Show that the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{Z}-3}{4}$ and $\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{Z}$ intersect each other. Also, find the point of their intersection.
Answer
Given: The equations of the two lines are $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$ and $\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{z}$
Io Prove: The two lines intersect and to find their point of intersection.
Formula Used: Equation of a line is
Vector form: $\vec{I}=\vec{a} \downarrow \overrightarrow{k b}$
Cartesian form: $\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{z}}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{k}_{\mathrm{j}}}=\lambda$
where $\hat{a}=\mathrm{x}_{1} \hat{\mathrm{I}}-\mathrm{y}_{1} \hat{l}+\mathrm{z}_{1} \hat{\mathrm{k}}$ is a point on the line and $\mathrm{b}_{1}: \mathrm{b}_{2}: \mathrm{b}_{3}$ is the direction ratios of the line.
Proof:
Let
So a point on the first line is $\left(2 \lambda_{1}+1,3 \lambda_{1}+2,4 \lambda_{1}+3\right)$
A point on the second line is $\left(5 \lambda_{2}+4,2 \lambda_{2}+1, \lambda_{2}\right)$
If they intersect they should have a common point.
$$
Therefore, $\lambda_{1}=-1$
Substituting for the $z$ coordinate, we get
$4 \lambda_{1}+3=-1$ and $\lambda_{2}=-1$
So, the lines intersect and their point of intersection is $(-1,-1,-1)$