3D Geometry
Plane Through Line of Intersection — Distance Condition
nta_pyq_2023_apr
Grade 12
Question:
Plane through intersection of $x+2y+az=2$ and $x-y+z=3$ is $5x-11y+bz=6a-1$. For $c\in\mathbb{Z}$, distance from $(a,-c,c)$ is $\frac{2}{\sqrt{a}}$. Then $a+b$ is equal to
Step-by-Step Solution
Key Concept: Family plane and given form determine $a,b$. Distance condition gives $c$.
$a+b=4$.
Correct Answer: 2