The Cartesian equations of a line are $\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{Z}-6}{4}$. Find the vector equation of the line.
Answer
Given: Cartesian equation of line

To find: equation of line in vector form
Formula Used: Equation of a line is
Vector form: $\overrightarrow{\mathrm{I}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$
Cartesian form: $\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{y}}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{~b}_{\mathrm{d}}}=\lambda$
where $_{\hat{A}}^{-\mathrm{g}}=\mathrm{x}_{1} \hat{\mathrm{l}}+\mathrm{y}_{1} \hat{l}+\mathrm{z}_{1} \hat{\mathrm{k}}$ is a point on the line and $\overrightarrow{\mathrm{b}}=\mathrm{b}_{1} \hat{1}+\mathrm{h}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{\mathrm{k}}$ is a vector parallel to the line.
Explanation:
From the Cartesian equation of the line, we can find $\vec{a}$ and $\vec{b}$
Here, $\vec{a}=3 \hat{\imath}-\hat{2}_{1}+6 \hat{k}$ and $\vec{b}=\overline{2}_{1}-5 \hat{\jmath}+4 \hat{k}$