Show that the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=z$ and $\frac{\mathrm{x}+1}{5}=\frac{\mathrm{y}-2}{1}, Z=2$ do not intersect each other.
Answer
Given: The equations of the two lines are $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\mathrm{z}$ and $\frac{\mathrm{x}+1}{5}=\frac{\mathrm{y}-2}{1}, \mathrm{z}=2$
To Prove: the lines do not intersect each other.
Formula Used: Equation of a line is
Vector form: $\overrightarrow{\vec{l}}=\vec{a}+\overrightarrow{k b}$
Cartesian form: $\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{y-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{y}}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{k}_{\mathrm{s}}}=\lambda$
where $\vec{a}=x_{1} \hat{1}+y_{1} \hat{l}+z_{1} k$ is a point on the line and $b_{1}: b_{2}: b_{3}$ is the direction ratios of the line.
Proof:
Let

$\frac{x–1}{3}=\frac{y+1}{3}=z={\lambda }_1$
\frac{x-1}{3}=\frac{y+1}{3}=z=\lambda_{1}
$\frac{\mathrm{x}+1}{\mathrm{5}}=\frac{\mathrm{y}–2}{}={\lambda }_2,\mathrm{z}=2$
\frac{\mathrm{x}+1}{\mathrm{5}}=\frac{\mathrm{y}-2}{\mathfrak{}}=\lambda_{2}, \mathrm{z}=2

So a point on the first line is $\left(2 \lambda_{1}+1,3 \lambda_{1}-1, \lambda_{1}\right)$
A point on the second line is $\left(5 \lambda_{2}-1, \lambda_{2}+1,2\right)$
If they intersect they should have a common point.

$\begin{array}{l}2{\lambda }_1+1=5{\lambda }_2–1\Rightarrow 2{\lambda }_1–5{\lambda }_2=–2\dots \\ 3{\lambda }_1–1={\lambda }_2+1\Rightarrow 3{\lambda }_1–{\lambda }_2=2\dots \end{array}$
\begin{array}{l}
2 \lambda_{1}+1=5 \lambda_{2}-1 \Rightarrow 2 \lambda_{1}-5 \lambda_{2}=-2 \ldots \\
3 \lambda_{1}-1=\lambda_{2}+1 \Rightarrow 3 \lambda_{1}-\lambda_{2}=2 \ldots
\end{array}

Solving (1) and (2),

$\begin{array}{l}–13{\lambda }_2=–10\\ {\lambda }_2=\frac{10}{13}\end{array}$
\begin{array}{l}
-13 \lambda_{2}=-10 \\
\lambda_{2}=\frac{10}{13}
\end{array}

Therefore, $\lambda_{1}=\frac{33}{65}$
Substituting for the z coordinate, we get
$\lambda_{1}=\frac{33}{\text { กร }}$ and $z=2$
So, the lines do not intersect.