Let P( , , ) be the point on the line x - 12 = y + 1-3 = z at a distance 4 14 from the point (1, -1, 0) and nearer to the origin. Then the shortest distance, between the lines x - 1 = y - 2 = z - 3 and x + 52 = y - 101 = z - 31, is equal to
- 4 75
- 7 54
- 4 57
- 2 74
Answer
(1) 4 75
Solution
Any point on the line: (1 + 2t, -1 - 3t, t)
Distance from (1, -1, 0): 4t2 + 9t2 + t2 = |t| 14 = 4 14
t = 4
t = 4: (9, -13, 4), distance2 from O = 81 + 169 + 16 = 266
t = -4: (-7, 11, -4), distance2 from O = 49 + 121 + 16 = 186 (nearer)
( , , ) = (-7, 11, -4)
Line 1: through A(-7, 11, -4), b1 = i + 2 j + 3 k
Line 2: through B(-5, 10, 3), b2 = 2 i + j + k
b1 b2 = vmatrix i & j & k \ 1 & 2 & 3 \ 2 & 1 & 1 vmatrix = - i + 5 j - 3 k, | b1 b2| = 35
AB = 2 i - j + 7 k
AB ( b1 b2) = -2 - 5 - 21 = -28
SD = 28 35
= 4 4935 = 4 75
Answer: (1)