Results 1 to 2 of 2

Math Help - Vectors and Lines

  1. #1
    Junior Member
    Joined
    May 2011
    Posts
    30

    Vectors and Lines

    Hi guys! Need help with this question please,

    The plane pi1 has vector EQN: r . (2, 2, 1) = 13. The points P and Q, which are not in pi1, have position vectors 4i + 5j + 7k and 10i + 8j + k respectively, and the line L1, which passes through P and Q, meets pi1 at point R. Find the position vector of R.
    The line L2, is contained in pi1, passes through R and is perpendicular to L1. Find a vector which is parallel to L2.

    I found OR = (-2, 2, 13)
    Since L2 is perpendicular to L1,
    Let F be the foot of perpendicular from L2 to L1.
    OF lies on L1 => OF = (4, 5, 7) + s(2, 1, -2) for some s E R
    OF = (4+2s, 5+s, 7-2s)
    RF = (4+2s, 5+s, 7-2s) - (-2, 2, 13) = (6+2s, 3+s, -6-2s)
    RF . (2, 1, -2) = 0 (since L1 and L2 are perpendicular)
    Solving, s = -3
    Then sub. s = -3 into RF which I ended up with a 0?

    Any idea why this method doesnt work?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123

    Re: Vectors and Lines

    Quote Originally Posted by Blizzardy View Post
    Hi guys! Need help with this question please,

    The plane pi1 has vector EQN: r . (2, 2, 1) = 13. The points P and Q, which are not in pi1, have position vectors 4i + 5j + 7k and 10i + 8j + k respectively, and the line L1, which passes through P and Q, meets pi1 at point R. Find the position vector of R.
    The line L2, is contained in pi1, passes through R and is perpendicular to L1. Find a vector which is parallel to L2.

    I found OR = (-2, 2, 13)
    Since L2 is perpendicular to L1,
    Let F be the foot of perpendicular from L2 to L1.
    OF lies on L1 => OF = (4, 5, 7) + s(2, 1, -2) for some s E R
    OF = (4+2s, 5+s, 7-2s)
    RF = (4+2s, 5+s, 7-2s) - (-2, 2, 13) = (6+2s, 3+s, -6-2s)
    RF . (2, 1, -2) = 0 (since L1 and L2 are perpendicular)
    Solving, s = -3
    Then sub. s = -3 into RF which I ended up with a 0?

    Any idea why this method doesnt work?
    Your calculations are correct. You proved that point R and point F are equal.

    1. The equation of the plane is

    2x+2y+z=13 with the normal vector \vec n = \langle 2,2,1 \rangle

    2. The line L_1: \langle x,y,z \rangle = \langle 4,5,7 \rangle +s \cdot \langle 2,1,-2 \rangle
    that means your result is correct.

    3. The line L_2 passes through R. The direction of this line must be perpendicular to \vec n so p_1 contains this line and this line must be perpendicular to the direction of L_1. Let \vec d denote the direction vector in question:

    \left|\begin{array}{l} \langle 2,1,-2 \rangle \cdot \vec d = 0 \\ \langle 2,2,1 \rangle \cdot \vec d = 0 \end{array}\right. ..... I've got \vec d = \langle 5,-6,2 \rangle
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vectors and Lines
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 17th 2011, 01:14 AM
  2. [SOLVED] Vectors of two skew lines
    Posted in the Geometry Forum
    Replies: 1
    Last Post: July 1st 2010, 11:13 PM
  3. Replies: 1
    Last Post: April 17th 2010, 11:52 PM
  4. Vectors and Lines
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 22nd 2009, 05:00 PM
  5. Vectors - Equations of Lines
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 31st 2008, 05:13 PM

Search Tags


/mathhelpforum @mathhelpforum