3D Geometry
Three Dimensional Geometry
nta_pyq_2025_jan
Grade 12
Question:
The square of the distance of the point ( 15 , 32 , 7) from the line x+1 = y+3 = z+5 in the direction of the vector 7 7 3 5 7 ^ ^ ^ i + 4 j + 7k is :
Step-by-Step Solution
Key Concept: Apply the core result for lines and planes in three dimensions and simplify using the given constraints.
(4) x + 1 y + 3 z + 5 L = = = 3 5 7 15 32 x - y - 7 7 z - 7 PQ = = = = \lambda 1 4 7 15 32 \Rightarrow Q (\lambda + , 4\lambda + , 7\lambda + 7) 7 7 Since Q lies on line L 15 \lambda + + 1 7 7\lambda + 7 + 5 So, = 3 7 \Rightarrow 7\lambda + 22 = 21\lambda + 36 \Rightarrow \lambda = -1 8 4 \therefore Point Q ( , , 0) 7 7 2 2 15 8 32 4 PQ = \sqrt( - ) + ( - ) + (7 - 0) 7 7 7 7 PQ = \sqrt66 2 \Rightarrow (PQ) = 66
Correct Answer: 4