Distance Between Two Skew Lines Using Scalar Triple Product

For two skew lines, the shortest distance formula is:
$$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$

From the given lines, identify:
- Point on Line 1: $$\vec{a_1} = \langle 1, 2, 3 \rangle$$, Direction: $$\vec{b_1} = \langle 2, 3, 4 \rangle$$
- Point on Line 2: $$\vec{a_2} = \langle 4, 5, 6 \rangle$$, Direction: $$\vec{b_2} = \langle 1, 2, 2 \rangle$$

Calculate the difference vector:
$$\vec{a_2} - \vec{a_1} = \langle 3, 3, 3 \rangle$$

Calculate the cross product $$\vec{b_1} \times \vec{b_2}$$:
$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 4 \\ 1 & 2 & 2 \end{vmatrix}$$

$$= \vec{i}(3 \cdot 2 - 4 \cdot 2) - \vec{j}(2 \cdot 2 - 4 \cdot 1) + \vec{k}(2 \cdot 2 - 3 \cdot 1)$$

$$= \vec{i}(-2) - \vec{j}(0) + \vec{k}(1) = \langle -2, 0, 1 \rangle$$

Calculate the dot product:
$$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = \langle 3, 3, 3 \rangle \cdot \langle -2, 0, 1 \rangle = -6 + 0 + 3 = -3$$

Calculate the magnitude of the cross product:
$$|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 0 + 1} = \sqrt{5}$$

Therefore, the shortest distance is:
$$d = \frac{|-3|}{\sqrt{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$$

Express in the form $$\frac{\sqrt{n}}{m}$$:
$$\frac{3\sqrt{5}}{5} = \frac{\sqrt{9 \cdot 5}}{5} = \frac{\sqrt{45}}{5}$$

So $$n = 45$$ and $$m = 5$$

Therefore: $$n + m = 45 + 5 = 50$$

$$\boxed{50}$$

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 :

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