Question

The vector equation of the line passing through the point $(-1,-1,2)$ and parallel to the line $2x-2=3y+1=6z-2,$ is

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

Answer 1

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

Let $\overrightarrow{a}$ be the position vector of the point $A(-1,-1,2)$
$\therefore \overrightarrow{a}=-\hat{i}-\hat{j}+2\hat{k}$
Given equation of line is
$2x-2=3y+1=6z-2$
$2(x-1)=3(y+\frac{1}{3})=6(z-\frac{1}{3})$
$\frac{x-1}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{3}}{\frac{1}{6}}$
$\therefore$ Direction ratios are
$\frac{1}{2},\frac{1}{3},\frac{1}{6}i.e.3,2,1$
Let $\overrightarrow{b}$ be the vector parallel to required line
$\overrightarrow{b}=3\hat{i}+2\hat{j}+\hat{k}$
$\therefore$ The vector equation of the line passing through A $\left(\overrightarrow{a}\right)$ and parallel to $\overrightarrow{b}$ is
$\overline{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
$\overline{r}=(-\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+2\hat{j}+\hat{k})$