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) \[\overrightarrow{r}=(\hat{i}-\hat{j}+3\hat{k})+\lambda (3\hat{i}+2\hat{j}+\hat{k})\]
Correct Answer: A
Solution :
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 \[\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\] \[=(1,-1,3)+\lambda (+3,+2,-1)\] \[=(\hat{i}-\hat{j}+3\hat{k})+\lambda (+3\hat{i}+2\hat{j}-\hat{k})\] \[\therefore \] \[\overrightarrow{r}=(\hat{i}-\hat{j}+3\hat{k})+\lambda (3\hat{i}+2\hat{j}-\hat{k})\]
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