Results 1 to 6 of 6

Math Help - finding the equation of the line which pasees..

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    53

    finding the equation of the line which pasees..

    through the point (4,-3.2) and is parallel to the line of intersection of the planes: i) 2x +3y + 2z = 6 and ii) 4x - 3y + z =11

    pre.s. i pologise if this does n't belong to this forum and would appreciate your help directing my question to the right forum in this case

    Points P(0,1,3/2), Q(2,0,1), Z(1,4/3,0) are on the first plane.
    vector PQ is <2,-1,-1/2> and QZ is <-1,4/3,-1> by forming their cross product I find the normal to the first plane to be <1,3,2> .

    Points P'(0,1,14), Q'(5/2,0,1), Z'(1,-7/3,0) are on the second plane.
    vector P'Q' is <5/2,-1,-13> and Q'Z' is <-3/2,-7/3,-1> by forming their cross product I find the normal to the second plane to be <-4,3,-1>

    By forming the cross product of the normals I find the direction vector of the line i am interested in to be <3,1,-9>

    and hence the cartesian equation of the wanted line to be
    (x-4)/3 = y + 3 = (z-2)/-9

    However the book gives (x-4)/3 = (y + 3) = (z-2)/-6

    I have gone through it a miriad times and can't find my mistake I would absolutely appreciate it if I can get some help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member enjam's Avatar
    Joined
    Jul 2009
    Posts
    26
    2x + 3y + 2z = 6
    X intercept occurs when y and z = 0, so 2x = 6, x = 3
    Y intercept occurs when x and z = 0, so 3y = 6, y = 2
    Z intercept occurs when x and y = 0, so 2z = 6, z = 3

    For the first question, in vector form, we have:
    r = ro + tv
    = (4i - 3j + 2k) + t (3i + 2j + 3k)
    And for the second question, the line in vector form would be:
    = (4i - 3j + 2k) + t (11/4i + -11/3j + 11z)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    53
    Emian, I don't know what you have done, but the question is asking for one equation... you have two... are you sure you have solved the same exercise
    ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1
    Quote Originally Posted by pepsi View Post
    through the point (4,-3.2) and is parallel to the line of intersection of the planes: i) 2x +3y + 2z = 6 and ii) 4x - 3y + z =11
    The book gives (x-4)/3 = (y + 3) = (z-2)/-6
    If you have given the correct equations for the planes, then the given answer is off.
    The direction vector of the line of intersection is <br />
\frac{{\left\langle {2,3,2} \right\rangle  \times \left\langle {4, - 3,1} \right\rangle }}<br />
{3} = \left\langle {3,{\color{blue}2}, - 6} \right\rangle .

    The direction number in blue is different.
    Last edited by Plato; August 26th 2009 at 08:17 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,572
    Thanks
    1412
    Quote Originally Posted by pepsi View Post
    through the point (4,-3.2) and is parallel to the line of intersection of the planes: i) 2x +3y + 2z = 6 and ii) 4x - 3y + z =11

    pre.s. i pologise if this does n't belong to this forum and would appreciate your help directing my question to the right forum in this case

    Points P(0,1,3/2), Q(2,0,1), Z(1,4/3,0) are on the first plane.
    vector PQ is <2,-1,-1/2> and QZ is <-1,4/3,-1> by forming their cross product I find the normal to the first plane to be <1,3,2> .
    You don't need to find two vectors in the plane and take their cross product to find a normal vector to a plane. A normal vector to the plane Ax+ By+ Cz= D is <A, B, C> so a normal to the plane 2x+3y+ 2z= 6 is <2, 3, 2>, not <1, 3, 2>. Your vectors PQ and QZ are correct but their cross product is <5/3, 5/2, 5/3>= (5/6)<2, 3, 2>.

    Points P'(0,1,14), Q'(5/2,0,1), Z'(1,-7/3,0) are on the second plane.
    vector P'Q' is <5/2,-1,-13> and Q'Z' is <-3/2,-7/3,-1> by forming their cross product I find the normal to the second plane to be <-4,3,-1>
    Similarly, a normal to 4x-3y+ z= 11 is <4, -3, 1>, equivalent to <-4, 3, -1> so that is correct.

    By forming the cross product of the normals I find the direction vector of the line i am interested in to be <3,1,-9> and hence the cartesian equation of the wanted line to be
    (x-4)/3 = y + 3 = (z-2)/-9

    However the book gives (x-4)/3 = (y + 3) = (z-2)/-6

    I have gone through it a miriad times and can't find my mistake I would absolutely appreciate it if I can get some help
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2009
    Posts
    53
    I much appreciate you help, its been real helpful..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding equation of a line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 26th 2009, 08:56 PM
  2. Finding the equation of the line.....
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: July 5th 2009, 03:55 AM
  3. Finding an equation of a line
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 14th 2007, 01:17 AM
  4. Finding the Equation of a Line
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 21st 2007, 04:30 AM
  5. Finding equation of a line
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 17th 2006, 01:38 PM

Search Tags


/mathhelpforum @mathhelpforum