Trace and reflect two rays inside a cuboid

$OA=(12,0,0)$, $OD=(0,10,5)$, and H=(0,4,5), so $AH=(-12,4,5)$. Blue:$\mathbf r_b=(12,0,0)+\lambda(-12,4,5)$; yellow:$\mathbf r_y=\mu(1,2,2)$. Use λ,μ≥0 for rays, while the printed source uses real parameters for the supporting lines.

An intersection requires $12-12\lambda=\mu$, $4\lambda=2\mu$, $5\lambda=2\mu$. The last two force λ=μ=0, contradicting the first equation. Even the supporting lines do not intersect, so the ray paths do not cross.

DEFG is z=5, with normal $(0,0,5)$ or its unit version $(0,0,1)$. Thus $\sin\theta=|(-12,4,5)\cdot(0,0,1)|/\sqrt{12^2+4^2+5^2}=5/\sqrt{185}$. Hence θ=21.6°(1d.p.) or 0.376rad(3s.f.).

$\mu(1,2,2)\cdot(0,0,1)=5$ gives 2μ=5, μ=2.5. Hence $OX=(2.5,5,5)$.

Let Z be the foot of X on OABC. Then $ZX=(0,0,5)$ and Z=(2.5,5,0). Reflection at horizontal z=5 reverses the vertical ray component, so Y=(5,10,0), with Z the midpoint of OY. Thus area (OXY)=$2\operatorname{area}(OXZ)=|ZX\times OX|=\sqrt{25^2+12.5^2}=25\sqrt5/2$ cm², or 28.0 cm²(3s.f.). The printed component matrix reverses the cross-product factors relative to its preceding magnitude expression; either order has the same magnitude.

The two right triangles share height XZ=5 and each base has length $\sqrt{2.5^2+5^2}$. Hence total area=$2[\frac12(5)\sqrt{2.5^2+5^2}]=5\sqrt{125/4}=25\sqrt5/2$ cm².