Vector Form from Cartesian Line

The incident ray

$$L_1:\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-1}{-2}$$

has direction vector

$$\vec d=(1,-2,-2)$$

The mirror plane is

$$x+y+z=0$$

whose normal vector is

$$\vec n=(1,1,1)$$

The reflected direction vector is

$$\vec r=\vec d-2\frac{\vec d\cdot\vec n}{|\vec n|^2}\vec n$$

Now, $$\vec d\cdot\vec n=1-2-2=-3$$

and $$|\vec n|^2=3$$

Therefore,

$$\vec r=(1,-2,-2)-2\left(\frac{-3}{3}\right)(1,1,1)$$

$$=(1,-2,-2)+2(1,1,1)$$

$$=(3,0,0)$$

Hence the reflected ray is parallel to the x-axis.

Next find the point of incidence.

Let $$\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-1}{-2}=t$$

Then $$x=1+t,\quad y=1-2t,\quad z=1-2t$$

Since the point lies on

$$x+y+z=0$$

$$1+t+1-2t+1-2t=0$$

$$3-3t=0$$

$$t=1$$

Thus the point of incidence is

$$(2,-1,-1)$$

Hence the reflected line is

$$L_2:(x,y,z)=(2,-1,-1)+\lambda(1,0,0)$$

The line

$$L_3$$

can be written as

$$(x,y,z)=(0,2,3)+\mu(2,1,2)$$

Its direction vector is

$$\vec b=(2,1,2)$$

For the shortest distance between skew lines,

$$D=\frac{|(\vec P_2-\vec P_1)\cdot(\vec a\times\vec b)|}{|\vec a\times\vec b|}$$

where

$$\vec a=(1,0,0)$$

and

$$\vec b=(2,1,2)$$

Compute the cross product:

$$\vec a\times\vec b=

\begin{vmatrix}

\hat i & \hat j & \hat k\\

1&0&0\\

2&1&2

\end{vmatrix}$$

$$=(0,-2,1)$$

Therefore,

$$|\vec a\times\vec b|=\sqrt5$$

Also,

$$\vec P_2-\vec P_1=(0,2,3)-(2,-1,-1)$$

$$=(-2,3,4)$$

Now,

$$(-2,3,4)\cdot(0,-2,1)$$

$$=-6+4$$

$$=-2$$

Hence,

$$D=\frac{|-2|}{\sqrt5}$$

$$=\frac{2}{\sqrt5}$$

Final Answer :

$$\frac{2}{\sqrt5}$$