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The angle between 2 planes 4x+8y+z−8=04x + 8y + z - 8 = 0 and y+z−4=0y + z - 4 = 0 is :

Solution

✅ Correct Option: 3

Normals: n1⃗=(4,8,1)\vec{n_1} = (4, 8, 1), n2⃗=(0,1,1)\vec{n_2} = (0, 1, 1).

n1⃗⋅n2⃗=0+8+1=9\vec{n_1} \cdot \vec{n_2} = 0 + 8 + 1 = 9.

∣n1⃗∣=16+64+1=81=9|\vec{n_1}| = \sqrt{16+64+1} = \sqrt{81} = 9.

∣n2⃗∣=2|\vec{n_2}| = \sqrt{2}.

cos⁡θ=992=12⇒θ=45∘\cos \theta = \frac{9}{9\sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45^\circ.

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