Let P₁ be the plane 3 x-y-7 z=11 and P₂ be the plane passing through the points (2,-1,0),(2,0,-1), and (5,1,1). If the foot of the perpendicular drawn from the point (7,4,-1) on the line of intersection of the planes P₁ and P₂ is (α, β, γ), then α+β+γ is equal to ___________.

Solution

Maths sir · 4 steps

Find the line of intersection first; the foot is where the joining vector is perpendicular to it.

1. Find plane P2. From (2,-1,0), two vectors in the plane are (0,1,-1) and (3,2,1). Their cross product is (3,-3,-3), so the normal is (1,-1,-1). Putting in (2,-1,0) gives P₂: x - y - z = 3.
2. Line of intersection. Its direction is perpendicular to both normals: (3,-1,-7) × (1,-1,-1) = (-6,-4,-2), which is parallel to (3,2,1). Putting z = 0 gives 3x - y = 11 and x - y = 3, so x = 4, y = 1. The line is (4+3t, 1+2t, t).
3. Perpendicular condition. From (7,4,-1) to a general point of the line, the vector is (3t-3, 2t-3, t+1). At the foot this vector is perpendicular to (3,2,1).

   3(3t-3) + 2(2t-3) + 1(t+1) = 14t - 14 = 0 ⇒ t = 1

4. So. The foot is (7, 3, 1), so α + β + γ = 7 + 3 + 1 = 11.

The idea to keep. For the line of two planes, cross the normals to get its direction and find one common point. For the foot, take a general point and make the joining vector perpendicular to the direction.

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