Results 1 to 5 of 5

Math Help - Vectors

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    Vectors

    Find the equation of the line through the point (-5,-4,2) that intersects the line at vector r=(7,-13,8)+ t(1,2,-2) at 90 degrees. Determine the point of intersection.


    So this is my work below, so far:

    Vector eqtn:
    Point A (-5,-4,2)
    Point B (7,-13,8)
    d=AB=(12, -9,6)

    therefore vector eqtn is:
    (x,y,z)=(-5,-4,2) + t(12,-9,6)

    So that is my attempt at part A, now part B:

    Point of Intersection
    l1: (x,y,z)=(-5,-4,2) + t(12,-9,6)
    l2: r=(x,y,z)=(7,-13,8)+t(1,2,-2)
    Parametric eqtn:
    l1: x=-5+12t, y=-4-9t, z=2+6t
    Symmetric eqtn:
    l2: (x-7)/1 = (y+13)/2 = (z-8)/-2
    Substitute parametric eqtn into symmetric eqtn
    l2: t=1 and t=1
    therefore point of intersection when t =1
    x=-5+12(1)=7
    y=-4-9(1)=-13
    z=2+6(1)=8
    final answer:
    (7,-13,8)

    O.K. Please let me know if I have my answers correct and method. If not, please let me know the correct one and thoroughly explain accordingly. Thank you for your help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    ..? can anyone help?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    64
    If the direction vector of your line goes from point A to point B then the intersection of the lines is obviously at point B. You have to find a line that intersects the other one making an angle of 90.

    To do so, the vectors that define directions have to be perpendicular, that is, their inner product must be 0:

    (v1,v2,v3) . (1,2,-2) = 0 <=> v1+2v2-2v3 = 0 (1)

    They must intersect, therefore:

    (-5,-4,2) + k(v1,v2,v3) = (7,-13,8) + t(1,2,-2) <=>

    -5 + kv1 = 7 + t
    -4 + kv2 = -13 + 2t
    2 + kv3 = 8 - 2t

    v1 = (12 + t) / k
    v2 = (-9 + 2t) / k
    v3 = (6 - 2t) / k

    Taking (1) we have, 12 + t + 2(-9 + 2t) - 2(6 - 2t) = 0 <=>

    -18 + 9t = 0 <=> t = 2

    picking k=1,

    v1 = 14
    v2 = -5
    v3 = 2

    Therefore the line has equation:

    (-5,-4,2) + k (14,-5,2)

    And the point of intersection is given when k=1 (and t=2):

    (9,-9,4)
    Last edited by pedrosorio; August 30th 2009 at 08:52 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    so should POI be (9,-9,4) and not (-9,-9,4)?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    64
    Yeah, sure, my mistake.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 15th 2011, 05:10 PM
  2. Replies: 3
    Last Post: June 30th 2011, 08:05 PM
  3. Replies: 2
    Last Post: June 18th 2011, 10:31 AM
  4. [SOLVED] Vectors: Finding coefficients to scalars with given vectors.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 23rd 2011, 12:47 AM
  5. Replies: 4
    Last Post: May 10th 2009, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum