Let a line L pass through the point P(2,3,1) and be parallel to the line x+3 y-2 z-2=0=x-y+2 z. If the distance of L from the point (5,3,8) is α, then 3 α² is equal to :

Solution

Maths sir · 4 steps

The line of two planes runs along the cross product of their normals. Then drop a perpendicular.

1. Direction of L. The given line lies in both planes, so it is perpendicular to both normals (1, 3, -2) and (1, -1, 2). Their cross product is (4, -4, -4), so take d⃗ = (1, -1, -1) with |d⃗|² = 3.
2. Join the points. From P(2,3,1) to Q(5,3,8): PQ⃗ = (3, 0, 7), so |PQ⃗|² = 9 + 49 = 58. Also PQ⃗ · d⃗ = 3 + 0 - 7 = -4.
3. Right triangle. PQ⃗ is the hypotenuse of a right triangle. Its projection on d⃗ is one leg, and the perpendicular distance α is the other leg, so Pythagoras gives α².

   α² = |PQ⃗|² - (PQ⃗·d⃗)²/|d⃗|² = 58 - 16/3 = 158/3

4. So. 3α² = 3 × 158/3 = 158.

The common slip: 126. Forgetting to divide (PQ⃗·d⃗)² by |d⃗|² = 3 gives α² = 58 - 16 = 42, and so 126 instead of 158.

The idea to keep. For point-to-line distance, square of the distance is |PQ⃗|² minus the squared projection, (PQ⃗·d⃗)²/|d⃗|².

Answer from the NTA final answer key. Solution by PrepWiser.