Find the vector equation of the line passing through the point with position vector and parallel to the vector Deduce the Cartesian equations of the line.

Home » Find the vector equation of the line passing through the point with position vector and parallel to the vector Deduce the Cartesian equations of the line.

Find the vector equation of the line passing through the point with position vector and parallel to the vector Deduce the Cartesian equations of the line.

Find the vector equation of the line passing through the point with position vector $(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}})$ and parallel to the vector $(\hat{i}+3 \hat{j}-\hat{k})$ . Deduce the Cartesian equations of the line.
Answer
Given: line passes through $2 \hat{\imath}+\hat{\imath}-5 \hat{k}$ and is parallel to $\hat{i}+3 \hat{\jmath}-\hat{k}$
To find: equation of line in vector and Cartesian forms
Formula Used: Equation of a line is
Vector form: $\overrightarrow{\mathrm{I}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$
Cartesian form: $\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{y}}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{k}_{\mathrm{d}}}=\lambda$
where $_{a}^{-y}=x_{1} \hat{i}+y_{1} \hat{l}+z_{1} \hat{k}$ is a point on the line and $\vec{b}=b_{1} \hat{i}+h_{2} \hat{\jmath}+b_{3} \hat{k}$ is a vector parallel to the line.
Explanation:
Here, $\vec{a}=\lambda \hat{\imath}+\hat{\jmath}-5 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{\jmath}-\hat{k}$
Therefore,
Vector form: