Final answer:
The planes x+y+z=1 and x-y+z=1 are neither parallel nor perpendicular. The angle between them is approximately 70.53°.
Step-by-step explanation:
The given equations are x+y+z=1 and x-y+z=1.
To determine if the planes are parallel or perpendicular, we can compare the normal vectors of the planes.
First, let's find the normal vector of the plane x+y+z=1:
Normal vector = (coefficient of x, coefficient of y, coefficient of z) = (1, 1, 1).
Next, let's find the normal vector of the plane x-y+z=1:
Normal vector = (coefficient of x, coefficient of y, coefficient of z) = (1, -1, 1).
Since the dot product of the normal vectors is not zero, the planes are neither parallel nor perpendicular. To find the angle between them, we can use the formula cosθ = (dot product of normal vectors) / (product of magnitudes of normal vectors).
Let's calculate:
Magnitude of the normal vector of the first plane = √(1² + 1² + 1²) = √3
Magnitude of the normal vector of the second plane = √(1² + (-1)² + 1²) = √3
Dot product of the normal vectors = (1)(1) + (1)(-1) + (1)(1) = 1.
Using the formula cosθ = 1 / (√3)(√3) = 1/3, we can find the angle θ:
θ = arccos(1/3) ≈ 70.53°
Therefore, the angle between the two planes is approximately 70.53°.