The plane passing through the line L: l x-y+3(1-l) z=1, x+2 y-z=2 and perpendicular to the plane 3 x+2 y+z=6 is 3 x-8 y+7 z=4. If θ is the acute angle between the line L and the y-axis, then 415 cos ² θ is equal to _____________.

Solution

Maths sir · 5 steps

The given plane must belong to the family through L, and matching coefficients pins down l.

1. Family of planes. Every plane through L has the form (lx - y + 3(1-l)z - 1) + μ(x + 2y - z - 2) = 0. So 3x - 8y + 7z - 4 = a(lx - y + 3(1-l)z - 1) + b(x + 2y - z - 2) for some constants a and b.
2. Match coefficients. Constant terms give a + 2b = 4, and the y terms give -a + 2b = -8. Adding them, b = -1 and a = 6. The x terms then give 6l - 1 = 3, so l = 2/3. Check with z: 6 × 3 × 1/3 + 1 = 7, which matches.
3. Direction of L. With l = 2/3, the first plane is 2/3x - y + z = 1, whose normal is parallel to (2, -3, 3). The line lies in both planes, so its direction is the cross product of the two normals.

   d⃗ = (2, -3, 3) × (1, 2, -1) = (-3, 5, 7), |d⃗|² = 9 + 25 + 49 = 83

4. Angle with y-axis. The y-axis has direction (0, 1, 0), so cos θ = |5|/√83 and cos² θ = 25/83.
5. So. 415 cos² θ = 415 × 25/83 = 5 × 25 = 125.

The common slip: 290. Using sin θ in place of cos θ for the line-axis angle gives 415 × 58/83 = 290. Between two lines the dot product gives the cosine.

The idea to keep. A plane through the line of intersection of two planes is P₁ + μ P₂ = 0. If the plane is already given, match coefficients to find the unknown parameter, then cross the two normals to get the line's direction.

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