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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 $\operatorname{gcd}(m, n)=1$, then the value of $m+n$ equals
Options:
$\begin{array}{ll}
L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\
L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \quad \mu \in \mathbb{R}
\end{array}$
is $\frac{m}{\sqrt{n}}$, where $\operatorname{gcd}(m, n)=1$, then the value of $m+n$ equals
Solution:
1923 Upvotes
Verified Answer
The correct answer is:
387
Shortes distance (CD) $=\left|\frac{\overline{\mathrm{AB}} \cdot \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|$
$\begin{aligned} & =\left|\frac{(0 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(-15 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})}{\sqrt{355}}\right| \\ & =\frac{0+14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\ & \therefore \mathrm{m}+\mathrm{n}=32+355=387\end{aligned}$
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