Let a line L passing through the point P(1, 1, 1) be perpendicular to the lines x-44 = y-11 = z-11 and x-171 = y-711 = z0. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point S(1, 0, -1) intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to .
Answer
6
Solution
Direction of L both (4, 1, 1) and (1, 1, 0):
vmatrix i & j & k \ 4 & 1 & 1 \ 1 & 1 & 0 vmatrix = - i + j + 3 k
L: (1 - t, 1 + t, 1 + 3t); on the yz-plane x = 0 t = 1
Q = (0, 2, 4)
Line through S parallel to L: (1 - t, t, -1 + 3t); x = 0 t = 1
R = (0, 1, 2)
PQ = - i + j + 3 k, PS = - j - 2 k (and SR = PQ, so PQRS is a parallelogram)
PQ PS = vmatrix i & j & k \ -1 & 1 & 3 \ 0 & -1 & -2 vmatrix = i - 2 j + k
(Area)2 = | PQ PS|2 = 1 + 4 + 1 = 6
Answer: 6