Final answer:
The cosine of the angle between the planes with normal vectors A = (1, 1, 1) and B = (1, 5, 4) is found using the dot product of A and B, divided by the product of their magnitudes, resulting in 10/√126. The angle can then be found by taking the inverse cosine of this value.
Step-by-step explanation:
To find the cosine of the angle between two planes represented by the equations x + y + z = 0 and x + 5y + 4z = 6, we must first identify the normal vectors of these planes. The normal vector of the first plane is A = (1, 1, 1) and of the second plane is B = (1, 5, 4).
The cosine angle between two vectors A and B is:
cos(θ) = (Ax × Bx + Ay × By + Az × Bz) / (|A| × |B|)
Calculating the dot product, we get:
A × B = 1*1 + 1*5 + 1*4 = 10
The magnitudes of A and B are:
|A| = √(1² + 1² + 1²) = √3
|B| = √(1² + 5² + 4²) = √(1 + 25 + 16) = √42
Substituting the values, we get:
cos(θ) = 10 / (√3 × √42) = 10 / √126
Thus, the angle between the planes is the inverse cosine of 10/√126, which can be calculated using a calculator.