Reflection of a Light Ray I found this problem to be very hard while studying for the exam: 

Let
  $$L: \vec r(t)=<1,-2,3>+t<-5,4,1>, \qquad t \in \mathbb{R}$$
  be a line. Light is traveling along the line $L$ in the direction of increasing $t$ value. Let $\Omega: x=0$ (the $yz$ plane) be a mirror. Find the vector equation of the line $\hat L$ that is the reflection of line $L$ by $\Omega$ 
  Any help would be helpful!

Any help would be helpful!

My solution 
  $$L: \vec r(t)= 1/5<0,-6,16>+t<5,4,1>, \qquad t \in \mathbb{R}$$

 A: Perhaps I’ve misunderstood something in the problem statement, but this seems rather straightforward. The problem talks about lines, not rays, so the point at which the given line intersects the mirror isn’t directly relevant.  
Forget about the line for the moment. What is the image of an arbitrary vector $\langle x,y,z \rangle$ when reflected in the $yz$-plane? It is simply $\langle -x,y,z \rangle$. To find the equation of the line’s reflection, apply this same transformation to the given equation, resulting in the line $$\langle -1,-2,3\rangle + t\langle 5,4,1\rangle.$$   
Now, if indeed you’re really supposed to find the incident and reflected rays, then you do need to reparametrize by finding where the line intersects the reflective plane (which is also the point of intersection of the two lines), but it seems to me that you then also have to specify that $t\ge 0$ so that you get a ray rather than a line.
A: In general, to find the equation of the reflection ray, you need to the find 
$1.$ the reflection vector and
$2.$ the intersection point of incident ray with the mirror.
Here, I discuss the general case. However, you may find simpler solution for special cases like the one in your example. 
Suppose the equation of the mirror which is a plane is given by
$$(X-P_1) \cdot n =0 \tag{1}$$
where $X$ is an arbitrary point on your plane and $P_1$ is a given special point and $n$ is also the unit normal of the plane. Next, consider that the equation of the incident ray is given by
$$X=P_2+tV_i \tag{2}$$
where $X$ is an arbitray point on the ray and $P_2$ is a given point on the ray. $V_i$ is a vector in the direction of the incident ray. To find the intersection point, we can simply combine $(1)$ and $(2)$ to get
$$t_0=\frac{(P_1-P_2) \cdot n}{V_i \cdot n} \tag{3}$$
and hence the intersection point $P_0$ will be
$$P_0=P_2+\frac{(P_1-P_2) \cdot n}{V_i \cdot n}V_i \tag{4}$$ 
also, we can decompose $V_i$ in the form 
$$\begin{align}
V_i &= V_{i,n}+V_{i,t} \\
&=(V_i \cdot n)n-(V_i \times n) \times n
\end{align} \tag{5}$$
which are the components normal and tangent to the mirror. When we have a reflection the tangential component of the ray remains the same while the normal component changes sign so the reflection vector will be
$$\begin{align}
V_r &= V_{r,n}+V_{r,t} \\
&=-V_{i,n}+V_{i,t} \\
&=-(V_i \cdot n)n-(V_i \times n) \times n \\
&=-2(V_i \cdot n) n + V_i
\end{align} \tag{6}$$
and hence the equation of the reflection line is
$$X=P_0+tV_r \tag{7}$$
or equivalently
$$\boxed{X=P_2+\frac{(P_1-P_2) \cdot n}{V_i \cdot n}V_i+t[V_i-2(V_i \cdot n)n]}\tag{8}$$
A: The point of intersection of the line with the yz plane is the same for both the lines L and L'.
The point of intersection is  $(0,-6/5,16/5)$ .
$L={x-1\over -5}={y+2\over 4}={z-3\over 1}$
Now after reflection the slope of the line will change. The inclination with the x axis will reverse but that of the y and z axis will remain same. You can see that. 
$L'={x-0\over 5} = {y+6/5\over 4} = {z-16/5\over 1}$
