Results 1 to 2 of 2

Math Help - Vector equation of line with angle to a plane

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    22

    Vector equation of line with angle to a plane

    Problem is: Find the vector equation of the line r that contains the point P (1, 1, 1), is parallel to the plane x + 2y - z = 0 and presents a pi/3 radians angle with the plane x - y + 2z - 1 = 0

    I let the direction vector of r be (a, b, c) and consider that (a, b, c).(1, 2, -1) = 0, since it's parallel to the first plane. Also, sqrt(3)/2 = (a, b, c).(1, -1, 2) / |a, b, c| times sqrt(6). But then, I'm stuck at a three-variable system with two equations and also a modulus.

    How am I supposed to do that? Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2009
    Posts
    95
    Consider the information that you are given carefully, I will help.

    The line r is parallel to the plane P_1 defined by \vec{x}\cdot (1,2,-1) = 0. This means that it lies in P_1 and as you have correctly deduced, if we let the direction vector of r be \vec{a}=(a,b,c) then we require that \vec{a} \cdot (1,2,-1) = 0. Let P_2 be the plane defined by \vec{x} \cdot(1,-1,2) = 1.

    Now consider the intersection of P_1 and P_2, P_1 \cap P_2. Clearly P_1 \cap P_2 is a line with direction vector \vec{b}=(1,2,-1) \times (1,-1,2), and furthermore P_1 \cap P_2 is parallel with both P_1 and P_2. For r to have \pi / 3 angle with P_2 we must have that r and P_1 \cap P_2 have angle \pi / 3 since r is parallel to P_1, and P_1 \cap P_2 is parallel with both P_1 and P_2.

    Thus, we require that \frac{\vec{a} \cdot \vec{b}}{\parallel \vec{a} \parallel \parallel \vec{b} \parallel} = \cos (  \pi / 3 ).

    You have two conditions so far:

    1) \vec{a} \cdot (1,2,-1) = 0
    2) \frac{\vec{a} \cdot \vec{b}}{\parallel \vec{a} \parallel \parallel \vec{b} \parallel} = \cos ( \pi / 3 ).

    The final (third) condition on \vec{a} is rather artificial. Lets put the condition on \vec{a} so that we find a unit vector, that is the 3 conditions are then

    1) \vec{a} \cdot (1,2,-1) = 0
    2) \frac{\vec{a} \cdot \vec{b}}{ \parallel \vec{b} \parallel} = \cos ( \pi / 3 ).
    3) \left\parallel \vec{a} \right\parallel = 1.

    I think that what I wrote above is correct; however, I am not sure why you are given the point through which the line r must pass through, I believe its a red herring.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Angle of intersection of the line and the plane
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: March 29th 2011, 11:58 AM
  2. Vector equation of line perpendicular to plane
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 4th 2010, 04:20 AM
  3. Angle between line and plane
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 17th 2009, 05:49 PM
  4. vector equation of intersection of line and plane
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 1st 2009, 01:58 AM
  5. Replies: 8
    Last Post: June 7th 2008, 10:09 PM

Search Tags


/mathhelpforum @mathhelpforum