3D Geometry
Image in Plane — Distance from Second Point
nta_pyq_2023_apr
Grade 12
Question:
Image of $(5,5,8)$ in $x-2y+z-2=0$ is $P$. Distance of $Q(6,-2,\alpha)$, $\alpha>0$ from $P$ is $\frac{13}{3}$. Then $\alpha$ is equal to _______
Step-by-Step Solution
Key Concept: Find image $P$ of $(5,5,8)$ in plane. Set $|PQ|=\frac{13}{3}$, solve for $\alpha>0$.
$\alpha=15$.
Correct Answer: 15