Answer:
The diagonals of a parallelogram bisect each other, so the point of intersection of the diagonals is the midpoint of each diagonal. To find the midpoint of a diagonal, we can average the coordinates of the endpoints.
The coordinates of the diagonal PR are:
P = (4, 2, -6)
R = (12, 4, 5)
The midpoint of PR is:
M_PR = [(4 + 12)/2, (2 + 4)/2, (-6 + 5)/2] = (8, 3, -1/2)
The coordinates of the diagonal QS are:
Q = (5, -3, 1)
S = (11, 9, -2)
The midpoint of QS is:
M_QS = [(5 + 11)/2, (-3 + 9)/2, (1 - 2)/2] = (8, 3, -1/2)
So the point of intersection of the diagonals is (8, 3, -1/2).
To find the equations of the diagonals, we need to find the vectors corresponding to each diagonal, since the diagonals are parallel to these vectors.
The vector corresponding to the diagonal PR is:
PR = R - P = (12, 4, 5) - (4, 2, -6) = (8, 2, 11)
The vector corresponding to the diagonal QS is:
QS = S - Q = (11, 9, -2) - (5, -3, 1) = (6, 12, -3)
To find the equation of the line passing through the midpoint of diagonal PR and parallel to PR, we can use the point-normal form of the equation of a line:
(x - 8)/8 = y/2 = (z + 1/2)/11
Simplifying this equation gives:
x/8 - 1 = y/2 = z/11 + 1/22
To find the equation of the line passing through the midpoint of diagonal QS and parallel to QS, we can use the same method:
(x - 8)/6 = y/12 = (z + 1/2)/(-3)
Simplifying this equation gives:
x/6 - 4/3 = y/12 = -z/3 - 1/6
Therefore, the equations of the diagonals of PQRS are:
8 - 1/8t = y/2 = z/11 + 1/22 for diagonal PR
x/6 - 4/3 - 1/6t = y/12 = -z/3 - 1/6 for diagonal QS, where t is a real number representing the parameter along the line.