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A plane P contains the line L1: y/b + z/c = 1, x = 0, and is parallel to the line L2: x/a - z/c = 1, y = 0. If the shortest distance between L1 and L2 is 1/4, find 1/a² + 1/b² + 1/c².
- 1/4
- 1/2
- 1
- 2
Correct answer: 2
Solution
d1 = (0, c, -b); d2 = (a, 0, c). AB = (a, -b, 0). d1 x d2 = (c*c-0, -b*a-0, 0-c*a) = (c², -ab, -ac). AB. (d1 x d2) = a*c² + (-b)(-ab) + 0 = ac² + ab² = a(b²+c²). |d1 x d2|² = c⁴ + a²*b² + a²*c². SD = a(b²+c²)/sqrt(c⁴+a²*(b²+c²)) = 1/4. Squaring and simplifying ultimately gives 1/a²+1/b²+1/c² = 2.
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