Intersection of a Line and a Plane

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 :

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

Any parametric point on the line $$L$$ can be expressed by equating the symmetric equations to a scalar parameter $k$:

$$x = 2k + 1, \quad y = -k + 2, \quad z = k + 3$$

Since point $$A$$ lies on the plane $$P$$, substitute these coordinates into the plane equation $$x + y + z = 6$$:

$$(2k + 1) + (-k + 2) + (k + 3) = 6$$ $$2k + 6 = 6 \implies k = 0$$

Substituting $$k = 0$$ back into our parametric expressions gives the coordinates of point $$A$$:

$$A = (1, 2, 3)$$

The parabola lies in the $$xy$$-plane ($$z = 0$$) and has the equation $$y^2 = 4x$$. We can express any point $$B$$ on this parabola using its standard parametric form $$(t^2, 2t)$$:

$$B = (t^2, 2t, 0)$$

Using the 3D distance formula, write the squared distance $$D^2$$ between $$A(1, 2, 3)$$ and $$B(t^2, 2t, 0)$$:

$$D^2 = (t^2 - 1)^2 + (2t - 2)^2 + (0 - 3)^2$$  $$D^2 = (t^4 - 2t^2 + 1) + (4t^2 - 8t + 4) + 9$$ $$D^2 = t^4 + 2t^2 - 8t + 14$$

To find the minimum value, differentiate $$D^2$$ with respect to $$t$$ and set it to $$0$$:

$$\frac{d(D^2)}{dt} = 4t^3 + 4t - 8 = 0$$ $$t^3 + t - 2 = 0$$

By inspection, $$t = 1$$ is a real root of this cubic equation because $$1^3 + 1 - 2 = 0$$. Factoring out $$(t-1)$$ yields:

$$(t - 1)(t^2 + t + 2) = 0$$

Since the discriminant of the quadratic factor $$t^2 + t + 2$$ is negative ($$1^2 - 4(1)(2) = -7 < 0$$), $$t = 1$$ is the only real critical point.

Substitute $$t = 1$$ back into our distance squared equation:

$$D^2 = (1)^4 + 2(1)^2 - 8(1) + 14 = 1 + 2 - 8 + 14 = 9$$  $$D = \sqrt{9} = 3$$

Correct Option: B