Derive the transformation on a position vector p (x,y,z) that results in its reflection across the plane?

Assume that n is unit and at end replace n by n/|n|

If perp from P to plane is PN then NP = ((r−r₀)•n)n = (nᵀ(r−r₀))n = n(nᵀ(r−r₀)) = (nnᵀ)(r−r₀)

∴ If reflected point is r’ then r’ = r + 2*PN = r – 2(nnᵀ)(r—r₀)

Replace n by n/|n| to get r’−r₀ = R(r−r₀) where reflection matrix R = I − 2(nᵀn)/|n|²

Note that R is its own inverse so r’−r₀ = R(r−r₀) and r−r₀ = R(r’−r₀)

These can be written as r = a+Rr’ or r’ = a+Rr where a = r₀−Rr₀

For 2x−y+3z=1 n=(2,−1,3), |n|²=14 and choose r₀=(0,−1,0)

nnᵀ = { {4,−2,6}, {−2,1,−3}, {6,−3,9} } so R =(1/14){ {6,4,−12}, {4,12,6}, {−12,6,−4} }

a = (0,−1,0) − R(0,−1,0) = (1/7)(2,−1,3)

For L : (1,1,1) → (1/7)(1,10−2) : (−1,0,1) → (−1,0,1)

So transformation of L is r = (1/7)(1,10,−2) + t(−1,0,1)

The plane mᵀr=p → mᵀ(a+Rr’) = p ≡ (mᵀR)r’ = p−mᵀa

With m=(5,1,1) & p=7 this gives mᵀR = (1/7)(11,19,−29) and p−mᵀa = 7−12/7 = 37/7

Equation of reflected plane is 11x+19y−29z = 37

Well,

the question is interesting, but i have to work... so maybe more this evening (french time !)

so what i can offer is :

n can be transformed in a unit vector (let' s admit this)

u is the unit vector according to (r-r0) : so (obviously) : r-r0 = ((r-r0) . u)u

then, let be: v = u x n (vector product) and w = n x v = n x (u x n)

then we have the decomposition:

r-r0 = ((r-r0).n)n + ((r-r0).w)w

and therefore, if S is the symmetry according to plane P passing through r0

S(r-r0) = - ((r-r0).n)n + ((r-r0).w)w

hope it' ll help !!