Question

The vector equation of the straight line $\frac{1-x}{3}=\frac{y+1}{-2}=\frac{3-z}{-1}$

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

Answer 1

Answer: (A)

Comparing $\frac{1-x}{3}=\frac{y+1}{-2}=\frac{3-z}{-1}$
with $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$
$\Rightarrow$ $x_1=1,y_1=-1,z_1=3$ and $l=-3,m=-2,z=1$
$\Rightarrow$ $l=3,m=2,z=-1$
$\therefore$ Vector equation of line is
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
$=(1,-1,3)+\lambda (+3,+2,-1)$
$=(\hat{i}-\hat{j}+3\hat{k})+\lambda (+3\hat{i}+2\hat{j}-\hat{k})$
$\therefore$ $\overrightarrow{r}=(\hat{i}-\hat{j}+3\hat{k})+\lambda (3\hat{i}+2\hat{j}-\hat{k})$