3D Geometry
Shortest Distance Line — Midpoint of PQ
nta_pyq_2024_jan
Grade 12
Question:
Let the line of the shortest distance between the lines $L_1:\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $L_2:\vec{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha,\beta,\gamma)$ is the midpoint of the line segment $PQ$, then $2(\alpha+\beta+\gamma)$ is equal to
Step-by-Step Solution
Key Concept: Find P on $L_1$ and Q on $L_2$ such that PQ is perpendicular to both direction vectors. Solve the system, find $\lambda,\mu$, then compute midpoint.
Midpoint $(5/2,2,6)$. $2(\alpha+\beta+\gamma)=21$.
Correct Answer: 21