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The shortest distance between the lines $\quad \bar{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t)…
The shortest distance between the lines $\quad \bar{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k}$ and
$\bar{r}=(p+1) \hat{\imath}+(2 p-1) \hat{\jmath}+(2 p+1) \hat{k} \quad$ is
$\frac{8}{\sqrt{29}}$ units $\frac{4}{\sqrt{29}}$ units $\frac{2}{\sqrt{5}}$ units $\frac{4}{\sqrt{19}}$ units Solution
$\begin{aligned} \ell_{1}: \overline{\mathrm{r}} &=(1-\mathrm{t}) \hat{\mathrm{i}}+(\mathrm{t}-2) \hat{\mathrm{j}}+(3-2 \mathrm{t}) \hat{\mathrm{k}} \\ &=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mathrm{t}(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \\ \ell_{2}: \overline{\mathrm{r}} &=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{p}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\ \text { Here } & \overline{\mathrm{a}}_{2}-\overline{\mathrm{a}}_{1}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \overline{\mathrm{b}}_{1} \times \overline{\mathrm{b}}_{2}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 1 & -2 \\ 1 & 2 & 2\end{array}\right|=\hat{\mathrm{i}}(2+4)-\hat{\mathrm{j}}(0)-3 \hat{\mathrm{k}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{k}} \end{aligned}$
$\therefore\left|\bar{b}_{1} \times \bar{b}_{2}\right| \quad=\sqrt{36+9}=3 \sqrt{5}$
shortest distance $=\left|\frac{\left(\bar{b}_{1} \times \bar{b}_{2}\right) \cdot\left(\bar{a}_{2}-\bar{a}_{1}\right)}{\left(\bar{b}_{1} \times \bar{b}_{2}\right)}\right|=\left|\frac{(6 \hat{i}-3 \hat{k}) \cdot(\hat{j}-2 \hat{k})}{3 \sqrt{5}}\right|=\frac{6}{3 \sqrt{5}}=\frac{2}{\sqrt{5}}$
Asked in: MHT CET 2020 (13 Oct Shift 1)
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