Results 1 to 7 of 7

Math Help - Intersection of a plane and a normal.

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    100

    Intersection of a plane and a normal.

    Problem: A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.

    I haven't solved a question like this before in my book, so I am not too sure on how to go about this question. I know that the normal vector to 3x-y+4z=12 is \vec{n} = (3, -1, 4). So, to find the point of intersection, do I take the cross product of the normal and the given point? Or, would I have to find another point on the plane (I'll let x = 2 and y = 1 to get
    3(2) - (1)+4z = 12 => 4z = 7 => z = \frac{7}{4}, this point will be B(2, 1,\frac{7}{4}) and then find \vec{AB} and take the cross product of \vec{n} and \vec{AB}?

    I just never encountered a question like this before, so I would appreciate some help on getting started. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,801
    Thanks
    1691
    Awards
    1
    Find where the line
    \ell (t) = \left\{ {\begin{array}{*{20}c}   {x = 5 + 3t}  \\   {y = 1 - t}  \\   {z = 6 + 4t}  \\ \end{array} } \right.
    intersects the plane.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    100
    Quote Originally Posted by Plato View Post
    Find where the line
    \ell (t) = \left\{ {\begin{array}{*{20}c}   {x = 5 + 3t}  \\   {y = 1 - t}  \\   {z = 6 + 4t}  \\ \end{array} } \right.
    intersects the plane.
    So, substituting the normal (3, -1, 4):
    3 = 5 + 3t => -2 = 3t => x = -\frac{2}{3}
    -1 = 1 - t => y = -2
    4 = 6 + 4t => -2 = 4t => z = -\frac{1}{2}

    Therefore, the line intersects the plane at the point (-\frac{2}{3}, -2, -\frac{1}{2}, correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,996
    Thanks
    1656
    No, that's not the point. The normal is a vector, not an (x, y, z) point in the plane. Also note that you have made no use at all of the "= 12" part of the equation of the plane.

    Instead, put x= 5+ 3t, y= 1- t, and z= 6+ 4t in for the x, y, and z in the equation of the plane. Then solve for t and use that to find x, y, and z.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2010
    Posts
    100
    Ah, I see. I'll reattempt this:
    3(5 + 3t) - (1-t) + 4 (6 + 4t) = 12
    15 + 9t - 1 + t +24 + 16t = 12
    26t = -26
    t = -1
    Substituting this back into the original parametric equation:
    x = 5 + 3(-1) = 2
    y = 1 - (-1) = 2
    z = 6 + 4(-1) = 2
    Point of intersection of the normal and the plane is (2, 2, 2). Correct?
    Last edited by Pupil; June 4th 2011 at 05:40 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,996
    Thanks
    1656
    Check it yourself. Does x=2, y= 2, z= 2 satisfy the equation of the plane, 3x- y+ 4z= 12 (so that (2, 2, 2) is on theplane)? Does there exist a single value of t such that x= 5+ 3t= 2, y= 1- t= 2, and z= 6+ 4t= 2 (so that (2, 2, 2) is on the normal line)?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2010
    Posts
    100
    Quote Originally Posted by HallsofIvy View Post
    Check it yourself. Does x=2, y= 2, z= 2 satisfy the equation of the plane, 3x- y+ 4z= 12 (so that (2, 2, 2) is on theplane)? Does there exist a single value of t such that x= 5+ 3t= 2, y= 1- t= 2, and z= 6+ 4t= 2 (so that (2, 2, 2) is on the normal line)?
    Thank you for all the help and effort, HallsofIvy. I now understand perfectly.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intersection of a plane 3D Geo
    Posted in the Geometry Forum
    Replies: 3
    Last Post: April 29th 2011, 11:38 PM
  2. intersection of a plane
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 21st 2010, 02:38 AM
  3. normal plane to the curve intersecting a plane
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 23rd 2009, 09:53 AM
  4. Intersection of two plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 13th 2009, 02:30 PM
  5. Intersection of a Normal and a plane
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 23rd 2009, 09:29 AM

Search Tags


/mathhelpforum @mathhelpforum