The Cartesian equations 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.
The Cartesian equations 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.

The Cartesian equations of a line are . Find the fixed point through which it passes, its direction ratios and also its vector equation.
Given: Cartesian equation of line are
Io find: fixed point through which the line passes through, its direction ratios and the vector equation.
Formula Used: Equation of a line is
Vector form:
Cartesian form:
where is a point on the line and is a vector parallel to the line and also its direction ratio.
Explanation:
The Cartesian form of the line can be rewritten as:

$\begin{array}{l}\frac{\mathrm{x}+\frac{1}{3}}{\frac{1}{3}}=\frac{\mathrm{y}–\frac{1}{3}}{\frac{1}{6}}=\frac{\mathrm{z}–1}{–1}=\lambda \\ ⇒\frac{\mathrm{x}+\frac{1}{3}}{2}=\frac{\mathrm{y}–\frac{1}{3}}{1}=\frac{\mathrm{z}–1}{–6}=\lambda \end{array}$
\begin{array}{l}
\frac{\mathrm{x}+\frac{1}{3}}{\frac{1}{3}}=\frac{\mathrm{y}-\frac{1}{3}}{\frac{1}{6}}=\frac{\mathrm{z}-1}{-1}=\lambda \\
\Rightarrow \frac{\mathrm{x}+\frac{1}{3}}{2}=\frac{\mathrm{y}-\frac{1}{3}}{1}=\frac{\mathrm{z}-1}{-6}=\lambda
\end{array}

Therefore, and
So, the line passes through and direction ratios of the line are and vector form is:

$–\frac{–2}{1}=\frac{–1}{3}\stackrel{^}{i}–1·\frac{1}{3}\stackrel{^}{j}–+\stackrel{^}{k}–1–\lambda \left(2\stackrel{^}{i}–1–\stackrel{^}{ȷ}–\xi \stackrel{^}{k}\right)$
-\frac{-2}{1}=\frac{-1}{3} \hat{i}-1 \cdot \frac{1}{3} \hat{j}-+\hat{k}-1-\lambda(2 \hat{i}-1-\hat{\jmath}-\xi \hat{k})