If the distance between the plane, 23x – 10y – 2z + 48 = 0 and the plane
containing the lines (x + 1)/2 = (y - 3)/4 = (z + 1)/3
and (x + 3)/2 = (y + 2)/6 = (z - 1)/λ( λ ∈ R )
is equal to k/√633, then k is equal to ______.

Solution

Maths sir · 4 steps

Two planes have a nonzero distance only if they are parallel, so borrow the given plane's normal.

1. Planes are parallel. Two planes that are not parallel meet, and their distance is 0. A nonzero distance k/√633 therefore means the required plane is parallel to the given one. So its normal is also (23, -10, -2).
2. Build the plane. It contains the first line, so it passes through (-1, 3, -1). As a check, (23)(2) + (-10)(4) + (-2)(3) = 46 - 40 - 6 = 0, so the line's direction is perpendicular to this normal. The plane is 23(x+1) - 10(y-3) - 2(z+1) = 0, which is 23x - 10y - 2z + 51 = 0.
3. Check the second line. The point (-3, -2, 1) fits: -69 + 20 - 2 + 51 = 0. The direction (2, 6, λ) must also be perpendicular to the normal: 46 - 60 - 2λ = 0, so λ = -7. Both lines lie in this plane.

   d = (|51 - 48|)/√(23² + 10² + 2²) = 3/√(529 + 100 + 4) = 3/√633

4. So. Comparing with k/√633 gives k = 3.

The common slip: 99. Writing the constant of the new plane as -51 instead of +51 gives |-51 - 48| = 99, so k = 99.

The idea to keep. When the question asks for a distance between planes, they must be parallel. Use the known normal, then apply (|d₁ - d₂|)/√(a²+b²+c²) with the same a, b, c in both equations.

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