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 $(-2-2):(4-3):(2-3)$

$\begin{array}{l}\Rightarrow –4:1:–1\\ \Rightarrow 4:–1:1\end{array}$
\begin{array}{l}
\Rightarrow-4: 1:-1 \\
\Rightarrow 4:-1: 1
\end{array}

So, $\overrightarrow{\mathrm{b}}=4 \hat{\mathrm{l}}-\hat{\mathrm{l}}+\hat{\mathrm{k}}$
Therefore,
Vector form:

$\overrightarrow{\mathrm{r}}=3\hat{\mathbf{i}}–2\hat{ı}+\hat{k}+\lambda (4\hat{ı}–\hat{ȷ}+\hat{k})$
\overrightarrow{\mathrm{r}}=3 \hat{\mathbf{i}}-2 \hat{\imath}+\hat{k}+\lambda(4 \hat{\imath}-\hat{\jmath}+\hat{k})

Cartesian form:

$\frac{x–3}{4}=\frac{y+2}{–1}=\frac{z–1}{1}$
\frac{x-3}{4}=\frac{y+2}{-1}=\frac{z-1}{1}