Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point ( 113, 113, 193 ) from the line L along the line 3x - 112 = 3y - 111 = 3z - 192 is equal to
- 6
- 3
- 5
- 4
Answer
(2) 3
Solution
L has d.r.'s (2 - 1, 3 - 2, 5 - 3) = (1, 1, 2), so any point of L is (1 + , 2 + , 3 + 2 ).
Let P be the given point. The given line through P has d.r.'s ( 23, 13, 23 ) (2, 1, 2), so we move from P along (2, 1, 2) till we meet L at Q:
[figure]
Q = ( 113 + 2t, 113 + t, 193 + 2t )
Q lies on L, so this equals (1 + , 2 + , 3 + 2 ).
Subtracting the y-coordinate from the x-coordinate: t = (1 + ) - (2 + ) = -1
Q = ( 53, 83, 133 ), which lies on L with = 23
Hence PQ = |t| 22 + 12 + 22 = 1 3 = 3
Answer: (2)