A line is drawn in the direction and it passes through a point with position vector Find the equations of the line in the vector as well as Cartesian forms.

Home » A line is drawn in the direction and it passes through a point with position vector Find the equations of the line in the vector as well as Cartesian forms.

A line is drawn in the direction and it passes through a point with position vector Find the equations of the line in the vector as well as Cartesian forms.

A line is drawn in the direction of $(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$ and it passes through a point with position vector $(2 \hat{i}-\hat{j}-4 \hat{k}) .$ Find the equations of the line in the vector as well as Cartesian forms.
Answer
Given: line passes through $2 \hat{\imath}-\hat{\jmath}-4 \hat{k}$ and is drawn in the direction of $\hat{i}+\hat{\jmath}-2 \hat{k}$
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 $\vec{a}=x_{1} \hat{\imath}+y_{1} \hat{l}+z_{1} \hat{k}$ is a point on the line and $\vec{b}=b_{1} \hat{l}+b_{2} \hat{\jmath}+b_{3} \hat{k}$ is a vector parallel to the line.
Explanation:
Since line is drawn in the direction of $\widehat{(1}+\hat{\jmath}-\widehat{2 \hat{k}})$ , it is parallel to $(\hat{\imath}+\hat{\jmath}-2 \hat{k})$
Here, and $\vec{b}=\hat{i}+-\hat{l}-2 \hat{k}$