Question

The vector equation of the line passing through the point $\hat{i}-2\hat{j}+\hat{k}$ and perpendicular to the vectors $2\hat{i}-3\hat{j}-1\hat{k},\hat{i}+4\hat{j}-2\hat{k}$ is

• A. $\overrightarrow{r}=(\hat{i}-2\hat{j}+\hat{k})+t(1\hat{i}-7\hat{j}+1\hat{k})$
• B. $\overrightarrow{r}=(\hat{i}-2\hat{j}+\hat{k})+t(3\hat{i}+\hat{j}-3\hat{k})$
• C. $\overrightarrow{r}=(\hat{i}-2\hat{j}+\hat{k})+t(10\hat{i}+3\hat{j}+11\hat{k})$
• D. $r=\hat{i}$

Answer 1

Answer: (C)

$\overrightarrow{r}=(\hat{i}-2\hat{j}+\hat{k})+t(10\hat{i}+3\hat{j}+11\hat{k})$

$\overrightarrow{a}$ which is to line
$\overrightarrow{a}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& -1\\ 1& 4& -2\end{array}\right|=10\hat{i}+3\hat{j}+11\hat{k}$
$\overrightarrow{r}=\hat{i}-2\hat{j}+\hat{k}+\lambda (10\hat{i}+3\hat{j}+11\hat{k})$