Answer:
Given vector equation of line is \[\vec{r}=(2\hat{i}+\hat{j})+\lambda (\hat{i}-\hat{j}+4\hat{k})\] \[\Rightarrow \] \[x\hat{i}+y\hat{j}+z\hat{k}=\hat{i}(2+\lambda )+\hat{j}(1-\lambda )+4\lambda \,\,\hat{k}\] On equating coefficients of \[\hat{i},\,\,\hat{j}\] and \[\hat{k}\] from both sides, we get \[x=2+\lambda ,\] \[y=1-\lambda \] and \[z=4\lambda \] \[\Rightarrow \] \[\frac{x-2}{1}=\lambda ,\] \[\frac{y-1}{-\,1}=\lambda \] and \[\frac{z}{4}=\lambda \] \[\therefore \] \[\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z}{4}\] Which is the required cartesian equation of line.
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