Find the vector equation of the line parallel to the line x-1/5=3-y/2=z+1/4 and passing through (3,0,-4). Also, find the distance between the 2 lines

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given} \mathrm{line}: \\ \frac{\mathrm{x}-1}{5}=\frac{3-\mathrm{y}}{2}=\frac{\mathrm{z}+1}{4} \\ \Rightarrow \frac{\mathrm{x}-1}{5}=\frac{\mathrm{y}-3}{-2}=\frac{\mathrm{z}+1}{4} \\ \mathrm{hence} \mathrm{dr}\text{'}\mathrm{s} \mathrm{of} \mathrm{the} \mathrm{line} \mathrm{are} 5,-2,4 \\ \mathrm{the} \mathrm{new} \mathrm{line} \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{the} \mathrm{given} \mathrm{line}, \mathrm{hence} \mathrm{its} \mathrm{dr}\text{'}\mathrm{s} \mathrm{will} \mathrm{be} \mathrm{same}. \\ \mathrm{And} \mathrm{the} \mathrm{new} \mathrm{line} \mathrm{passes} \mathrm{through} \left(3,0,-4\right) \\ \stackrel{\rightharpoonup }{\mathrm{a}}=3\hat{\mathrm{i}}-4\hat{\mathrm{k}} \\ \stackrel{\rightharpoonup }{\mathrm{b}}=5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \mathrm{vector} \mathrm{equation} \mathrm{of} \mathrm{new} \mathrm{line} \mathrm{be}: \\ \stackrel{\rightharpoonup }{\mathrm{r}}=\stackrel{\rightharpoonup }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\rightharpoonup }{\mathrm{b}} \\ \stackrel{\rightharpoonup }{\mathrm{r}}=3\hat{\mathrm{i}}-4\hat{\mathrm{k}}+\mathrm{\lambda }\left(5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right) \\ \mathrm{now}, \\ \mathrm{a} \mathrm{point} \mathrm{lying} \mathrm{on} \mathrm{first} \mathrm{line} \mathrm{is}- \\ \left(1,3,-1\right) \\ \stackrel{\rightharpoonup }{{\mathrm{a}}_1}=\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \mathrm{Shortest} \mathrm{distance} \mathrm{between} \mathrm{two} \mathrm{parallel} \mathrm{lines} \mathrm{is}: \\ \mathrm{d}=\frac{\left|\overrightarrow{\mathrm{b}}\times \left(\overrightarrow{{\mathrm{a}}_1}-\overrightarrow{\mathrm{a}}\right)\right|}{\left|\overrightarrow{\mathrm{b}}\right|} \\ \mathrm{where} \stackrel{\rightharpoonup }{\mathrm{b}}=5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \mathrm{d}=\frac{\left|\left(5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\times \left(\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}}-3\hat{\mathrm{i}}+4\hat{\mathrm{k}}\right)\right|}{\left|\overrightarrow{\mathrm{b}}\right|} \\ \mathrm{d}=\frac{\left|\left(5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\times \left(-2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)\right|}{\sqrt{5^2+{\left(-2\right)}^2+4^2}} \\ \mathrm{d}=\frac{\left|\left(5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\times \left(-2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)\right|}{\sqrt{25+16+4}}=\frac{\left|\left(5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\times \left(-2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)\right|}{\sqrt{45}} \\ \mathrm{d}=\frac{\left|-18\hat{\mathrm{i}}-23\hat{\mathrm{j}}+11\hat{\mathrm{k}}\right|}{\sqrt{45}}=\frac{\sqrt{974}}{\sqrt{45}}=4.65 \mathrm{units} \end{gathered}$$

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