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, \[{{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}=\hat{i}\] \[\therefore \] \[({{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}).({{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}})=-1\]You need to login to perform this action.
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