question_answer

The shortest distance between skew-lines \[\overrightarrow{r}=s\text{ }\hat{k}\]and\[\overrightarrow{r}=(1-t)\hat{i}+t\hat{j}\]is:

A)  0                                            

B)  \[\sqrt{2}\]

C)  1                            

D)         \[\frac{1}{\sqrt{2}}\]

E)  2

Correct Answer: D

Solution :

\[\overrightarrow{r}=s\hat{k}\]if\[\overrightarrow{r}={{\overrightarrow{a}}_{1}}+\lambda {{\overrightarrow{b}}_{1}}\] \[\Rightarrow \]               \[{{\overrightarrow{a}}_{1}}=0,\lambda =s,{{\overrightarrow{b}}_{1}}=\hat{k}\] \[\overrightarrow{r}=\hat{i}+t(-\hat{i}+\hat{j})\]if\[\overrightarrow{r}={{\overrightarrow{a}}_{2}}+\mu {{\overrightarrow{b}}_{2}}\] \[\Rightarrow \]               \[{{\overrightarrow{a}}_{2}}=\hat{i},\mu =t,{{\overrightarrow{b}}_{2}}=-\hat{i}+\hat{j}\] \[\therefore \]  \[SD=\left| \frac{({{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}).({{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}})}{{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}} \right|\]                 \[{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    0 & 0 & 1  \\    -1 & 1 & 0  \\ \end{matrix} \right|\]                 \[=\hat{i}(-1)-\hat{j}(1)+k(0)=-\hat{i}-\hat{j}\]                 \[|{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}|=\sqrt{2}\] Now,     \[{{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}=\hat{i}\] \[\therefore \]  \[({{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}).({{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}})=-1\]