The angle between the planes whose equations are $4x+8y + z= 8 $ and $y+z=4 $ is :
Answer & explanation
Correct answer: option 2
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
$\vec{n_1}.\vec{n_2}=|\vec{n_1}||\vec{n_2}|\cos θ$
$9=\sqrt{4^2+8^2+1}\sqrt{1^2+1^2}\cos θ$
$\cos θ=\frac{1}{\sqrt{2}}$
$θ=45°$