# Show that the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda \;and\;\frac{x-4}{5}=\frac{y-1}{2}=z=\mu\;$intersect. Also,find thier point of intersection.

$\begin{array}{1 1} \text{The point of intersection=(-1,-1,-1)} \\ \text{The point of intersection=(-1,-1,1)} \\ \text{The point of intersection=(-1,1,-1)} \\ \text{The point of intersection=(1,-1,-1)} \end{array}$

Toolbox:
• Any point on the line $\large\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}=\lambda$ is given by $(l\lambda +x_1,m\lambda +y_1,n\lambda +z_1)$
• If this point satisfy any equation of line then the lines intersect.
Given equations of the lines are
$\large\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$.........(i)
$\large\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}=\mu$........(ii)
We know that any point on the line $\large\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}=\lambda$ is given by $(l\lambda +x_1,m\lambda +y_1,n\lambda +z_1)$
Any point in (i) is $A(2\lambda+1,3\lambda+2,4\lambda+3)$
Any point in (ii) is $B(5\mu +4,2\mu+1,\mu+0)$
For point of intersection A=B,
$\Rightarrow 2\lambda+1=5\mu+4$.........(iii)
$3\lambda+2=2\mu+1$..........(iv)
$4\lambda+3=\mu$..............(v)
Solving (iii) and (iv) we get
$2\lambda -5\mu -3=0\:\:and\:\:3\lambda -2\mu +1=0$
$\Rightarrow 6\lambda -15\mu -9=0\:\:and\:\:6\lambda -4\mu +2=0$
subtracting both we get $-11\lambda -11=0$
$\lambda=-1 \:\:and\:\:\mu=-1$
These values of $\lambda\:and\:\mu$ satisfy (v)
$\therefore$ The lines intersect.
The point of intersection A is obtained by

substituting the value of $\lambda=-1$ in A

A(-1,-1,-1)