Find the vector equation of a line passing through the point having the position vector $(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to the line joining the points with position vectors $(\hat{i}-\hat{j}+5 \hat{k})$ and $(2 \hat{i}+3 \hat{j}-4 \hat{k})$
Also, find the Cartesian equivalents of this equation.
Answer
$\underline{\text { Given: }}$ line passes through the point with position vector $\hat{\mathbf{i}}+2 \hat{\jmath}-3 \hat{\mathbf{k}}$ and parallel to the line joining the points with position vectors $\hat{i}-\hat{\jmath}+5 \hat{k}$ and $2 \hat{i}+3 \hat{\jmath}-4 \hat{k}$.
Io find: equation of line in vector and Cartesian forms
Formula Used: Equation of a line is
Vector form: $\overrightarrow{\mathrm{I}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{k}} \overrightarrow{\mathrm{b}}$
Cartesian form: $\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{y}}}=\frac{\mathrm{z}-\mathrm{z}_{\mathrm{l}}}{\mathrm{h}_{\mathrm{s}}}=\lambda$.
where $\vec{a}=x_{1} \hat{l}+y_{1} \hat{l}+z_{1} \hat{k}$ is a point on the line and $\vec{b}=b_{1} \hat{l}+b_{2} \hat{l}+b_{3} \vec{k}$ with $b_{1}: b_{2}: b_{3}$ being the direction ratios of the line.
Explanation:
Here, $\overrightarrow{\hat{A}}=\hat{i}+\hat{2} \hat{\jmath}-\overline{3} \hat{k}$
The direction ratios of the line are $(1-2):(-1-3):(5+4)$

$\begin{array}{l}\Rightarrow –1:–4:9\\ \Rightarrow 1:4:–9\\ \text{ So, }\overrightarrow{\mathrm{b}}=\hat{\mathrm{l}}+4\hat{ȷ}–9\overrightarrow{\mathrm{k}}\end{array}$
\begin{array}{l}
\Rightarrow-1:-4: 9 \\
\Rightarrow 1: 4:-9 \\
\text { So, } \overrightarrow{\mathrm{b}}=\hat{\mathrm{l}}+4 \hat{\jmath}-9 \overrightarrow{\mathrm{k}}
\end{array}

Therefore,
Vector form:

$\overrightarrow{r}=\hat{i}+2\hat{ı}–3\hat{k}–+–\lambda (\hat{i}+4–4\hat{ı}–9\hat{k})$
\vec{r}=\hat{i}+2 \hat{\imath}-3 \hat{k}-+-\lambda(\hat{i}+4-4 \hat{\imath}-9 \hat{k})

Cartesian form:

$\frac{x–1}{1}=\frac{y–2}{1}=\frac{z+3}{–9}$
\frac{x-1}{1}=\frac{y-2}{1}=\frac{z+3}{-9}