Cartesian Equation of a Line

Since line $$L$$ is perpendicular to both $$L_1$$ and $$L_2$$, its direction vector $$\vec{v}$$ is the cross product of the direction vectors of $$L_1$$ and $$L_2$$.

• $$\vec{d_1} = \langle 2, 1, -2 \rangle$$
• $$\vec{d_2} = \langle 1, 3, 4 \rangle$$

$$\vec{v} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}$$

$$\vec{v} = \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = \langle 10, -10, 5 \rangle$$.

We can simplify this direction to $$\langle 2, -2, 1 \rangle$$.

The equation of line $$L$$ passing through $$P(2, -1, 3)$$ is:

$$\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda$$

General point on $$L$$: $$(2\lambda + 2, -2\lambda - 1, \lambda + 3)$$.

Point $$Q$$ lies on the $$yz$$-plane, where $$x = 0$$:

$$2\lambda + 2 = 0 \implies \lambda = -1$$.

Substitute $$\lambda = -1$$ into the coordinates:

$$Q = (0, -2(-1)-1, -1+3) = \mathbf{(0, 1, 2)}$$.

$$P = (2, -1, 3)$$ and $$Q = (0, 1, 2)$$.

$$PQ = \sqrt{(0-2)^2 + (1-(-1))^2 + (2-3)^2}$$  $$PQ = \sqrt{(-2)^2 + (2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = \mathbf{3}$$

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 :

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