A plane E is perpendicular to the two planes 2 x-2 y+z=0 and x-y+2 z=4, and passes through the point P(1,-1,1). If the distance of the plane E from the point Q(a, a, 2) is 3 √2, then (P Q)² is equal to :

Solution

Maths sir · 4 steps

A plane perpendicular to two planes has its normal along their cross product, so find that normal first.

1. Normal of E. E is perpendicular to both planes, so its normal is perpendicular to both of their normals (2,-2,1) and (1,-1,2). That makes it their cross product.

   n⃗ = |î, ĵ, k̂; 2, -2, 1; 1, -1, 2| = -3î - 3ĵ + 0k̂

2. Equation of E. Take the normal as (1,1,0). Through P(1,-1,1) the plane is (x-1)+(y+1)=0, that is, x+y=0.
3. Use the distance. The distance of Q(a,a,2) from x+y=0 is (|a+a|)/√(1²+1²) = √2 |a|. Setting this equal to 3√2 gives |a| = 3, so a² = 9.
4. So. (PQ)² = (a-1)² + (a+1)² + (2-1)² = 2a² + 3 = 18 + 3 = 21, the same for a = 3 and a = -3. The answer is (c).

The idea to keep. A plane perpendicular to two given planes has its normal along the cross product of their normals. In the distance formula, always divide by the length of the normal.

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