Question

If the shortest distance between the lines \[\frac{x-\mathrm{k}}{2}=\frac{y-4}{3}=\frac{z-3}{4}\] and \[\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}\] is \[\frac{13}{\sqrt{29}}\] then k =

विकल्प

• 1

• -1

• 2

• -2

Answer 1

1

Explanation:

\[\mathbf{\bar{r}_1}=\left(\mathbf{k}\mathbf{\hat{i}}+4\mathbf{\hat{j}}+3\mathbf{\hat{k}}\right)+\alpha\left(2\mathbf{\hat{i}}+3\mathbf{\hat{j}}+4\mathbf{\hat{k}}\right)\]

\[\bar{\mathrm{r}}_2=2\hat{\mathrm{i}}+4\hat{\mathrm{j}}+7\hat{\mathrm{k}}+\beta\left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\]

\[\therefore\quad\mathrm{\bar{b}=2\hat{i}+3\hat{j}+4\hat{k}}\]

\[\mathbf{\bar{a}_1}=\mathbf{k}\mathbf{\hat{i}}+4\mathbf{\hat{j}}+3\mathbf{\hat{k}}\]

\[\mathbf{a}_2=2\mathbf{\hat{i}}+4\mathbf{\hat{j}}+7\mathbf{\hat{k}}\]

\[\text{Shortest distance}=\frac{\left|\overline{\mathbf{b}}\times\left(\overline{\mathbf{a}}_2-\overline{\mathbf{a}}_1\right)\right|}{\left|\mathrm{b}\right|}\]

\[\therefore\quad\frac{13}{\sqrt{29}}=\frac{\left|\left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\times\left[\left(2-\mathrm{k}\right)\hat{\mathrm{i}}+4\hat{\mathrm{k}}\right]\right|}{\sqrt{29}}\]

\[\therefore\quad\left|12\hat{\mathrm{i}}-4\hat{\mathrm{kj}}+(3\mathrm{k}-6)\hat{\mathrm{k}}\right|=13\]

\[\therefore\quad144+16\mathrm{k}^2+(3\mathrm{k}-6)^2=169\]

\[\therefore\quad16\mathrm{k}^2+(3\mathrm{k}-6)^2=25\]

∴ k = 1