0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Please do not reask questions; instead edit or comment. $\endgroup$
    – quid
    Dec 26, 2018 at 14:19

3 Answers 3

2
$\begingroup$

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 $r=-2i-8j+t(88i+103j-13k)$

$\endgroup$
5
  • $\begingroup$ doing this I got an answer of $(-2i-8j+0k)+\mu(92/21i + 206/21j +26/21k)$ which is apparently the wrong answer? $\endgroup$
    – H.Linkhorn
    Dec 20, 2018 at 12:03
  • $\begingroup$ Looks like you need to check your calculations $\endgroup$ Dec 20, 2018 at 22:49
  • $\begingroup$ Would it be possible for someone to solve the problem and give me a work solution for me to compare against. $\endgroup$
    – H.Linkhorn
    Dec 21, 2018 at 16:28
  • $\begingroup$ I have added an answer $\endgroup$ Dec 21, 2018 at 23:10
  • $\begingroup$ Would you be able to include a method for me to see where ive gone wrong? $\endgroup$
    – H.Linkhorn
    Dec 22, 2018 at 14:13
1
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ 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! $\endgroup$
    – dwb
    Oct 6, 2020 at 18:25
  • $\begingroup$ 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 $\Pi_1$. $\endgroup$
    – Cesareo
    Oct 6, 2020 at 18:41
0
$\begingroup$

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 $l_{2}$.

$\endgroup$
2
  • $\begingroup$ what do you mean by "normal component of $(4,5,−1)$ related to the place $π_1$" $\endgroup$
    – H.Linkhorn
    Dec 24, 2018 at 17:54
  • $\begingroup$ I mean the projection of $(4,5,-1)$ in the direction of $(2,-1,4)$. Precisely speaking, I am talking about the vector: $$\langle(4,5,-1),(2,-1,4)\rangle (2,-1,4)/\lVert(2,-1,4)\rVert^{2}$$ $\endgroup$
    – user0102
    Dec 24, 2018 at 17:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .