The largest value of a, for which the perpendicular distance of the plane containing the lines r⃗=(î+ĵ)+λ(î+a ĵ-k̂) and r⃗=(î+ĵ)+μ(-î+ĵ-a k̂) from the point (2,1,4) is √3, is _________.

Solution

Maths sir · 4 steps

Both lines share (1,1,0), so cross their directions for the normal and apply the distance formula.

1. Normal to plane. Both lines start at (1,1,0), so the plane passes through it. Its normal is perpendicular to both directions, so take their cross product.

   (î+aĵ-k̂) × (-î+ĵ-ak̂) = (1-a²)î + (1+a)ĵ + (1+a)k̂ = (1+a)[(1-a)î+ĵ+k̂]

2. Equation of plane. For a ≠ -1 the normal is along (1-a, 1, 1). The plane is (1-a)(x-1) + (y-1) + z = 0. At a = -1 the two lines are parallel, so they do not fix a unique plane.
3. Use the distance. Put in (2,1,4): the numerator is (1-a) + 0 + 4 = 5-a. So (|5-a|)/√((1-a)²+2) = √3. Squaring gives 25 - 10a + a² = 3a² - 6a + 9, that is a² + 2a - 8 = 0.
4. So. (a+4)(a-2) = 0, so a = 2 or a = -4. The largest value is 2. Check: at a = 2 the plane is -x + y + z = 0, and the distance is 3/√3 = √3.

The common slip: -4. Picking the other root of a² + 2a - 8 = 0 gives -4, but the question asks for the largest value.

The idea to keep. Two lines through a common point fix a plane: the common point lies on it, and the cross product of the directions is its normal. Pull out common factors before using the normal.

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