Results 1 to 6 of 6

Math Help - Plane....

  1. #1
    Member
    Joined
    May 2007
    Posts
    150

    Plane....

    Write an equation for the plane that contains the point A = (4,5,-3) and that is perpendicular to the line through B = (5,-2,-2) and C = (7,1,4)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1582
    Awards
    1
    Quote Originally Posted by stones44 View Post
    Write an equation for the plane that contains the point A = (4,5,-3) and that is perpendicular to the line through B = (5,-2,-2) and C = (7,1,4)
    Write the equation of the plane with normal \overrightarrow {BC} and contains the point A.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by stones44 View Post
    Write an equation for the plane that contains the point A = (4,5,-3) and that is perpendicular to the line through B = (5,-2,-2) and C = (7,1,4)
    one way to write the equation of a plane is:

    ax + by + cz + d = 0

    here, \bold{n} = \left< a,b,c \right> is the normal vector to the plane. here, your normal vector is the vector \overrightarrow {BC}, do you see why?

    we begin with a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

    where (x_0,y_0,z_0) is a point the plane passes through. then we expand and simplify
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2007
    Posts
    150
    Quote Originally Posted by Jhevon View Post
    one way to write the equation of a plane is:

    ax + by + cz + d = 0

    here, \bold{n} = \left< a,b,c \right> is the normal vector to the plane. here, your normal vector is the vector \overrightarrow {BC}, do you see why?

    we begin with a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

    where (x_0,y_0,z_0) is a point the plane passes through. then we expand and simplify
    2(x-5) + 3(y-4) + 2(z+3) = 0
    2x-10+3y-12+2z-6 = 0
    2x + 3y + 2z = 28
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by stones44 View Post
    2(x-5) + 3(y-4) + 2(z+3) = 0
    2x-10+3y-12+2z-6 = 0
    2x + 3y + 2z = 28
    here you have n = <2,3,2>, that is incorrect
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1582
    Awards
    1
    There is a minor mistake.
    \overrightarrow {BC}  = \left\langle {2,3,6} \right\rangle
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 10:59 AM
  2. Replies: 3
    Last Post: December 5th 2010, 05:46 PM
  3. Replies: 2
    Last Post: May 9th 2009, 10:35 AM
  4. Replies: 2
    Last Post: December 3rd 2008, 06:26 PM
  5. Replies: 0
    Last Post: November 19th 2008, 06:04 AM

Search Tags


/mathhelpforum @mathhelpforum