ParallelogramΒΆ
Problem
A straight line \(g\) goes through the points A(0|1|2) and B(2|5|6).
Demonstrate that the distance between points A and B is 6. The points C and D lie on \(g\) and have each the distance 12 from A. Determine the coordinates of C and D.
The points A, B and E(1|2|5) together with one more point shall form the vertices of a parallelogram. There exist several possibilities for the position of the fourth vertex. State the coordinates of two of the possible fourth vertices.
Solution of part a
The vector connecting points A and B has the coordinates (2, 4, 4). Its length is therefore given by \(\sqrt{2^2+4^2+4^2}=\sqrt{36}=6\). The points C and D can be obtained by adding or subtracting twice the vector from A to B to the position vector of A. We thus obtain the points C (4|9|10) and D(-4|-7|-6).
We now implement this reasoning in Sage. First we calculate the distance between points A and B. then we determine the coordinates of points C and D. Finally, we verify that the distance between points C and D on the one hand and the point A on the other hand equals indeed 12.
Solution of part b
Choosing two of three possible vectors between the given points, one adds one vector to the end of the other one to obtain the fourth point.
We start by using the vector from A to B and from A to E:
The two ways to obtain the fourth vertex F yield the same result as it should be. One possible fourth vertex therefore is given by F(3|6|9).
Another parallelogram is obtained, if point B as being diagonally opposite to the new point.
For the sake of completeness we also determine the third possible vertex.
