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

30° 45° 60° 90°

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 $\vec{n_1}.\vec{n_2}=|\vec{n_1}||\vec{n_2}|\cos θ$ $9=\sqrt{4^2+8^2+1}\sqrt{1^2+1^2}\cos θ$ $\cos θ=\frac{1}{\sqrt{2}}$ $θ=45°$