

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    221
    Of course!

    Got it, thanks Soroban.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Ideasman View Post
    I have to find the eq. of the plane which contains the following lines:

    x = 1 - t
    y = 2 + 3t
    z = 2t

    and

    x = 1 - s
    y = 5
    z = 4 - 2s

    So we have the vectors <-1, 3, 2> and <-1, 0, -2>

    I found the cross product which is <-6, -4, 3> which is my normal vector.

    Now what?

    I know a_1(x - x_1) + a_2(x - x_2) + a_3(x - x_3) = 0; I have my a_1, ..., a_3, but how do I get the points? Apparently the point is (1, 2, 0), but I don't see how to get that.
    You have two (non-linearly dependent) vectors, which lie in the plane if it
    exists (1,3,2)' and (-1,0,-2)'. Then in vector form any point in plane can be writen:

    (x,y,z)'=(a,b,c)' + lambda(1,3,2)' + mu(-1,0,-2),

    where (a,b,c)' is any point in the plane, so we can use (1,2,0), so our plane
    in parametric form is:

    (x,y,z)'=(1,2,0)' + lambda(1,3,2)' + mu(-1,0,-2).

    So from the y-component we have:

    lambda=(y-2)/3.

    From the z-component we have:

    mu=lambda-z/2=(y-2)/3 - z/2.

    Hence, from the x-component we have:

    x=1+lambda-mu=1+(y-2)/3-(y-2)/3+z/2=1+z/2

    or:

    2x-z=1.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 1st 2011, 07:57 PM
  2. Replies: 3
    Last Post: December 5th 2010, 05:46 PM
  3. Replies: 3
    Last Post: October 12th 2010, 05:06 AM
  4. Replies: 4
    Last Post: May 26th 2010, 10:48 AM
  5. Replies: 2
    Last Post: May 23rd 2010, 10:46 AM

Search Tags


/mathhelpforum @mathhelpforum