3D Geometry
Three Dimensional Geometry
nta_pyq_2025_jan
Grade 12
Question:
Let P be the foot of the perpendicular from the point Q(10, -3, -1) on the line x-3 y-2 z+1 7 = -1 = -2 . Then the area of the right angled triangle P QR, where R is the point (3, -2, 1), is
9\sqrt15
\sqrt30
8\sqrt15
3\sqrt30
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 - 3 y - 2 z + 1 = = = \lambda 7 -1 -2 \Rightarrow 7\lambda + 3, -\lambda + 2, -2\lambda - 1 dr's of QP \Rightarrow 7\lambda - 7, -\lambda + 5, -2\lambda Now (7\lambda - 7) ⋅ 7 - (-\lambda + 5) + (2\lambda) ⋅ 2 = 0 54\lambda - 54 = 0 \Rightarrow \lambda = 1 \therefore P = (10, 1, -3) - -\to ^ ^ PQ = -4 j + 2k - -\to ^ ^ ^ PR = -7 i - 3 j + 4k i j k∣ ∣ 1 ∣ ∣ Area = ∣ ∣ 0 -4 2 = 3\sqrt30 ∣ 2 ∣ ∣ -7 -3 4∣
Correct Answer: 4