Results 1 to 12 of 12

Math Help - Intersecting Planes

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    60

    Intersecting Planes

    I seem to be having a problem here finding the intersection of the 2 planes and I know these planes intersect at a line. Here are the equations of the 2 planes:

    3x + 5y + 6z = 21

    2x - y + 4z = 14

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

    Now, the answer I get is:

    x = 7 - 2t

    y = 0

    z = \frac{7}{4} + 2t

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

    This is apparently incorrect. Could someone please show me how to get the correct answer - which is:

    x = 3 - 2t

    y = 0

    z = 2 + t

    ====================================

    Thank you. All help is appreciated.
    Last edited by sqleung; August 11th 2008 at 01:03 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Are you sure that answer is correct?

    The line certainly lies in the second plane but not necessarily the first. Example, set t = 2:

    3(3-2(2)) + 5(0) + (2 + (2)) = 1 \neq 21

    Works for the second plane though: 2(3-2(2))  - (0) + 4(2+(2)) = 14
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    60
    Quote Originally Posted by o_O View Post
    Are you sure that answer is correct?

    The line certainly lies in the second plane but not necessarily the first. Example, set t = 2:

    3(3-2(2)) + 5(0) + (2 + (2)) = 1 \neq 21

    Works for the second plane though: 2(3-2(2))  - (0) + 4(2+(2)) = 14
    I thought the given answer was a little strange because it didn't seem correct...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    In any case, let's try to find the correct answer.

    The normals of both planes are: \vec{n}_{1} = (3,5,1) and \vec{n}_{2} = (2, -1, 4).

    Now, the cross product of these two normals will give the direction vector of the line. Any vector perpendicular to the normal of the first plane will lie in the first plane and any vector perpendicular to the normal of the second plane will lie in the second plane. Since the cross product of these two normals is perpendicular to both, it will lie in the line of intersection of both planes.

    So, (3, 5, 1) \times (2, -1, 4) = (a,b,c) (Leaving it up to you to find a, b, and c). This will be our direction vector.

    Now we have to find a point on this line (recalling that a line can be defined by a point and a direction vector; we have the latter). One way to do this is set one of the variables to a specific constant, say y = 0, and solve for the remaining system of equations:

    \begin{array}{ccccc} 3x & + & z  & = & 21 \\ 2x & + & 4z& = &14\end{array}

    After solving for the remaining your variables, you will have a point (call it (x_{0}, y_{0}, z_{0})).

    Now put it together and we have your equation of a line: \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix} + t\begin{pmatrix} a \\ b \\ c\end{pmatrix}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2008
    Posts
    60
    Quote Originally Posted by sqleung View Post
    I seem to be having a problem here finding the intersection of the 2 planes and I know these planes intersect at a line. Here are the equations of the 2 planes:

    3x + 5y + z = 21

    2x - y + 4z = 14

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

    Now, the answer I get is:

    x = 7 - 2t

    y = 0

    z = \frac{7}{4} + 2t

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

    This is apparently incorrect. Could someone please show me how to get the correct answer - which is:

    x = 3 - 2t

    y = 0

    z = 2 + t

    ====================================

    Thank you. All help is appreciated.
    Ah darn...looks like I made a slight mistake:

    It should be 3x + 5y + 6z = 21, NOT 3x + 5y + z = 21

    So the equations are:

    3x + 5y + 6z = 21
    2x - y + 4z = 14

    Sorry about that.

    ---------------------------
    Last edited by sqleung; August 11th 2008 at 01:12 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by sqleung View Post
    Ah darn...looks like I made a slight mistake:

    It should be 3x + 5y + 6z = 21, NOT 3x + 5y + z = 21

    So the equations are:

    3x + 5y + 6z = 21
    2x - y + 4z = 14

    Sorry about that.
    It doens't change much of o_O's setup.

    It changes the outcome for \bold n_1\times \bold n_2=\left<a,b,c\right>..

    The system of equations would become

    3x+6z=21
    2x+4z=14

    the values you get here for x and z, and the point y=0 will give you the point on the line P(x_0,y_0,z_0)

    Thus, the line will have the form \left<x_0,y_0,z_0\right>+t\left<a,b,c\right>

    Does this make sense?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2008
    Posts
    60
    The original question was to find the equation of the plane through A(3,0,2) , B(1,0,3) , C(2,-3,5) and from that, I got 3x + 5y + 6z = 21. I got the direction vector as [3,5,6] which is DEFINITELY correct.

    Next I'm supposed to find where that plane meets with another plane given the equation of the other plane is 2x - y + 4z = 14

    And I seem to get the wrong answer...does there seem to be something wrong with the equation I got?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    I'm guessing you got (3,5,6) from taking the cross product of AB and AC or something similar (this is the normal of the plane btw)? Well unfortunately, early on in your steps:

    \vec{AB} \times \vec{AC} = (-2, 0, 1) \times (-1, -3, 3) = \ldots = (3, {\color{red}-5}, 6)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Feb 2008
    Posts
    60
    Quote Originally Posted by o_O View Post
    I'm guessing you got (3,5,6) from taking the cross product of AB and AC or something similar (this is the normal of the plane btw)? Well unfortunately, early on in your steps:

    \vec{AB} \times \vec{AC} = (-2, 0, 1) \times (-1, -3, 3) = \ldots = (3, {\color{red}-5}, 6)
    No offence but -5 does not seem correct:

    [3 , {\color{red}1(-1) - (-2)(3)} , 6]

    [3 , {\color{red}5} , 6]

    I guess I can just reach a conclusion that the answers provided to me on the sheet are wrong...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by o_O View Post
    (-2, 0, 1) \times (-1, -3, 3) = \left( \left| \begin{matrix}0 & 1 \\ -3 & 3\end{matrix}\right|, {\color{red}-}\left| \begin{matrix}-2 & 1 \\ -1 & 3\end{matrix}\right|, \left| \begin{matrix}-2 & 0 \\ -1 & -3\end{matrix}\right|\right)

    = \bigg( (0)(3) - (1)(-3), \: \:{\color{red}-[}(-2)(3) - (-1)(1){\color{red}]}, \: \: (-2)(-3) - (-1)(0)\bigg) \: = \: (3, {\color{red}5}, 6)
    You forgot the negative...

    --Chris
    Follow Math Help Forum on Facebook and Google+

  11. #11
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Ah rookie mistake
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,956
    Thanks
    1780
    Awards
    1
    Quote Originally Posted by sqleung View Post
    I seem to be having a problem here finding the intersection of the 2 planes and I know these planes intersect at a line. Here are the equations of the 2 planes:
    3x + 5y + 6z = 21
    2x - y + 4z = 14
    If this is the correct question, the the line of intersection is:
    \left\{ {\begin{array}{lll}   {x = 7 + 2t}  \\   {y = 0}  \\<br />
   {z = - t}  \\\end{array}} \right.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] 2 Intersecting 3-D planes
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 7th 2011, 06:27 PM
  2. Intersecting planes
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 27th 2011, 10:19 AM
  3. Parallel planes and intersecting planes
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 8th 2010, 09:25 AM
  4. 3D: 2 planes intersecting
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 25th 2010, 08:24 PM
  5. Intersecting Planes
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 2nd 2008, 01:08 AM

Search Tags


/mathhelpforum @mathhelpforum