Find the vector equation of a line passing through the point A(3, -2, 1) and parallel to the line joining the points B(-2, 4, 2) and C(2, 3, 3). Also, find the Cartesian equations of the line.

Home » Find the vector equation of a line passing through the point A(3, -2, 1) and parallel to the line joining the points B(-2, 4, 2) and C(2, 3, 3). Also, find the Cartesian equations of the line.

Find the vector equation of a line passing through the point A(3, -2, 1) and parallel to the line joining the points B(-2, 4, 2) and C(2, 3, 3). Also, find the Cartesian equations of the line.

Find the vector equation of a line passing through the point $A(3,-2,1)$ and parallel to the line joining the points $\mathrm{B}(-2,4,2)$ and $\mathrm{C}(2,3,3) .$ Also, find the Cartesian equations of the line.
Answer
Given: line passes through the point $(3,-2,1)$ and is parallel to the line joining points $B(-2,4,2)$ and $\mathrm{C}(2,3,3)$
To find: equation of line in vector and Cartesian forms
Formula Used: Equation of a line is
Vector form: $\overrightarrow{\mathrm{I}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{B}} \overrightarrow{\mathrm{b}}$
Cartesian form: $\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{y-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{y}}}=\frac{z-\mathrm{z}_{1}}{\mathrm{~h}_{\mathrm{J}}}=\lambda$
where $\vec{a}=x_{1} \hat{\imath}+y_{1} \hat{l}+z_{1} \hat{k}$ is a point on the line and $\vec{b}=b_{1} \hat{i}+b_{2} \hat{l}+b_{3} \hat{k}$ with $b_{1}: b_{2}: b_{3}$ being the direction ratios of the line.
Explanation:
Here, $_{a}^{-\hat{\lambda}}=\overline{3} \hat{i}-\overline{2} \hat{\mathrm{l}}+\hat{\mathrm{k}}$
The direction ratios of the line are $(-2-2):(4-3):(2-3)$