If the image of the point P(1, 2, a) in the line x - 63 = y - 72 = 7 - z2 is Q(5, b, c), then a2 + b2 + c2 is equal to
- 298
- 264
- 293
- 283
Answer
(1) 298
Solution
The line is x - 63 = y - 72 = z - 7-2 = t, with direction ratios (3, 2, -2).
The mid-point M of PQ, (3, 2 + b2, a + c2 ), lies on the line, so 3 - 63 = t, i.e. t = -1, and
M = (6 - 3, 7 - 2, 7 + 2) = (3, 5, 9)
2 + b2 = 5, a + c2 = 9
b = 8, a + c = 18
PQ is perpendicular to the line and PQ = (4, b - 2, c - a) = (4, 6, c - a), so
3(4) + 2(6) - 2(c - a) = 0
c - a = 12
c = 15, a = 3
a2 + b2 + c2 = 9 + 64 + 225 = 298
Answer: (1)