Question

Find the vector equation of the line which is parallel to the vector $3\hat{i}-2\hat{j}+6\hat{k}$ and which passes through the point (1,-2,3).

Answer 1

Let $\overrightarrow{a}=3\hat{i}-2\hat{j}+6\hat{k}and\overrightarrow{b}=\hat{i}-2\hat{j}+3\hat{k}$
So, vector equation of the line, which is parallel to the vector $\overrightarrow{a}=3\hat{i}-2\hat{j}+6\hat{k}$ and passes through the vector $\overrightarrow{b}=\hat{i}-2\hat{j}+3\hat{k}is\overrightarrow{b}+\lambda \overrightarrow{a}.\therefore \overrightarrow{r}=(\hat{i}-2\hat{j}+3\hat{k})+\lambda (3\hat{i}-2\hat{j}+6\hat{k})\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})=\lambda (3\hat{i}-2\hat{j}+6\hat{k})\Rightarrow (x-1)\hat{i}+(y+2)\hat{j}+(z-3)\hat{k}-=\lambda (3\hat{i}-2\hat{j}+6\hat{k})$