Question

If the roots of the equation $ 5{{x}^{2}}-7x+k=0 $ are reciprocal of each other, then value of k is

• A. 5
• B. 2
• C. 3
• D. 1

Answer 1

Answer: (A)

Comparing the given equations of lines $ \overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k}) $ and $ \overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k}) $ with the general equations of lines $ \overrightarrow{r}={{\overrightarrow{a}}_{1}}+\lambda {{\overrightarrow{b}}_{1}} $ and $ \overrightarrow{r}={{\overrightarrow{a}}_{2}}+\mu {{\overrightarrow{b}}_{2}} $ we get $ {{\overrightarrow{a}}_{1}}=4\hat{i}-\hat{j},\text{ }{{\overrightarrow{a}}_{2}}=\hat{i}-\hat{j}+2\hat{k} $ $ {{\overrightarrow{b}}_{1}}=\hat{i}+2\hat{j}-3\hat{k},{{\overrightarrow{b}}_{2}}=2\hat{i}+4\hat{j}-5\hat{k}, $ We know that the shortest distance between the lines $ \overrightarrow{r}={{\overrightarrow{a}}_{1}}+\lambda {{\overrightarrow{b}}_{1}} $ and $ \overrightarrow{r}={{\overrightarrow{a}}_{2}}+\mu {{\overrightarrow{b}}_{2}} $ is given by $ d=\left| \frac{({{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}).({{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}})}{|{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}|} \right| $ where $ {{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}=(\hat{i}-\hat{j}-2\hat{k})-(4\hat{i}-\hat{j}) $ $ =-3\hat{i}+2\hat{k} $ $ =-3\hat{i}+0\hat{j}+2\hat{k} $ and $ {{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \\ \end{matrix} \right| $ $ =2\hat{i}-\hat{j}+0\hat{k}, $ $ \Rightarrow $ $ |{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}|=\sqrt{4+1+0}=\sqrt{5} $ $ \therefore $ $ d=\left| \frac{(-3\hat{i}+0\hat{j}+2\hat{k}).(2\hat{i}-\hat{j}+0k)}{\sqrt{5}} \right| $ $ =\left| -\frac{6}{\sqrt{5}} \right| $ $ \Rightarrow $ $ d=\frac{6}{\sqrt{5}} $