Let the area of the triangle formed by the lines x + 2 = y - 1 = z, x - 35 = y-1 = z - 11 and x-3 = y - 33 = z - 21 be A. Then A2 is equal to .
Answer
56
Solution
L1: (p - 2, p + 1, p); L2: (5t + 3, -t, t + 1); L3: (-3s, 3s + 3, s + 2)
L1 L2: p = t + 1 and p + 1 = -t t = -1, p = 0 ⇒ P(-2, 1, 0) (x: -2 = -5 + 3 ✓)
L1 L3: p = s + 2 and p - 2 = -3s s = 0, p = 2 ⇒ Q(0, 3, 2) (y: 3 = 3 ✓)
L2 L3: t + 1 = s + 2 and -t = 3s + 3 s = -1, t = 0 ⇒ R(3, 0, 1) (x: 3 = 3 ✓)
PQ = 2 i + 2 j + 2 k, PR = 5 i - j + k
PQ PR = vmatrix i & j & k \ 2 & 2 & 2 \ 5 & -1 & 1 vmatrix = 4 i + 8 j - 12 k
| PQ PR| = 16 + 64 + 144 = 224
A = 12 224 A2 = 2244 = 56
Answer: 56