In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angle between them:
$4x + 8y + z - 8 = 0$ and $y + z - 4 = 0$

(D) The direction ratios of the normal to the planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ are $(A_1, B_1, C_1)$ and $(A_2, B_2, C_2)$ respectively.
The planes are parallel if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.
The planes are perpendicular if $A_1A_2 + B_1B_2 + C_1C_2 = 0$.
The angle $\theta$ between the planes is given by $\cos \theta = \left| \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} \right|$.
For the given planes $4x + 8y + z - 8 = 0$ and $0x + y + z - 4 = 0$:
$A_1 = 4, B_1 = 8, C_1 = 1$ and $A_2 = 0, B_2 = 1, C_2 = 1$.
Check for perpendicularity: $A_1A_2 + B_1B_2 + C_1C_2 = (4)(0) + (8)(1) + (1)(1) = 0 + 8 + 1 = 9 \neq 0$. Thus,they are not perpendicular.
Check for parallelism: $\frac{A_1}{A_2} = \frac{4}{0}$ (undefined),$\frac{B_1}{B_2} = 8$,$\frac{C_1}{C_2} = 1$. Since the ratios are not equal,they are not parallel.
Calculate the angle: $\cos \theta = \left| \frac{9}{\sqrt{4^2 + 8^2 + 1^2} \sqrt{0^2 + 1^2 + 1^2}} \right| = \left| \frac{9}{\sqrt{16 + 64 + 1} \sqrt{2}} \right| = \left| \frac{9}{9 \times \sqrt{2}} \right| = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}$ or $\frac{\pi}{4}$ radians.