Let PQR be a triangle such that PQ = -2 i - j + 2 k and PR = a i + b j - 4 k, a, b Z. Let S be the point on QR, which is equidistant from the lines PQ and PR. If | PR| = 9 and PS is parallel to i - 7 j + 2 k, then the value of 3a - 4b is .
Answer
37
Solution
S lies on QR and is equidistant from lines PQ and PR PS is the internal bisector of QPR.
| PQ| = 4 + 1 + 4 = 3, | PR| = 9
Direction of the internal bisector = PQ3 + PR9 = (a - 6) i + (b - 3) j + 2 k9
it is parallel to i - 7 j + 2 k:
a - 61 = b - 3-7 = 22 = 1
a = 7, b = -4
Check: | PR|2 = 49 + 16 + 16 = 81 ✓
3a - 4b = 21 + 16 = 37
Answer: 37