Find the angle in radians between the normal vectors of the two planes 5x+z=1 and -1y+1=1.

Step 1: Understand the problem We are asked to find the angle between the normal vectors of two given planes.

Step 2: Recall the equation of a plane The general equation of a plane in 3D space is given by Ax + By + Cz = D, where A, B, C, and D are constants, and x, y, and z are the coordinates of any point on the plane. The normal vector of a plane is (A, B, C).

Step 3: Find the normal vectors of the given planes The first plane is 5x + z = 1. Comparing this with the general equation of a plane, we can see that the normal vector of the first plane is (5, 0, 1).

The second plane is -1y + 1 = 1. Comparing this with the general equation of a plane, we can see that the normal vector of the second plane is (0, -1, 0).

Step 4: Find the angle between the normal vectors The angle θ between two vectors a and b is given by the formula:

cos(θ) = (a . b) / (||a|| ||b||)

where a . b is the dot product of a and b, and ||a|| and ||b|| are the magnitudes of a and b, respectively.

The dot product of the two normal vectors is:

(5, 0, 1) . (0, -1, 0) = 50 + 0(-1) + 1*0 = 0

The magnitudes of the normal vectors are:

|| (5, 0, 1) || = sqrt(5^2 + 0^2 + 1^2) = sqrt(26) || (0, -1, 0) || = sqrt(0^2 + (-1)^2 + 0^2) = 1

Substituting these values into the formula, we get:

cos(θ) = 0 / (sqrt(26) * 1) = 0

Therefore, θ = arccos(0) = π/2 radians.

So, the angle between the normal vectors of the two planes is π/2 radians.