Finding the point of incidence of a light ray on a plane

Question

I have that a ray of light is emitted from the point $(3,-2,-1)$ and reflected off the plane $x-2y-2z=0$. The reflected ray passes through the point $(4,-1,-6)$. I have to find the point at which the ray hit the plane.

My approach was to realise that the angle to the normal of the reflected and incident ray will be the same. Then letting the required point of incidence be $P(a,b,c)$, the given points be $A(3,-2,-1)$ and $B(4,-1,-6)$, and the normal to the plane be $n=(1,-2,-2)$ If $w$ is the angle of incidence, then $|AP|cosw = (n|AP)$ Also, $|BP|cosw = (n|BP)$ To get a third equation, we have that $P$ is on the plane so $P$ satisfies $a-2b-2c=0$

This gave me a terrible system of equations. There must be a better way around this problem. What is it?

Answer 1

Here's how I'd do it.

For each point $A,B$ find the intersection with the plane as you travel along $\vec{n}.$

Call these $A_p = (a_1, a_2, a_3)$ and $B_p = (b_1, b_2, b_3)$. Call the distances to the plane $d_A, d_B,$ respectively.

This is eight equations in eight unknowns, but it's not that messy. For $A_p, d_A$:

$$3 + d_A = a_1 \\ -2 - 2d_A = a_2 \\ -1 - 2d_A = a_3 \\ a_1 - 2a_2 - 2a_3 = 0$$

And for $B_p, d_B$:

$$4 + d_B = b_1 \\ -1 - 2d_B = b_2 \\ -6 - 2d_B = b_3 \\ b_1 - 2b_2 - 2b_3 = 0$$

Now, you're in a position to use similar triangles $\triangle AA_pP, \triangle BB_pP$.

The proportionality factor is $f = \left|d_A/d_B\right|$, and then find $P$ by starting at $A_p$ and going fraction $f$ of the distance to $B_p$ along the direction from $A_p$ to $B_p$.

Answer 2

Let $\vec{n}=\langle1,-2,-2\rangle$ and let the point $P=(a,b,c)$.

If $Q=(3,-2,-1)$ and $R=(4,-1,-6)$, $\;\;\vec{q}=\langle3,-2,-1\rangle$ and $\vec{r}=\langle4,-1,-6\rangle$, and

$\vec{u}=\vec{PQ}=\langle3-a,-2-b,-1-c\rangle$ and $\vec{v}=\vec{PR}=\langle4-a,-1-b,-6-c\rangle$ ,

then the projection of $Q$ in the plane is given by $Q^{*}=(2,0,1)$ since $\vec{q}-\frac{\vec{q}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}=\langle2,0,1\rangle$

and the projection of $R$ in the plane is given by $R^{*}=(2,3,-2)$ since $\vec{r}-\frac{\vec{r}\cdot\vec{n}}{\vec{n}\cdot\vec{n}}=\langle2,3,-2\rangle$ .

Since the line between $Q^{*}$ and $R^{*}$ has equations $x=2, \;\;y=t,\;\; z=1-t$ and P is on this line,

$\hspace{.4 in}a=2,\;\; b=t, \;\;c=1-t$

If $\theta_{1}$ and $\theta_2$ are the angles between $\vec{n}$ and $\vec{u}$ and $\vec{v}$, respectively, then

$\displaystyle\cos\theta_{1}=\frac{\vec{u}\cdot\vec{n}}{|\vec{u}||\vec{n}|}=\frac{3}{\sqrt{2t^2+9}}$ $\;\;\;$ and $\;\;\;$ $\displaystyle\cos\theta_{2}=\frac{\vec{v}\cdot\vec{n}}{|\vec{v}||\vec{n}|}=\frac{6}{\sqrt{2t^2-12t+54}}$.

Therefore $\displaystyle\frac{3}{\sqrt{2t^2+9}}=\frac{6}{\sqrt{2t^2-12t+54}}\implies2t^2-12t+54=4(2t^2+9)\implies$

$6t^2+12t-18=0\implies t^2+2t-3=0\implies (t+3)(t-1)=0$, so $t=1$ or $t=-3$.

Since $P$ is between $Q^{*}$ and $R^{*}$, $\;$ $0\le t\le 3;\;\;$ so $\;t=1$ gives $P=(2,1,0)$.

Answer 3

Let $P = (3, -2, -1)$, $ Q = (4, -1, -6)$, and N = $(1, -2, -2) $

And the required point of incident be $I(x, y, z) $

Since the ray $PA$ reflects and passes through $Q$, then the the ray $PA$ passes through the reflection of $Q$ about the given plane.

So, first find the reflection $Q'$ of point $Q$.

There is a formula for that. Note that the plane passes through the origin. Then,

$Q' = T Q = (I - 2 \ \dfrac{N N^T}{N^T N} ) Q$

Now,

$T = (I - 2 \ \dfrac{N N^T}{N^T N} ) = I - \dfrac{2}{9} \begin{bmatrix} 1 && -2 && -2 \\ -2 && 4 && 4 \\ -2 && 4 && 4 \end{bmatrix} $

And this evaluates to

$T = (I - 2 \dfrac{N N^T}{N^T N} ) = \dfrac{1}{9} \begin{bmatrix} 7 && 4 && 4 \\ 4 && 1 && -8 \\ 4 && -8 && 1 \end{bmatrix} $

Premultiplying $Q$ with T, gives

$Q' = T Q = (0, 7, 2)

This point lies on $PA$ therefore, the incidence point is the intersection of the segment $PQ'$ with the plane

The parametric equation of $PQ'$ is

$ PQ' = ( 3, -2, -1) + t (-3, 9 , 3 ) = (3 - 3 t , -2 + 9 t, -1 + 3 t) $

Substitute this point into the equation of the plane

$x - 2 y - 2 z= 0$

Hence,

$ (3 - 3 t) - 2 (-2 + 9t) - 2 (-1 + 3 t) = 0 $

Therefore,

$ 9 - 27 t = 0$

So that,

$ t = \dfrac{1}{3} $

Thus the intersection point (which is the incidence point) is

$ I = ( 3 - 1 , - 2 + 3 , -1 + 1 ) = (2, 1, 0)$

Answer 4

This solution is based on this video here https://www.youtube.com/watch?v=qS8MG7jwmc8.

Let $Q(4,-1,-6)$ be the point on the reflected ray, $S(3,-2,-1)$ the point on the ray and $P$ the point of intersection of the ray, reflected ray and plane.

If we let a line $L$ contain the reflected ray we just need another point on $L$ to obtain $P$. This can easily be done since the reflection of $S$ with respect to the plane lies on $L$ (watch the video), call this point $S'$.

Let $L_2$ be the line through $S$ and $S'$. Then $L_2:S+tn=(3,-2,-1)+t(1,-2,-2)\iff \begin{cases}x=3+t\\ y=-2-2t \\ z=-1-2t\end{cases}$.

Inserting this in the plane $x−2y−2z=0$, we get $3+t+4+4t+2+4t=0\iff t=-1$.

Thus $S'$ is given by $S+2tn=(3,−2,−1)-2(1,−2,−2)=(1,2,3)$. It follows that $L:Q+(S'-Q)t_L=(4,-1,-6)+t_L(-3,3,9)=(4,-1,-6)+s_L(-1,1,3)\iff\begin{cases}x=4-s_L\\ y=-1+s_L\\ z=-6+3s_L\end{cases}$

Then we insert it the plane equation to get $s_L$ so we can get $P$, $4-s_L-2(-1+s_L)-2(-6+3s_L)=-9s_L+18=0\iff s_L=2.$ Finally $P(2,1,0)$.