maths --plse help me..

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prove that the lines through a(0,-1,-1) and b(4,5,1) intersects the line through c (3,9,4) and d (-4,4,4). also find their point of intersection

  • maths --plse help me.. -

    L1:
    x = 0 + 4t
    y = -1 + 6t
    z = -1 + 2t

    L2:
    x = 3 - 7s
    y = 9 - 5s
    z = 4 + 0s

    L1 intersects L2 if

    4t = 3-7s
    -1+6t = 9-5s
    -1+2t = 4

    (t,s) = (5/2, -1) is the solution, so
    (x,y,z) = (10,14,4)

  • maths --plse help me.. -

    direction vector of 1st line = (4,6,2) or reduced to (2,3,1)
    parametric equation of 1st:
    x = 0 + 2t
    y = -1 + 3t
    z = -1 + t

    direction vector of 2nd line = (7, 5, 0)
    parametric equation of 2nd:
    x = 3 + 7k
    y = 9 + 5k
    z = 4

    solving for x's
    0+2t = 3 + 7k
    2t = 3+ 7k (#1)

    solving for y's
    3t-1 = 5k+9
    3t - 5k = 10 (#2)

    using the z's :
    t-1 = 4
    t = 5

    sub that into #1:
    10 = 3 + 7k
    k = 1

    so t=5 and k = 1
    check for consistency in #2 (the equation not used to solve)
    3t - 5k = 10
    LS = 15 - 5 = 10 = RS

    so the point of intersection is: (Using the 1st, could use either one)
    x = 0+2t = 10
    y = -1+3t = 14
    z = -1 + t = 4

    They intersect at (10,14,4)

  • maths --plse help me.. -

    L(ab): (0,-1,-1)+(4u,6u,2u)
    L(cd): (3,9,4)+(-7v,-5v,0)

    If they intersect at P, then corresponding x,y,z values match, hence
    2u-1 = 0 => u=2.5
    where
    P=(0+4*2.5, -1+6*2.5, -1+2*2.5)=(10,14,4)
    For line L(cd),
    we have P(3-7v, 9-5v, 4+0v)
    =>
    3-7v=10 => v=-1
    or
    9-5v=14 => v=-1
    or
    4+0v=4 => v=-1 (works).
    Therefore they intersect at P, since P(10,14,4) lies on both lines.

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