Let P be an arbitrary point having sum of the squares of the distances from the planes x + y + z = 0, lx - nz = 0 and x - 2y + z = 0, equal to 9. If the locus of the point P is x² + y² + z² = 9, then the value of l - n is equal to _________.

Solution

Maths sir · 4 steps

Add the two known distance-squares first, and the leftover tells you what the third plane must be.

1. Set up. For P(x, y, z), the square of the distance from ax + by + cz = 0 is (ax+by+cz)²/(a²+b²+c²). Adding the three squares and setting the sum equal to 9 gives the locus.

   (x+y+z)²/3 + (lx-nz)²/(l²+n²) + (x-2y+z)²/6 = 9

2. Two known terms. Expanding, (x+y+z)²/3 + (x-2y+z)²/6 = x²/2 + y² + z²/2 + xz. The xy and yz terms cancel.
3. What is missing. To get x² + y² + z², the middle term must supply x²/2 + z²/2 - xz = (x-z)²/2. So (lx-nz)²/(l²+n²) = (x-z)²/2 for every point. That needs lx - nz to be a multiple of x - z, that is l = n.
4. So. With l = n, the plane is x - z = 0 and the locus is x² + y² + z² = 9. Hence l - n = 0.

The idea to keep. Squared distances from three mutually perpendicular planes through the origin always add up to x²+y²+z². So find the unknown plane by making its normal perpendicular to the other two.

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