Let the image of the point (5/3, 5/3, 8/3) in the plane x-2 y+z-2=0 be P. If the distance of the point Q(6,-2, α), α > 0, from P is 13 , then α is equal to ___________.

Solution

Maths sir · 4 steps

First find the image P by stepping across the plane along its normal, then the distance gives α directly.

1. Image formula. For a point (x₁, y₁, z₁) and the plane ax + by + cz + d = 0, the image moves along the normal (a, b, c) by twice the distance to the foot.

   (x - x₁)/a = (y - y₁)/b = (z - z₁)/c = -2(a x₁ + b y₁ + c z₁ + d)/(a² + b² + c²)

2. Find P. Here a x₁ + b y₁ + c z₁ + d = 5/3 - 10/3 + 8/3 - 2 = -1 and a² + b² + c² = 6, so the common ratio is 2/6 = 1/3. Then P = (5/3 + 1/3, 5/3 - 2/3, 8/3 + 1/3) = (2, 1, 3).
3. Use the distance. PQ² = (6-2)² + (-2-1)² + (α - 3)² = 16 + 9 + (α - 3)² = 169. So (α - 3)² = 144, which gives α - 3 = ± 12.
4. So. α = 15 or α = -9. Since α > 0, the answer is 15.

The idea to keep. For an image in a plane, the multiplier is -2(ax₁+by₁+cz₁+d)/(a²+b²+c²). Without the 2 you get the foot of the perpendicular instead.

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