# Reflection of a line in a plane

The line $$l_1$$ has the equation $$r=(6i+2j-2k)+\lambda(4i+5j-k)$$ and the plane $$\pi_1$$ has the equation $$2x-y+4z=4$$. The line $$l_2$$ is the reflection of $$l_1$$ in the plane $$\pi_1$$. Find the exact vector equation of line $$l_2$$.

So the line intersects the plane when $$\lambda=-2$$, giving the point $$(-2,-8,0)$$ which will be common on $$l_1$$ and $$l_2$$. But I am unsure on how to find the direction vector for $$l_2$$. Any help would be appreciated.

– quid
Dec 26, 2018 at 14:19

Hint...Construct the line through $$(6,2,-2)$$ with direction the same as the normal to the plane, and find where this line meets the plane. This point will be the midpoint of the line joining $$(6,2,-2)$$ and its reflection in the plane. Once you have this reflection point you can form the line of reflection because you now have two points.

the answer I get is $$r=-2i-8j+t(88i+103j-13k)$$

• doing this I got an answer of $(-2i-8j+0k)+\mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer? Dec 20, 2018 at 12:03
• Looks like you need to check your calculations Dec 20, 2018 at 22:49
• Would it be possible for someone to solve the problem and give me a work solution for me to compare against. Dec 21, 2018 at 16:28
• Would you be able to include a method for me to see where ive gone wrong? Dec 22, 2018 at 14:13

Given a plane $$\Pi_1$$ and a line $$L$$

$$\Pi_1\to (p-p_1)\cdot \vec n_1 = 0\\ L\to p = p_0 +\lambda \vec n_2$$

first we determine the intersection point

$$p^* = \Pi_1\cap L$$

by making

$$(p_0-p_1+\lambda\vec n_2)\cdot \vec n_1 = 0\Rightarrow \lambda = -\frac{(p_0-p_1)\cdot\vec n_1}{\vec n_1\cdot\vec n_2}$$

then

$$p^* = p_0-\frac{(p_0-p_1)\cdot\vec n_1}{\vec n_1\cdot\vec n_2}\vec n_2$$

After that the reflection for $$\vec n_2$$ regarding $$\Pi_1$$ is obtained as follows:

$$\vec n_2 = \alpha \vec n_1+\vec m\Rightarrow \vec n_1\cdot\vec n_2 = \alpha||\vec n_1||^2$$

then

$$\alpha = \frac{\vec n_1\cdot\vec n_2}{||\vec n_1||^2}\Rightarrow \vec m = \vec n_2- \frac{\vec n_1\cdot\vec n_2}{||\vec n_1||^2}\vec n_1$$

and then the reflected line is

$$L_R\to p = p^* +\lambda \vec n_R$$

with

$$\vec n_R = \vec m - \alpha \vec n_1 = \vec n_2- 2\frac{\vec n_1\cdot\vec n_2}{||\vec n_1||^2}\vec n_1$$

• Could you please expand on your working after the line introdicing p*? I don't quite understand the role of alpha and m and the working following on from that - especially how you manage to use ||n1||^2 rather than ||n1|| ||n2||. Cheers!
– dwb
Oct 6, 2020 at 18:25
• In $\vec n_2 = \alpha \vec n_1 +\vec m$ we have $\vec n_1\cdot \vec m = 0$ because this is an orthogonal decomposition. We can consider roughly as if $\vec m \in \Pi_1$ and $\alpha \vec n_1$ is parallel to $\vec n_1$ hence perp to $\Pi_1$. Oct 6, 2020 at 18:41

HINT

In the first place, I advise you to obtain the normal component of $$(4,5,-1)$$ related to the place $$\pi_{1}$$. After so, all you have to do is to substract twice it from the original direction of $$l_{1}$$ in order to obtain the direction of $$l_{2}$$.

• what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$" Dec 24, 2018 at 17:54
• I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$\langle(4,5,-1),(2,-1,4)\rangle (2,-1,4)/\lVert(2,-1,4)\rVert^{2}$$ Dec 24, 2018 at 17:58