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Finding the coordinates to a point on a normal triangle

  1. Feb 24, 2008 #1
    Hello everyone, I have been trying this problem for the last couple of days, and I am stuck! it is as follows:

    the point Q(3,4,3) lies on the plane . The line L passes through the midpoint of [PQ]. Point S is on L such that |PS| = |QS| = 3, and the triangle PQS is normal to the plane. Given that there are two possible positions for S, find their cooridinates.

    This is part C of a problem and for part and A and B I found that the plane is -2x + y - z = -5, and the point P is (1, -1, 2). and if it is any help at all, I found that the distance between the plane and the point S is the squareroot of 3/2


    First of all, I realized that |PS| = the square root of (X - 1)² + (Y +1)² + (Z - 2)² with X,Y,Z representing the cooridinates of S. I did the same thing for |QS| = the square root of (X - 3)² + (Y - 4)² + (Z- 3)², since both of these equal 3, it is possible to set them equal to eachother and work out that algebraically. For that I believe I calculated -4x - 10y - 2z = -28. So, from what I know, this equation is the plane on which the 2 coordinates of point S lie. Now this is where I am stuck. while I have tried many other ways, so far I think this may be the one that could lead me to an answer, but I do not know where to go from this point.
     
    Last edited: Feb 24, 2008
  2. jcsd
  3. Feb 24, 2008 #2
    Perhaps?

    Maybe give this a try:

    Find the coordinates of the midpoint between P and Q. Since you have the equation of the plane, you know the normal to the plane ax+by+cz+d=0 is the vector v={a,b,c}.

    If I understood the geometry, s should lie along some multiple of the normal leaving from the midpoint between P and Q. Thus, start the normalized normal from the midpoint between P and Q, then multiply by the distance between s and the plane. To get the other point, simply take the opposite direction of the normal, then repeat procedure.
     
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