If the shortest distance between the lines L_1: r =(2+ ) i +(1-3 … - Vidyadip

Question

If the shortest distance between the lines

$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $

$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$

is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.

✓

Answer

b
Shortes distance $(\mathrm{CD})=\left|\frac{\overline{\mathrm{AB}} \cdot \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|$

$=\left|\frac{(0 \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-15 \hat{i}+7 \hat{j}+9 \hat{k})}{\sqrt{355}}\right|$

$=\frac{0+14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}}$

$\therefore \mathrm{m}+\mathrm{n}=32+355=387$

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