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

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.

• I apologize. I re-edited the equation. I put plus signs instead of equals signs. It is meant to be the cartesian form of a line. Commented May 2, 2020 at 7:41

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.
• Wow! That's a great solution! But, I have one question. Isn't $2(a-4, b-1, c-5)$ a point and not a direction vector? Then how did you put it in place of the direction vector for the final vector equation of the line? I am assuming you meant to put $2(a-4, b-1, c-5)-(1, 2, 3)$ in place of the direction vector? Commented May 2, 2020 at 9:35
• @SanjayChintapally There's a duality between points and their position vectors. That is, if $(p,q,r)$ represents a position vector relative to some fixed arbitrary point, then it also represents the coordinates of that point relative to the fixed point. Commented May 2, 2020 at 11:52
$$\def\v{\vec v}\def\n{\vec n}$$ Hint:
The direction vector of the reflected line is: $$\v-2\frac{\v\cdot\n}{|n|^2}\n.$$