Question

The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.

Answer 1

The cartesian equation of the given line is

3x + 1 = 6y − 2 = 1 − z

It can be re-written as

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

Thus, the given line passes through the point $\left(-\frac{1}{3},\frac{1}{3},1\right)$ and its direction ratios are proportional to 2, 1, $-$6. It is parallel to the vector $\overrightarrow{b}=2\hat{i}+\hat{j}-6\hat{k}$.

We know that the vector equation of a line passing through a point with position vector $\overrightarrow{a}$ and parallel to the vector $\overrightarrow{b}$ is $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$.

Vector equation of the required line is
$$\begin{gathered} \overrightarrow{r}=\left(-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k}\right)+\lambda \left(2\hat{i}+\hat{j}-6\hat{k}\right) \\ \mathrm{Here},\lambda \mathrm{is}\mathrm{a}\mathrm{parameter}. \end{gathered}$$