Question

Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.

Answer 1

The equation of the line 5x − 25 = 14 − 7y = 35z can be re-written as

$$\begin{gathered} \frac{x-{\displaystyle 5}}{{\displaystyle \frac{1}{5}}}=\frac{y-2}{{\displaystyle \frac{-1}{7}}}=\frac{z}{{\displaystyle \frac{1}{35}}} \\ \Rightarrow \frac{x-{\displaystyle 5}}{{\displaystyle 7}}=\frac{y-2}{-5}=\frac{z}{1} \end{gathered}$$

Since the required line is parallel to the given line, so the direction ratios of the required line are proportional to $7,-5,1$.

The vector equation of the required line passing through the point (1, 2, $-$1) and having direction ratios proportional to $7,-5,1$ is

$\overrightarrow{r}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\lambda \left(7\hat{i}-5\hat{j}+\hat{k}\right)$.