Question

Find the equation of a line which passes through $(2,-1,3)$ and is perpendicular to the line;
$\overline{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ and $\overline{r}=(\widehat{2i}-\hat{j}+3\hat{k})+\mu (\hat{i}+2\hat{j}+2\hat{k})$

• A. $\overline{r}=(\widehat{2i}-\hat{j}+3\hat{k})+\lambda (\widehat{-6i}-3\hat{j}+6\hat{k})$
• B. $\overline{r}=(\hat{i}+2\hat{j}+2\hat{k})+\lambda (\widehat{-6i}-3\hat{j}+6\hat{k})$
• C. $\overline{r}=(2\hat{i}-2\hat{j}+\hat{k})+\lambda (\widehat{-6i}-3\hat{j}+6\hat{k})$
• D. None of these

Answer 1

The correct option is A $\overline{r}=(\widehat{2i}-\hat{j}+3\hat{k})+\lambda (\widehat{-6i}-3\hat{j}+6\hat{k})$

To find a straight line perpendicular to given lines, $\overrightarrow{r}=\overrightarrow{a_1}+\overrightarrow{{\lambda b}_1}$ and $\overrightarrow{r}=\overrightarrow{a_2}+\overrightarrow{{\lambda b}_2}$ has dr's pro-portional to $\overrightarrow{b}=\overrightarrow{b_1}\times \overrightarrow{b_2}$
Now, $\overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& 2& 2\end{array}\right|=-6\hat{i}-3\hat{j}+6\hat{k}$
Thus, the required line passes through the point $(2,-1,3)$ and is parallel to the vector,
$\overrightarrow{b}=-\hat{6}i-3\hat{j}+6\hat{k}$
$\therefore$ its equation is $\overrightarrow{r}=(2\hat{i}-\hat{j}+\widehat{3k})+\lambda (-6\hat{i}-3\hat{j}+6\hat{k})$