3D Geometry
Three Dimensional Geometry
nta_pyq_2025_jan
Grade 12
Question:
Let P be the foot of the perpendicular from the point (1, 2, 2) on the line L : x-1 = y+1 = z-2 . Let the line 1 -1 2 \to ^ ^ ^ ^ ^ ^ r = (- i + j - 2k) + \lambda( i - j + k), \lambda \in R , intersect the line L at Q . Then 2(PQ) is equal to : 2
Step-by-Step Solution
Key Concept: Apply the core result for lines and planes in three dimensions and simplify using the given constraints.
General point on line L : x-1 y+1 z-2 = = 1 -1 2 (4) is (\lambda + 1, -\lambda - 1, 2\lambda + 2) DR's of PM are (\lambda, -\lambda - 3, 2\lambda) PM \perp L \Rightarrow \lambda + (-1)(-\lambda - 3) + 2(2\lambda) = 0 \Rightarrow 6\lambda + 3 = 0 1 -1 P ( , , 1) 2 2 x+1 y-1 z+2 Let another line L : ′ 1 = -1 = 1 General point on line L is (\mu - 1, -\mu + 1, \mu - 2) ′ Point of intersection of line L and L is ′ \lambda + 1 = \mu - 1 2\lambda + 2 = \mu - 2 \Rightarrow \mu - \lambda = 2 \ldots (1) \Rightarrow 2\lambda = \mu - 4 Q(-1, 1, -2) 2 2 1 -1 2 2 2(P Q) = 2 (( + 1) + ( - 1) + (1 + 2) ) 2 2 9 9 = 2( + + 9) 4 4 = 27
Correct Answer: 4