3D Geometry
Plane through line of intersection; distance from a point
nta_pyq_2023_jan
Grade None
Question:
Let the plane containing the line of intersection of the planes P1: $x + (\lambda+4)y + z = 1$ and P2: $2x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane P2 is
$5\sqrt{6}$
$4\sqrt{6}$
$2\sqrt{6}$
$3\sqrt{6}$
Step-by-Step Solution
Key Concept: Write plane as $P_1 + kP_2 = 0$, substitute given points to find $k$ and $\lambda$, then compute distance from P2.
$k=-1$, $\lambda=-4$. Point $(2\lambda,\lambda,-\lambda) = (-8,-4,4)$. Distance from $2x+y+z=2$: $d = \frac{|-16-4+4-2|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6}$. Answer: (4)
Correct Answer: $3\sqrt{6}$