Question

The line through $\hat{i}+3\hat{j}+2\hat{k}$ and $\perp$ to the line $\hat{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda (2\hat{i}+\hat{j}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}+6\hat{j}+\hat{k})+\mu (\hat{i}+2\hat{j}+3\hat{k})$ is:

• A. $\overrightarrow{r}=(\hat{i}+2\hat{k}j-\hat{k})-\lambda (-\hat{i}+5\hat{j}-3\hat{k})$
• B. $\overrightarrow{r}=\hat{i}+3\hat{j}+2\hat{k}+\lambda (\hat{i}-5\hat{j}+3\hat{k})$
• C. $\overrightarrow{r}=\hat{i}+3\hat{j}+2\hat{k}+\lambda (\hat{i}+5\hat{j}+3\hat{k})$
• D. $\overrightarrow{r}=\hat{i}+3\hat{j}+2\hat{k}+\lambda (-\hat{i}-5\hat{j}-3\hat{k})$

Answer 1

Answer: (B)

$\overrightarrow{r}=\hat{i}+3\hat{j}+2\hat{k}+\lambda (\hat{i}-5\hat{j}+3\hat{k})$

Since the line is perpendicular to the both given lines
$\therefore$ direction ratio of required line is
$\left|\begin{array}{ccc}i& j& k\\ 2& 1& 1\\ 1& 2& 3\end{array}\right|$
$i(3-2)-j(6-1)+k(4-1)$
$=i-5j+3k$
therefore equation line is
${\overrightarrow{r}}^{}=i+3j+2k+\lambda (i-5j+3k)$