3D Geometry
Three Dimensional Geometry
nta_pyq_2025_jan
Grade 12
Question:
Let L 1 : x-1 1 = y-2 y-2 -1 = 2 and L 2 : -1 = 2 z-1 x+1 = z1 be two lines. Let L 3 be a line passing through the point (\alpha, \beta, \gamma) and be perpendicular to both L 1 and L 2 . If L 3 intersects L 1 , then ∣5\alpha - 11\beta - 8\gamma∣ equals :
Step-by-Step Solution
Key Concept: Apply the core result for lines and planes in three dimensions and simplify using the given constraints.
∣ ^ i ^ j ^∣ k \to \to ∣ ∣ (3) DR's of L3 = m \times n = ∣ ∣ 1 -1 2∣ ∣ ∣ -1 2 1∣ ^ ^ ^ = -5 i - 3 j + k x - \alpha y - \beta z - \gamma L3 : = = = \lambda -5 -3 1 A(\alpha - 5\lambda, \beta - 3\lambda, \gamma + \lambda) x - 1 y - 2 z - 1 L1 : = = = k 1 -1 2 B(k + 1, -k + 2, 2k + 1) Now \alpha - 5\lambda = k + 1 \Rightarrow \alpha = 5\lambda + k + 1 \beta - 3\lambda = -k + 2 \Rightarrow \beta = 3\lambda - k + 2 \gamma + \lambda = 2k - 1 \Rightarrow \gamma = -\lambda + 2k + 1 |5\alpha - 11\beta - 8\gamma| = | - 25| = 25
Correct Answer: 3