Let the lines
L₁:r⃗ = λ ( î + 2ĵ + 3k̂ ), λ ∈ R
L₂:r⃗ = ( î + 3ĵ + k̂ ) + μ ( î + ĵ + 5k̂ ); μ ∈ R,
intersect at the point S. If a plane ax + by - z + d = 0 passes through S and is parallel to both the lines L₁ and L₂, then the value of a + b + d is equal to ____________.

Solution

Maths sir · 4 steps

Parallel to both lines means the normal is their cross product. Then S fixes d.

1. Find S. Equate the two lines: λ = 1 + μ, 2λ = 3 + μ, 3λ = 1 + 5μ. The first two give μ = 1, λ = 2, and the third checks out (6 = 6). So S = (2, 4, 6).
2. Normal to plane. The plane is parallel to both lines, so its normal is perpendicular to both direction vectors (1, 2, 3) and (1, 1, 5). Take their cross product.

   n⃗ = |î, ĵ, k̂; 1, 2, 3; 1, 1, 5| = 7î - 2ĵ - k̂

3. Match and find d. The z-coefficient is already -1, as in ax + by - z + d = 0, so a = 7 and b = -2. Putting in S: 14 - 8 - 6 + d = 0, so d = 0.
4. So. a + b + d = 7 - 2 + 0 = 5.

The common slip: -5. Taking the normal as (-7, 2, 1) without scaling it to match the -z term gives a = -7, b = 2 and a total of -5.

The idea to keep. A plane parallel to two lines has its normal along the cross product of their directions. Scale that normal to match the given coefficient, then use one point to get d.

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