Question

Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.

Answer 1

The equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as

$$\begin{gathered} \frac{x-{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{6}}}=\frac{y+{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{3}}}=\frac{z-1}{{\displaystyle \frac{1}{2}}} \\ =\frac{x-{\displaystyle \frac{1}{3}}}{{\displaystyle 1}}=\frac{y+{\displaystyle \frac{1}{3}}}{2}=\frac{z-1}{3} \end{gathered}$$

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to $1,2,3$.

The vector equation of the required line passing through the point (2, $-$1, $-$1) and having direction ratios proportional to $1,2,3$ is $\overrightarrow{r}=\left(2\hat{i}-\hat{j}-\hat{k}\right)+\lambda \left(\hat{i}+2\hat{j}+3\hat{k}\right)$.