Reflection of a plane in a plane. The question is:
The reflection of the plane $2x+3y+4z-3=0$ in the plane $x-y+z-3=0$ is the plane:
I tried to find the equation of the normal to the plane and then tried putting in some values, but I couldn't do it. I know how to find the reflection of a point about a plane, but I have no idea how to proceed in this one.
Thanks.
 A: Start with reflecting the normal of $2x+3y+4z-3=0$ in $x-y+z-3=0$ to get a new normal vector $(a,b,c)$. Write an equation $ax+by+cz+d=0$ for the new plane, and reflect a point of $2x+3y+4z-3=0$ in $x-y+z-3=0$ to get a point $(x,y,z)$. Insert this point into $ax+by+cz+d=0$ to find out $d$.
A: I propose this strategy ( but I don't know if it's the simpler):
1) find a translation that brings the given planes to parallel planes passing through the origin. This is done taking a point on the straight line common to the two planes.
If I've not made some mistake we can take the translation 
$
T(x,y,z)\rightarrow \left(x,y+\dfrac{15}{7},z-\dfrac{6}{7}\right)
$
and we have the two planes:
$$
a)\qquad 2x+3y+4z=0
$$
$$
b)\qquad x-y+z=0
$$
the first orthogonal to:
$\mathbf{v}=(2,3,4)^T$
and the second orthogonal to:
$\mathbf{u}=(1,-1,1)^T$
2) Now the reflection of $\mathbf{v}$   through the plane $b)$ is given by (see here):
$$
R(\mathbf{v})=\mathbf{v}-2\dfrac{\langle\mathbf{v},\mathbf{u} \rangle}{\langle\mathbf{u},\mathbf{u}\rangle}\mathbf{u}
=(v_1',v_2',v_3')^T
$$
and the reflected plane has equation
$$
v_1'x+v_2'y+v_3'z=0
$$
3) use the inverse translation $T^{-1}(x,y,z)\rightarrow \left(x,y-\dfrac{15}{7},z+\dfrac{6}{7}\right)
$ to find the request plane.
A: The most basic, easy to understand and straightforward method according to me is this:-


*

*Consider any point on the plane that has to be reflected, here $P_1:2x+3y+4z−3=0$. Let $(0,0,\frac 34)$ (or any other of your choice)

*Now find the reflection of the point in the mirror plane, here $P_2:x−y+z−3=0$.

*We know that any plane in space is a linear combination of two distinct planes, let the reflected plane is $P_3=P_1 +λP_2$. As the reflection lies in the reflected plane, put the coordinates of the reflected point in $P_3$ and get $λ$, hence $P_3$.
A: You can use vector method for this question.
x-y+z-3=0 is the mirror plane. So equation of normal will be 
i-j+k.
Equation of normal of original plane is 2i+3j+4k.
Now find parallel component of 2i+3j+4k w.r.t. to i-j+k,
Which is equal to {(2i+3j+4k)•(i-j+k)}•(i-j+k)/|i-j+k|^2 = 
i-j+k.
Therefore perpendicular component = 2i+3j+4k-(i-j+k) =
i+4j+3k.
We know perpendicular component will get reversed in direction upon reflection. 
Therefore perpendicular component = -(i+4j+3k)=
-i-4j-3k.
Net reflection vector= perpendicular vector +  parallel vector  = i-j+k +(-i-4j-3k)= -5j-2k.
Now equation of reflected plane is x(α+2) + y(-α+3) + z(α+4) -3-3α=0.
Comparing with this equation we get, 
α+2/0=-α+3/-5=-α+4/-2
Which implies α=-2.
Putting this value  in the equation of plane we get 
5y+2z+3=0
A: There is also a second method of solving this question.
We know that the reflection of $Ax+By+Cz+D=0$ in the plane $ax+by+cz+d=0$ is given by
$$2(Aa+Bb+Cc)(ax+by+cz+d)=(a^2+b^2+c^2)(Ax+By+Cz+D) \ .$$
In the question, $A=2,B=3,C=4,D=-3$ and $a=1,b=-1,c=1,d=-3$, so using these values we get
$$2(2-3+4)(x-y+z-3)=(1+1+1)(2x+3y+4z-3)$$
which simplifies to $5y+2z+3=0$, which is the required answer.
