If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :

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Aicharya · Accepted Answer

$\left.\begin{array}{l}\overline{\mathrm{r}}_1=(\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\alpha(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \ \overline{\mathrm{r}}_2=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+\beta(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\end{array}ight\} \begin{gathered}\overline{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \ \overline{\mathrm{a}}_2+\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \ \mathrm{a}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\end{gathered}$ Shortest dist. $=\frac{\left|\overline{\mathrm{b}} 	imes\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1ight)ight|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}}$ $\begin{aligned} & \left\lvert\, \frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) 	imes((2-\lambda) \hat{i}+4 \hat{k})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}ight. \ & |-8 \hat{\mathrm{j}}-3(2-\lambda) \hat{k}+12 \hat{i}+4(2-\lambda) \hat{j}|=13 \ & |12 \hat{i}-4 \lambda \hat{j}+(3 \lambda-6) \hat{k}|=13\end{aligned}$ $\begin{aligned} & 144+16 \lambda^2+(3 \lambda-6)^2=169 \ & 16 \lambda^2+(3 \lambda-6)^2=25=\lambda \Rightarrow=1\end{aligned}$

If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and…

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