Let the lines L1: r = i + 2 j + 3 k + (2 i + 3 j + 4 k), R and L2: r = (4 i + j) + (5 i + 2 j + k), R, intersect at the point R. Let P and Q be the points lying on lines L1 and L2, respectively, such that | PR| = 29 and | PQ| = 473. If the point P lies in the first octant, then 27(QR)2 is equal to
- 360
- 348
- 320
- 340
Answer
(1) 360
Solution
General points: on L1: (1+2 , 2+3 , 3+4 ); on L2: (4+5 , 1+2 , )
For intersection: 3+4 = and 2+3 = 1+2
2+3 = 1+2(3+4 )
= -1, = -1 (check: 1+2 = -1 = 4+5 )
R = (-1, -1, -1)
Point P on L1: P = (-1+2t, -1+3t, -1+4t)
| PR| = |t| 4+9+16 = |t| 29 = 29
t = 1
t = 1 P(1, 2, 3) (first octant); t = -1 (-3, -4, -5), rejected
Point Q on L2: Q = (-1+5s, -1+2s, -1+s)
PQ = (5s-2) i + (2s-3) j + (s-4) k
| PQ|2 = 30s2 - 40s + 29 = 473
90s2 - 120s + 40 = 0
9s2 - 12s + 4 = 0
(3s-2)2 = 0 s = 23
(QR)2 = s2(25+4+1) = 49 30 = 403
27(QR)2 = 360
Answer: (1)