Find the angle in radians between the planes!?

Question

-2x +z = 1, and -1y + z =1

Answer 1

The angle between the planes is equal to the angle between the normal vectors of the planes.

-2x + z - 1 = 0
-y + z - 1 = 0

The normal vectors for the planes are:

n1 = <-2, 0, 1>
n2 = <0, -1, 1>

The magnitude of the normal vectors is:

|| n1 || = √[(-2)² + 0² + 1²] = √(4 + 0 + 1) = √5
|| n2 || = √[0² + (-1)² + 1²] = √(0 + 1 + 1) = √2

The dot product of the normal vectors is:

n1 • n2 = <-2, 0, 1> • <0, -1, 1> = 0 + 0 + 1 = 1

Let θ = the angle between the two vectors. Another way the dot product can be expressed is:

n1 • n2 = || n1 || || n2 || cosθ
cosθ = (n1 • n2) / (|| n1 || || n2 ||) = 1 / [(√5)(√2)] = 1/√10

θ = arccos(1/√10) ≈ 1.2490458 radians

The angle between the planes is 1.2490458 radians.

Answer 2

that's no longer a bad question. unsure how severe in math you're, yet you could desire to be kinda up there. the assumption is to locate the attitude between the unit normals of the planes. as quickly as you do this, only locate the dot made of those 2 normals and that result would be the cosine of the attitude you're finding for. ultimately only take the arc cosine of that kind and you will have your attitude ;o) Me being a math junkie, i do no longer want to wreck the whole concern so I wont provide you all of the numbers, yet i visit tell you that the artwork required isn't puzzling or long. Oh yeah, make beneficial your calculator is desperate to radians and not ranges!