Find the point of intersection. r=(1,2,-3)+t(1,-1,1) and r=(5,-2,-3)+s(2,-2,-2)?

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  • Ian H
    Lv 7
    1 year ago
    Favorite Answer

    r = (1, 2, -3) + t(1, -1 ,1) = (5, -2, -3) + s(2, -2, -2)

    The point of intersection satisfies both equations for r, and for each coordinate.

    x: 1 + t = 5 + 2s

    z: -3 + t = - 3 – 2s add those

    4 + 0 = 8 + 4s, so, s = -1

    r = (5, -2, -3) – 1*(2, -2 ,-2) = (3, 0, -1)

    Check: t = 4 + 2s = 2

    r = (1, 2, -3) +2(1, -1, 1) = = (3, 0, -1)

  • 1 year ago

    r = (1,2,-3) + t(1,-1,1) = (5,-2,-3) + s(2,-2,-2)

    t(1,-1,1) - s(2,-2,-2) = (5,-2,-3) - (1,2,-3)

    t(1,-1,1) + s(-2,2,2) = (4,-4,0)

    Solve t - 2s = 4 (from x or y) and t + 2s = 0 (from z)

    s = -1, t = 2

    Intersection: r = (1,2,-3) + 2(1,-1,1) = (5,-2,-3) - (2,-2,-2) = (__,__,__)

  • alex
    Lv 7
    1 year ago

    Hint:

    (1,2,-3)+t(1,-1,1) = (5,-2,-3)+s(2,-2,-2)

  • ted s
    Lv 7
    1 year ago

    1 + t = 5 + 2 s ; 2 - t = -2 - 2s ; - 3 + t = - 3 - 2s...add #2&3 to get -1 = - 5 - 4 s ===> s = -1 and t = 2...you finish

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