Find Flux Density On One Side of Dielectric Boundary Given Boundary Conditions

Bizkit

1. The problem statement, all variables and given/known data
A dielectric interface is defined as 4x + 3y = 10 m. The region including the origin is free space, where D₁ = 2a_x - 4a_y + 6.5a_z nC/m². In the other region, ε_r2 = 2.5. Find D₂ given the previous conditions.


2. Relevant equations
a_n12 = ± grad(f)/|grad(f)|

D_2n = D_1n = a_n(D₁ · a_n)

D_1t = D₁ - D_1n

ε = ε₀ε_r

D_2t = (D_1t)(ε₂)/ε₁

D₂ = D_2n + D_2t


3. The attempt at a solution
f = 4x + 3y - 10 = 0

a_n12 = ± grad(4x + 3y - 10)/|grad(4x + 3y - 10)| = ± (4a_x + 3a_y)/5 = ± (.8a_x + .6a_y) Since the vector points in the positive x and y directions, I choose the plus sign to get: a_n12 = .8a_x + .6a_y

D_2n = D_1n = (.8a_x + .6a_y)((2a_x - 4a_y + 6.5a_z) · (.8a_x + .6a_y)) nC/m² = (.8a_x + .6a_y)(-.8) nC/m² = -.64a_x - .48a_y nC/m²

D_1t = (2a_x - 4a_y + 6.5a_z nC/m²) - (-.64a_x - .48a_y nC/m²) = 2.64a_x - 3.52a_y + 6.5a_z nC/m²

ε₁ = ε₀ε_r1 = ε₀ = 8.854 pF/m (since the region is free space)

ε₂ = ε₀ε_r2 = (8.854 pF/m)(2.5) = 22.135 pF/m

D_2t = (2.64a_x - 3.52a_y + 6.5a_z nC/m²)(22.135 pF/m)/(8.854 pF/m) = (2.64a_x - 3.52a_y + 6.5a_z nC/m²)(2.5) = 6.6a_x - 8.8a_y + 16.25a_z nC/m²

D₂ = (-.64a_x - .48a_y nC/m²) + (6.6a_x - 8.8a_y + 16.25a_z nC/m²) = 5.96a_x - 9.28a_y + 16.25a_z nC/m²

The answer in the back of the book, however, is given as D₂ = .416a_x - 1.888a_y + 2.6a_z nC/m², which is completely different than what I got. I'm not sure where I went wrong. I followed one of my teacher's examples that he has posted (you can find it here), which is very similar to this question, but I still come up with the wrong answer. Can someone please show me where I went wrong? Thanks.