# Thread: Reflection of a point

1. ## Reflection of a point

How can we get a reflection of a point $\displaystyle$A$$threw a plane \displaystyle \Pi$$?

A(1,-1,2)
plane : x + y - z + 3 = 0

I was thinking to find a point P on a plane, and then find a vector AP, and then find an opposite vector. Or is this wrong?

2. You need to find the vector normal to the plane going through A. It will intersect the plane at point P, and have its tip at point A. Take the negative of this vector. Put its tail at P, and then the tip will be at the reflection point.

The only difference between what I have said and what you have said is that I claim you need a normal vector. I don't think this will work if the vector you find is not normal to the plane.

3. And remember that if a plane is given by Ax+ By+ Cz= D then $\displaystyle A\vec{i}+ B\vec{j}+ C\vec{k}$ is a vector perpendicular to the plane. A line through point $\displaystyle (x_0, y_0, z_0)$, perpendicular to plane Ax+ By+ Cz= D, is given by $\displaystyle x= At+ x_0$, $\displaystyle y= Bt+ y_0$, $\displaystyle z= Ct+ z_0$. You can now determine where that line intersects the plane, then go an equal distance on the other side of the plane.

4. Now I am confused. I was trying to solve this but I don't know how.

So how I understand it now is:
We have plane x + y - z = -3, and A(1,-1,2)

I insert it in x = At + x0...
so:

x = 0t + 1
y = 0t - 1
z = 0t + 2

Now what? We get back those coordinates of a point A.

5. Um... no. You need to use the normal vector of the plane that is:

$\displaystyle \left(\begin{array}{c}1 \\ 1 \\ -1 \end{array}\right)$

Then, your line through A and the plane becomes:

$\displaystyle \vec{l} = \left(\begin{array}{c}1 \\ -1 \\ 2 \end{array}\right) + \lambda\left(\begin{array}{c}1 \\ 1 \\ -1 \end{array}\right)$

Which gives:

$\displaystyle x = 1 + \lambda$

$\displaystyle y =-1 + \lambda$

$\displaystyle z = 2 - \lambda$

Now, you equate the line with the plane to find the value of lambda. Now that you know the value of lambda, you can double it and add it to point A as a shortcut instead of finding the point of intersection of the line and the plane.

6. The normal vector to the plane is $\displaystyle <1,1,-1>$.

So the line through A with that direction is $\displaystyle L1,-1,2>+t<1,1,-1>$.

Where does that line meet the plane? Call that point P.

Then A's reflection is A' such that $\displaystyle \overrightarrow {AA'} = 2\overrightarrow {AP}$.

7. Oh yes I understand it now, so I have to find t (or lambda), just insert back x,y,z to get the vertex of a plane and the line. Now I have the point P and A, and i can calculate the vector AP.

But how did you get the normal vector (1,1,-1)?

8. Originally Posted by Nforce
Oh yes I understand it now, so I have to find t (or lambda), just insert back x,y,z to get the vertex of a plane and the line. Now I have the point P and A, and i can calculate the vector AP.

But how did you get the normal vector (1,1,-1)?
From the equation of the plane. If you have a plane:

ax + by + cz = d

<a, b, c> is the normal vector of the plane.

9. Originally Posted by Nforce
But how did you get the normal vector (1,1,-1)?
If $\displaystyle ax+by+cz=d$ is a plane, its normal is $\displaystyle <a,b,c>$

10. Originally Posted by Plato
Then A's reflection is A' such that $\displaystyle \overrightarrow {AA'} = 2\overrightarrow {AP}$.
Right and I get the double vector of AP and for getting just the point I just add A, so 2AP + A = A'

Great, thanks a lot.