Results 1 to 4 of 4

Math Help - analytic geometry helpp !! plz it is urgent

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    26

    analytic geometry helpp !! plz it is urgent

    well i have some problems i have tried hard to solve them but wasnt succeed

    1- find the equation of the plane throw (2,-1,3) , (-3,4,3) and parallel to the vector <-5,-3,2>


    2- find the distance between the two planes 4x-3y+2z-9=0 .
    8x-6y+4z=0

    3-find the equation of the plane through (1,2,3) and perpendicular to
    3x-y+2z+5=0 , 4x+2y-3z+8=0

    help plz my exams is next week
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by mohamedsafy View Post
    ...

    1- find the equation of the plane throw (2,-1,3) , (-3,4,3) and parallel to the vector <-5,-3,2>


    ...
    to #1: You know one point of the plane and one vector (parallel to the plane). You need a second vector such that the 2 known vectors span the plane. The general equation is:

    \vec r = \vec a + s\cdot \vec u + t\cdot \vec v

    With your question:

    \vec a = (2, -1, 3)

    \vec u = (-5, -3, 2)

    \vec v = (2, -1, 3) - (-3, 4, 3) = (5, -5, 0) = k\cdot (1, -1, 0)

    Therefore the equation of the plane is:

    \vec r = (2, -1, 3) + s\cdot (-5, -3, 2) + t\cdot (1, -1, 0)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by mohamedsafy View Post
    ...


    2- find the distance between the two planes 4x-3y+2z-9=0 .
    8x-6y+4z=0

    ...
    The equations of both planes are in normal form. They are parallel because (8, -6, 4) = 2\cdot (4,-3,2)

    The second plane passes through the origin. You only have to calculate the distance of the origin to the first plane:

    Re-write the first equation in Hesse normal form, that means divide the equation by |\vec n| = \sqrt{29}

    Then the distance of the origin to the first plane is d = \dfrac{9}{\sqrt{29}}\approx 1.67
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by mohamedsafy View Post
    ...

    3-find the equation of the plane through (1,2,3) and perpendicular to
    3x-y+2z+5=0 , 4x+2y-3z+8=0

    ...
    You need a vector which is perpendicular to both planes:

    \vec u = (3, -1, 2) \times (4, 2, -3) = (-1, 17, 10)

    Write the equation of the plane in normal form:

    (-1, 17, 10)\cdot ((x, y, z) - (1,2,3))=0~\implies~-(x-1)+17(y-2) + 10(z-3)=0 ~\implies~-x+17y+10z-63=0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Analytic Geometry
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 18th 2010, 09:49 AM
  2. Analytic Geometry Q10
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: March 4th 2009, 05:05 AM
  3. analytic geometry
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 6th 2009, 02:19 PM
  4. Analytic geometry
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 11th 2008, 05:56 AM
  5. please help on my analytic geometry
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 25th 2006, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum