3D Geometry
Midpoint of Intersection Segment — Distance from Plane
nta_pyq_2023_apr
Grade 12
Question:
Line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ meets $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ at $A$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at $B$. Distance of midpoint of $AB$ from $2x-2y+z=14$ is
3
$\dfrac{11}{3}$
4
$\dfrac{10}{3}$
Step-by-Step Solution
Key Concept: Find $A$ and $B$ by solving intersection conditions. Midpoint, then distance formula.
Distance $=4$.
Correct Answer: 3