The angle between the planes whose equations are $4x+8y + z= 8 $ and $y+z=4 $ is :

45°

The correct answer is Option (2) → 45°

$\vec{n_1}$ ⊥ plane 1 = $4\hat i+8\hat j+\hat k$

$\vec{n_2}$ ⊥ plane 2 = $\hat j+\hat k$

angle between planes = angle between normals

$\vec{n_1}.\vec{n_2}=|\vec{n_1}||\vec{n_2}|\cos θ$

$9=\sqrt{4^2+8^2+1}\sqrt{1^2+1^2}\cos θ$

$\cos θ=\frac{1}{\sqrt{2}}$

$θ=45°$