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}) $