
Shortes distance (CD) =∣p×q∣AB⋅p×q =355(0i^+2j^+2k^)⋅(−15i^+7j^+9k^)=3550+14+18=35532∴m+n=32+355=387
JEE Main 2024 — Mathematics Vectors & 3D Geometry
If the shortest distance between the lines $\begin{array}{ll}
L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \
L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \quad \mu \in \mathbb{R}
\end{array}is\frac{m}{\sqrt{n}},where\operatorname{gcd}(m, n)=1,thenthevalueofm+n$ equals
Held on 8 Apr 2024 · Verified 6 Jul 2026.
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