Consider the lines L1: x - 1 = y - 2 = z and L2: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L1 and L2 be Q and R respectively. If the area of the triangle PQR is A, then 4A2 is equal to :
- 151
- 147
- 139
- 143
Answer
(2) 147
Solution
L1 passes through (1, 2, 0) and L2 through (2, 0, 1); both have direction ratios (1, 1, 1).
Foot Q on L1: take Q = (1 + t, 2 + t, t).
PQ = (t - 4, t + 1, t + 3) and PQ (1, 1, 1) = 0
3t = 0 t = 0, so Q = (1, 2, 0)
Foot R on L2: take R = (2 + s, s, 1 + s).
PR = (s - 3, s - 1, s + 4) and PR (1, 1, 1) = 0
3s = 0 s = 0, so R = (2, 0, 1)
PQ = -4 i + j + 3 k, PR = -3 i - j + 4 k
PQ PR = vmatrix i & j & k \ -4 & 1 & 3 \ -3 & -1 & 4 vmatrix = 7 i + 7 j + 7 k
A = 12 | PQ PR | = 7 32
4A2 = 4 1474 = 147
Answer: (2)