Question

The vector equation of a line passing through the point $\overline{i}+2\overline{j}+3\overline{k}$ and perpendicular to the vectors $\overline{i}+\overline{j}+\overline{k}$ and $2\overline{i}-\overline{j}+\overline{k}$

• A. $\overline{r}=(1+2\lambda )\overline{i}+(2+\lambda )\overline{j}-(3\lambda -3)\overline{k}$
• B. $\overline{r}=(1-2\lambda )\overline{i}+(2+\lambda )\overline{j}+(3-3\lambda )\overline{k}$
• C. $\overline{r}=(1+2\lambda )\overline{i}-(2+\lambda )\overline{j}+(3-3\lambda )\overline{k}$
• D. $\overline{r}=(1+2\lambda )\overline{i}+(2+\lambda )\overline{j}-(3-3\lambda )\overline{k}$

Answer 1

The correct option is A $\overline{r}=(1+2\lambda )\overline{i}+(2+\lambda )\overline{j}-(3\lambda -3)\overline{k}$

Given line is passing through

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$

So eq of line become

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$-------------(1)

line is perpendicular to vectors

$\overrightarrow{n_1}=\hat{i}+\hat{j}+\hat{k}$

and

$\overrightarrow{n_2}=2\hat{i}-\hat{j}+\hat{k}$

line is perpendicular to both lines so

$\overrightarrow{b}=\overrightarrow{n_1}\times \overrightarrow{n_2}$

$\overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& -1& 1\end{array}\right|$

$\overrightarrow{b}=\hat{i}(1+1)-\hat{j}(1-2)+\hat{k}(-1-2)$

$\overrightarrow{b}=2\hat{i}+\hat{j}-3\hat{k}$

putting a and b in eq (1)

So required line of eq

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

$\overrightarrow{r}=(1+2\lambda )\hat{i}+(2+\lambda )\hat{j}+(3-3\lambda )\hat{k}$

$\overrightarrow{r}=(1+2\lambda )\hat{i}+(2+\lambda )\hat{j}-(3\lambda -3)\hat{k}$

This is required eq