3D Geometry
Points at Distance 6 on a Line — Centroid
nta_pyq_2024_jan
Grade 12
Question:
Let $P$ and $Q$ be the points on the line $\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}$ which are at a distance of 6 units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha,\beta,\gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is:
Step-by-Step Solution
Key Concept: Parametrize: points on line are $(8\lambda-3,2\lambda+4,2\lambda-1)$. Distance from $R(1,2,3)$: $(8\lambda-4)^2+(2\lambda+2)^2+(2\lambda-4)^2=36$. Solve for $\lambda$.
Centroid $(1,4,1)$. $\alpha^2+\beta^2+\gamma^2=18$.
Correct Answer: 3