Question

The equation of the plane containing the line
$\overrightarrow{r}=\hat{i}+\hat{j}+t\left(2\hat{i}+\hat{j}+4\hat{k}\right)$, is

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

Answer 1

The correct option is A $\overrightarrow{r}.\left(\hat{i}+2\hat{j}-\hat{k}\right)=3$

The line

$\overrightarrow{r}=\hat{i}+\hat{j}+t(2\hat{i}+\hat{j}+4\hat{k})$

passes through the point $(1,1,0)$ and parallel to vector

$\overrightarrow{b}=2\hat{i}+\hat{j}+4\hat{k}$

taking plane

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

putting

$\overrightarrow{r}=(\hat{i}+\hat{j}+0\hat{k})$ in equation (i)

$(\hat{i}+\hat{j}+0\hat{k})\cdot (\hat{i}+2j\hat{k}-k)=3$

$1+2+0=3$

$3=3$

therefore the plane (1) contains point $(1,1,0)$

also

$2\times 1+1\times 2+4\times (-1)$

$=2+2-4$

$=4-4$

$=0$

i.e perpendicular to plans(1) is perpendicular to the given line

therefore the plane is $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}-\hat{k})=3$