The line of shortest distance between the lines (x-2)/0=(y-1)/1=z/1 and (x-3)/2=(y-5)/2=(z-1)/1 makes an angle of cos ⁻¹(√(2/27)) with the plane P: a x-y-z=0, (a>0). If the image of the point (1,1,-5) in the plane P is (α, β, γ), then α+β-γ is equal to _________________.

Solution

Maths sir · 4 steps

The shortest-distance line is perpendicular to both lines, so a cross product gives its direction and then a.

1. Direction of SD line. The line of shortest distance is perpendicular to both d₁⃗ = (0, 1, 1) and d₂⃗ = (2, 2, 1). So its direction is d₁⃗ × d₂⃗ = (-1, 2, -2), with magnitude 3.
2. Set up for a. The normal of P is n⃗ = (a, -1, -1), so n⃗ · d⃗ = -a + 2 - 2 = -a. Using sin of the angle as 5/√27 would need a/√(a²+2) = 5/√3 > 1, which is impossible. So the given value √(2/27) is (|n⃗ · d⃗|)/|n⃗||d⃗|, as the question intends.

   a/3√(a²+2) = √2/3√3 ⇒ 3a² = 2a² + 4 ⇒ a = 2

3. Image in the plane. The plane is 2x - y - z = 0. For (1, 1, -5), 2 - 1 + 5 = 6 and |n⃗|² = 6. So the image is (1, 1, -5) - 2 × 6/6 × (2, -1, -1) = (-3, 3, -3).
4. So. α + β - γ = -3 + 3 + 3 = 3.

The common slip: 5. Stopping at the foot of the perpendicular (-1, 2, -4) instead of going twice as far to the image gives 5.

The idea to keep. The direction of the shortest-distance line is always d₁⃗ × d₂⃗. For a mirror image, subtract twice the projection onto the normal. Subtracting it once only gives the foot.

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