Finding the coordinates to a point on a normal triangle

In summary, the problem involves finding the coordinates of point S, which lies on the line passing through the midpoint of [PQ], making triangle PQS normal to the plane with point Q(3, 4, 3) on it. The plane is given by -2x + y - z = -5 and point P is (1, -1, 2). Using the equations |PS| = |QS| = 3, we can set them equal to each other and find the equation of the plane on which the two coordinates of point S lie. To find the coordinates of point S, we can use the normalized normal to the plane starting from the midpoint between P and Q and multiply it by the distance between
  • #1
rosalinde4711
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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 each other 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.
 
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  • #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.
 

Related to Finding the coordinates to a point on a normal triangle

1. What is a normal triangle?

A normal triangle, also known as a right triangle, is a triangle with one 90-degree angle.

2. How do I find the coordinates of a point on a normal triangle?

The coordinates of a point on a normal triangle can be found using the Pythagorean theorem and basic trigonometric functions. It involves calculating the lengths of the sides of the triangle and then using the appropriate ratios to find the coordinates.

3. What is the Pythagorean theorem?

The Pythagorean theorem is a mathematical formula that states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

4. What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, and tangent. These functions are used to relate the angles of a right triangle to the lengths of its sides.

5. Can I find the coordinates of a point on a normal triangle without using trigonometry?

Yes, it is possible to find the coordinates of a point on a normal triangle without using trigonometry. This can be done using the slope formula and the equation of a line. However, using trigonometry is a more straightforward and efficient method.

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