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Find the point on line x+2/3=y+1/2=z-3/2 at a distance 5 units from the point P(1,3,3)....pllz help me out.

  • maths -

    Let the new point be (x1, y1, z1)

    Then, using the distance formula:

    ((x1-1)^2 + (y1-3)^2 + (z1-3)^2)^0.5 = 5

    (x1-1)^2 + (y1-3)^2 + (z1-3)^2 = 25

    Put all units in terms of one variable. I'm choosing x1 first:

    x1+2/3 = y1+1/2; y1 = x1 + 2/3 - 1/2
    = x1 + 1/6

    x1 + 2/3 = z1-3/2; z1 = x1 + 2/3 + 3/2 = x1 + 4/6 + 9/6 = x1 + 13/6

    (x1-1)^2 + (x1+1/6-3)^2 + (x1+13/6-3)^2 = 25

    Multiply this out; solve for x1; then plug these back into the other equations to solve for y1 and z1

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