Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$ is…

Question

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$ is $\frac{1}{\sqrt{6}}$, be $\mathrm{a}, \mathrm{b}$, $(a \lt b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is :-

Aicharya · Accepted Answer

$	ext { shortest distance }=\frac{|(\overline{\mathrm{a}}-\overline{\mathrm{b}})| \cdot(\overline{\mathrm{p}} 	imes \overline{\mathrm{q}})}{|\overline{\mathrm{p}} 	imes \overline{\mathrm{q}}|}$ where $\begin{aligned} & \overline{\mathrm{a}}=-\hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \ & \overline{\mathrm{~b}}=\mathrm{p} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ & \overline{\mathrm{p}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}-\overline{\mathrm{b}}=(-1-\mathrm{p}) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \ & \overline{\mathrm{q}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \ & \frac{1}{16}=\frac{|-1-\mathrm{p}+4-1|}{\sqrt{6}} \ & |-\mathrm{p}+2|=1 \ & \mathrm{p}=3 \ & \frac{\mathrm{x}^2}{1^2}+\frac{\mathrm{y}^2}{3^3}=1 \ & \mathrm{~L} \cdot \mathrm{R}=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 	imes 1}{3}=\frac{2}{3} \end{aligned}$ option (3) ^

Let the values of p , for which the shortest distance between the lines…

Solution