A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: C
Solution :
Given, lines can be written in the following form \[\mathbf{r}=(3\mathbf{i}+8\mathbf{j}+3\mathbf{k})+{{\mu }_{1}}(3\mathbf{i}-\mathbf{j}+\mathbf{k})\] and \[\mathbf{r}=(-3\mathbf{i}+7\mathbf{j}+6\mathbf{k})+{{\mu }_{2}}(-3\mathbf{i}+2\mathbf{j}+4\mathbf{k})\] \[\left[ \begin{align} & \text{If}\,\,\text{two}\,\,\text{given}\,\,\text{lines}\,\,\text{are} \\ & \mathbf{r}={{\mathbf{a}}_{1}}+{{\mu }_{1}}{{\mathbf{b}}_{1}} \\ & and\mathbf{r}={{\mathbf{a}}_{2}}+{{\mu }_{2}}{{\mathbf{b}}_{2}} \\ & \text{then}\,\,\text{shortest}\,\,\text{distance}\,\,\text{is}\,\,\text{given}\,\,\text{by} \\ & \left| \frac{({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})\cdot ({{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}})}{|{{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}}|} \right| \\ \end{align} \right]\] \[\therefore \]Shortest distance between two lines \[=\left| \frac{\begin{align} & \{(-3\mathbf{i}-7\mathbf{j}+6\mathbf{k})-(3\mathbf{i}+8\mathbf{j}+3\mathbf{k})\} \\ & \,\,\,\,\,\,\{(3\mathbf{i}-\mathbf{j}+\mathbf{k})\times (-3\mathbf{i}+2\mathbf{j}+4\mathbf{k})\} \\ \end{align}}{|(3\mathbf{i}-\mathbf{j}+\mathbf{k})\times (-3\mathbf{i}+2\mathbf{j}+4\mathbf{k})} \right|\] \[=\left| \frac{(-6\mathbf{i}-15\mathbf{j}+3\mathbf{k})\cdot (6\mathbf{i}-15\mathbf{j}+3\mathbf{k})}{|-6\mathbf{i}-15\mathbf{j}+3\mathbf{k}|} \right|\] \[\left[ \because (3\mathbf{i}-\mathbf{j}+\mathbf{k})\times (-3\mathbf{i}\times 2\mathbf{j}+4\mathbf{k})=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \\ \end{matrix} \right| \right.\] \[\left. \begin{align} & =\mathbf{i}(-4-2)-\mathbf{j}(12+3)+\mathbf{k}(6-3) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-6\mathbf{i}+15\mathbf{j}+3\mathbf{k} \\ \end{align} \right]\] \[=\frac{(36+225+9)}{\sqrt{36+225}+9}\] \[=\frac{270}{\sqrt{270}}=\sqrt{270}-3\sqrt{30}\] Hence,\[\lambda =3\]
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