Question

Find the cosine of the angle between the planes 4x−1y+2z=−4 and the plane 1x+1y−4z=3.

Answer 1

Try this solution:

1. Given: α:4x-y+2z=-4 and β:x+y-4z=3.

if vector n₁⊥α and n₂⊥β, then required cosine is:

$cos \gamma =|\frac{n_1*n_2}{|n_1|*|n_2|}|;$

where n₁*n₂=4*1+(-1)*1+2*(-4) and |n₁|*|n₂|=√(4²+1²+2²)*√(1²+1²+4²).

2. Using the formula of cosine:

$cos \gamma=|\frac{4-1-8}{\sqrt{21}*\sqrt{18}}|=\frac{5}{3\sqrt{42}}=0.2572$

answer: ≈0.2572