The distance of the point (7, 10, 11) from the line x-41 = y-40 = z-23 along the line x-92 = y-133 = z-176 is
- 16
- 12
- 18
- 14
Answer
(4) 14
Solution
Let P(7, 10, 11). The required distance is measured along a line through P parallel to the line with direction ratios 2, 3, 6.
∵ 22 + 32 + 62 = 7
⇒ any point on this line: Q(7 + 2t, 10 + 3t, 11 + 6t), with PQ = 7|t|
Q lies on x-41 = y-40 = z-23, so its y-coordinate is 4:
⇒ 10 + 3t = 4
⇒ t = -2, giving Q(3, 4, -1)
Check: 3-41 = -1 and -1-23 = -1, so Q lies on the line.
∴ PQ = 7 |-2| = 14
Answer: (4)