3D Geometry
Intersecting Lines — 27(QR)²
nta_pyq_2026_jan
Grade 12
Question:
Let the lines $L_1:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ and $L_2:\vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k})$ intersect at the point $R$. Let $P$ and $Q$ be points on $L_1$ and $L_2$ respectively such that $|\overrightarrow{PR}|=\sqrt{29}$ and $|\overrightarrow{PQ}|=\sqrt{47/3}$. If $P$ lies in the first octant, then $27(QR)^2$ is equal to
Step-by-Step Solution
Key Concept: Solve $L_1=L_2$: $\lambda=-1$, $\mu=-1$, $R=(-1,-1,-1)$. $P$ on $L_1$: $|\overrightarrow{PR}|^2=29(1+\lambda_P)^2=29\Rightarrow\lambda_P=0$. $P=(1,2,3)$.
$27(QR)^2=360$.
Correct Answer: 4