Let the image of the point P(1,2,6) in the plane passing through the points A(1,2,0), B(1,4,1) and C(0,5,1) be Q(α, β, γ). Then (α²+β²+γ²) is equal to :

Solution

Maths sir · 4 steps

Get the plane's normal from a cross product, then slide P twice its distance along that normal.

1. Find the normal. Two vectors in the plane are AB⃗ = (0, 2, 1) and AC⃗ = (-1, 3, 1). Their cross product is (-1, -1, 2), so take the normal as n⃗ = (1, 1, -2).
2. Write the plane. Through A(1, 2, 0) the plane is x + y - 2z = 3. Check with B: 1 + 4 - 2 = 3, and with C: 0 + 5 - 2 = 3.

   (α - 1)/1 = (β - 2)/1 = (γ - 6)/-2 = (-2 (1 + 2 - 12 - 3))/(1² + 1² + (-2)²) = (-2 × (-12))/6 = 4

3. Read off Q. So α = 1 + 4 = 5, β = 2 + 4 = 6, γ = 6 - 8 = -2, giving Q(5, 6, -2).
4. So. α² + β² + γ² = 25 + 36 + 4 = 65, which is option (d).

The idea to keep. Image of a point in ax + by + cz = d: move along (a, b, c) by -2 (ax₁ + by₁ + cz₁ - d)/(a² + b² + c²). Without the 2 you only get the foot of the perpendicular.

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