Question

If the roots of the equation $5x^2-7x+k=0$ 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 ${\overrightarrow{a}}_1=4\hat{i}-\hat{j},{\overrightarrow{a}}_2=\hat{i}-\hat{j}+2\hat{k}$ ${\overrightarrow{b}}_1=\hat{i}+2\hat{j}-3\hat{k},{\overrightarrow{b}}_2=2\hat{i}+4\hat{j}-5\hat{k},$ 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 ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ 2& 4& -5\end{array}\right|$ $=2\hat{i}-\hat{j}+0\hat{k},$ $\Rightarrow$ $|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|=\sqrt{4+1+0}=\sqrt{5}$ $\therefore$ $d=\left|\frac{(-3\hat{i}+0\hat{j}+2\hat{k}).(2\hat{i}-\hat{j}+0k)}{\sqrt{5}}\right|$ $=\left|-\frac{6}{\sqrt{5}}\right|$ $\Rightarrow$ $d=\frac{6}{\sqrt{5}}$