Vector equation of a line that is symmetrical to another line L with respect to plane $\Pi$

Question

The plane $\Pi$ is defined as - $$4x - 3y + z = 1$$

The line $L$ is defined as - $$\frac{x-4}{3} = \frac{y-1}{-1} = \frac{z-5}{2}$$

I am trying to find the vector equation of a line that is essentially a reflection of line $L$ across the plane $\Pi$.

Using -

$$\sin(θ)=\frac{v\cdot n}{|v||n|}$$

Where $v$ is the direction vector of the line $L$, and $n$ is the normal vector of plane $\Pi$, I found the angle between both planes to be $1.1 $ $ rad$ or $63^∘$. Furthermore, by substituting the parametric equation of line $L$ into the equation of plane $\Pi$, I found the point of intersection of the line $L$ and plane $\Pi$ to be $(1, 2, 3)$. However, I am unable to find the vector equation of the symmetrical or reflected line, as I am not sure how to go about finding its direction vector.

Answer 1

$\def\v{\vec v}\def\n{\vec n}$ Hint:

The direction vector of the reflected line is: $$ \v-2\frac{\v\cdot\n}{|n|^2}\n. $$

One can memorize the expression using analogy with reflection of light: the tangential (wrt. to the mirror) component of the light ray conserves whereas the normal component changes sign.

Answer 2

Hint. Here is a recipe for solving such problems:

Check if $L$ and $\Pi$ intersect. In this case, you have seen that they do, in $(1,2,3).$ Then pick a different point on your line, for example $(4,1,5).$ You want to find a line passing through this point and orthogonal to the plane. So find the vector defining the plane, in this case which is $(4,-3,1).$ Then the sought line (not the reflection yet) is given by $(4,1,5)+t(4,-3,1)$ for all real $t.$ Find where this line intersects the plane, say $(a,b,c).$ Then you want the point which is determined by the tip of twice the the vector $(a,b,c)-(4,1,5).$ Thus the line you want (finally!) passes through the points $2(a-4,b-1,c-5)$ and $(1,2,3),$ and is therefore given by $$(1,2,3)+2\lambda(a-4,b-1,c-5),$$ where $\lambda$ is real.