If the shortest distance between the linesL_1 | vedclass

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(B) The lines are given by:$L_1: \overrightarrow{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 4\hat{k})$$L_2: \overrightarrow{r

Vedclass · Accepted Answer

$387$

Vedclass · Answer

(B) The lines are given by:$L_1: \overrightarrow{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 4\hat{k})$$L_2: \overrightarrow{r} = (2\hat{i} + 3\hat{j} + 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$Let $\overrightarrow{a_1} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\overrightarrow{a_2} = 2\hat{i} + 3\hat{j} + 5\hat{k}$.Let $\overrightarrow{p} = \hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{q} = 2\hat{i} + 3\hat{j} + \hat{k}$.Then $\overrightarrow{a_2} - \overrightarrow{a_1} = (2-2)\hat{i} + (3-1)\hat{j} + (5-3)\hat{k} = 0\hat{i} + 2\hat{j} + 2\hat{k}$.The cross product $\overrightarrow{p} 	imes \overrightarrow{q}$ is:$\overrightarrow{p} 	imes \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 4 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3-12) - \hat{j}(1-8) + \hat{k}(3+6) = -15\hat{i} + 7\hat{j} + 9\hat{k}$.The magnitude $|\overrightarrow{p} 	imes \overrightarrow{q}| = \sqrt{(-15)^2 + 7^2 + 9^2} = \sqrt{225 + 49 + 81} = \sqrt{355}$.The shortest distance is given by $d = \left| \frac{(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{p} 	imes \overrightarrow{q})}{|\overrightarrow{p} 	imes \overrightarrow{q}|} ight|$.$d = \left| \frac{(0\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-15\hat{i} + 7\hat{j} + 9\hat{k})}{\sqrt{355}} ight| = \left| \frac{0 + 14 + 18}{\sqrt{355}} ight| = \frac{32}{\sqrt{355}}$.Comparing with $\frac{m}{\sqrt{n}}$,we have $m = 32$ and $n = 355$.Since $\operatorname{gcd}(32, 355) = 1$,the value of $m+n = 32 + 355 = 387$.

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