A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.
A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

A line passes through the point (2,1,-3) and is parallel to the vector (\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}). Find the equations of the line in vector and Cartesian forms.
Answer
Given: line passes through (2,1,-3) and is parallel to \hat{\imath}-2 \hat{\jmath}+3 \hat{k}
To find: equation of line in vector and Cartesian forms
Formula Used: Equation of a line is
Vector form: \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}
Cartesian form: \frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{2}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{~b}_{3}}=\lambda
where \vec{a}=x_{1} \hat{\imath}+y_{1} \hat{\jmath}+z_{1} \hat{k} is a point on the line and \vec{b}=b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k} is a vector parallel to the line.
Explanation:
Here, \overrightarrow{\mathrm{a}}=2 \hat{\imath}+\hat{\jmath}-3 \hat{\mathrm{k}} and \overrightarrow{\mathrm{b}}=\hat{\mathrm{\imath}}-2 \hat{\mathrm{\jmath}}+3 \hat{\mathrm{k}}
Therefore,
Vector form:

r=2ı^+ȷ^3k^+λ(ı^2ȷ^+3k^)
\overrightarrow{\mathrm{r}}=2 \hat{\imath}+\hat{\jmath}-3 \hat{k}+\lambda(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})

Cartesian form:

x21=y12=z+33
\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-1}{-2}=\frac{\mathrm{z}+3}{3}