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Math Help - straight passes by point

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    straight passes by point

    Finding the equation of the straight (l1), which passes in point P(2,1,-1) and perpendicular to the straight (l2): (x,y,z)=(2,0,0)+t(3,1,-1)


    Answer:
    (x,y,z)=(2,1,-1)+h(2,-3,3)
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    Finding the equation of the straight (l1), which passes in point P(2,1,-1) and perpendicular to the straight (l2): (x,y,z)=(2,0,0)+t(3,1,-1)


    Answer:
    (x,y,z)=(2,1,-1)+h(2,-3,3)
    There are 2 completely different ways to do this problem:

    1. By calculus: Let Q denote a point on l_2 : Q \in l_2
    Then calculate the distance |\overline{PQ}| which will give you an equation of a function with respect to t. Calculate the minimum distance using the first derivative.

    (d(t))^2=2-4t+11t^2

    If d(t) has an extreme value then (d(t))^2 has an extreme value too. Therefore it is sufficient if you examine (d(t))^2.
    Solve ((d(t))^2)' = 0 for t.
    Calculate the coordinates of Q.
    Calculate the equation of the line PQ.

    2. By analytic geometry: Calculate the equation of an auxiliar plane a perpendicular to l_2 and containing P.
    Then calculate the intersection of l_2 and a which is the point Q.
    Calculate the equation of the line PQ.
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