Let a line passing through the point (4, 1, 0) intersect the line L1 : x-12 = y-23 = z-34 at the point A( , , ) and the line L2 : x - 6 = y = -z + 4 at the point B(a, b, c). Then vmatrix 1 & 0 & 1 \ & & \ a & b & c vmatrix is equal to
- 16
- 6
- 8
- 12
Answer
(3) 8
Solution
A on L1: A(1 + 2s, 2 + 3s, 3 + 4s); B on L2: B(6 + t, t, 4 - t)
With P(4, 1, 0): PA = (2s - 3, 3s + 1, 4s + 3), PB = (t + 2, t - 1, 4 - t)
P, A, B collinear ⇒ PA = k PB:
2s - 3 = k(t + 2) … (i), 3s + 1 = k(t - 1) … (ii), 4s + 3 = k(4 - t) … (iii)
(i) − (ii): -s - 4 = 3k; (ii) + (iii): 7s + 4 = 3k
⇒ 7s + 4 = -s - 4 ⇒ s = -1, k = -1
From (i): -5 = -(t + 2) ⇒ t = 3
⇒ A(-1, -1, -1), B(9, 3, 1)
∴ vmatrix 1 & 0 & 1 \ -1 & -1 & -1 \ 9 & 3 & 1 vmatrix = 1(-1 + 3) - 0 + 1(-3 + 9)
= 2 + 6 = 8
Answer: (3)