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Let P 1 : 3 y + z + 1 = 0 and P 2 : 2 x - y + 3 z - 7 = 0 and the equation of line A B is x - 1 2 = y - 3 - 1 = z - 4 3 in 3 D space. Shortest distance between the line of intersection of planes P 1 and P 2 and the line A B is equal to

Options

  1. A7 10 units
  2. B7 2 5 units
  3. C6 10 units
  4. D2 2 5 units

Correct answer

B. 7 2 5 units

Step-by-step solution

Plane P 1 is parallel to line A B ⇒ Shortest distance = Perpendicular distance of any point on line to plane P 1 = 0 = 0 + 9 + 4 + 1 9 + 1 = 14 10 = 7 2 5 units

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