finding the coordinates to a point on a normal triangle
[b]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.