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(C) The shortest distance between two lines $\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{p}$ and $\overrightarrow{r} = \overri

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) The shortest distance between two lines $\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{p}$ and $\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{q}$ is given by $d = \frac{|(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{p} 	imes \overrightarrow{q})|}{|\overrightarrow{p} 	imes \overrightarrow{q}|}$.Here,$\overrightarrow{a_1} = -\hat{i}$,$\overrightarrow{p} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\overrightarrow{a_2} = p\hat{i} + 2\hat{j} + \hat{k}$,and $\overrightarrow{q} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.$\overrightarrow{a_2} - \overrightarrow{a_1} = (p+1)\hat{i} + 2\hat{j} + \hat{k}$.$\overrightarrow{p} 	imes \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8) = \hat{i} - 2\hat{j} + \hat{k}$.$|\overrightarrow{p} 	imes \overrightarrow{q}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.Shortest distance $d = \frac{|((p+1)\hat{i} + 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})|}{\sqrt{6}} = \frac{|p+1 - 4 + 1|}{\sqrt{6}} = \frac{|p-2|}{\sqrt{6}}$.Given $d = \frac{1}{\sqrt{6}}$,so $|p-2| = 1$,which implies $p-2 = 1$ or $p-2 = -1$.Thus,$p = 3$ or $p = 1$. Since $a The ellipse is $\frac{x^2}{1^2} + \frac{y^2}{3^2} = 1$. Here $a^2 = 1$ and $b^2 = 9$,so $b > a$.The length of the latus rectum for $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $b > a$ is $\frac{2a^2}{b} = \frac{2(1)}{3} = \frac{2}{3}$.

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