If the plane P passes through the intersection of two mutually perpendicular planes 2 x+k y-5 z=1 and 3 k x-k y+z=5, k<3 and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is :

Solution

Maths sir · 4 steps

Use the perpendicular condition to fix k, then make the x-intercept 1 to pick the right plane in the family.

1. Find k. Two planes are perpendicular when the dot product of their normals is zero. Here (2)(3k) + (k)(-k) + (-5)(1) = 0 gives k² - 6k + 5 = 0, so k = 1 or k = 5. Since k < 3, we take k = 1.
2. Family of planes. With k = 1 the planes are 2x + y - 5z = 1 and 3x - y + z = 5. Every plane through their line of intersection has the form below.

   (2x + y - 5z - 1) + λ(3x - y + z - 5) = 0

3. Use x-intercept. Put y = z = 0: (2 + 3λ)x = 1 + 5λ. For x = 1 we need 1 + 5λ = 2 + 3λ, so λ = 1/2. Multiplying through by 2 gives the plane P: 7x + y - 9z = 7.
4. So. Put x = z = 0 in 7x + y - 9z = 7 to get y = 7. The intercept on the y-axis is 7, option (d).

The idea to keep. A plane through the line of two planes is P₁ + λ P₂ = 0. One extra condition, like an intercept or a point, fixes λ. Always use the given restriction to pick the right root first.

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