ind the angle between the following surfaces at the point (−2, 1, 1):x2 + y2 + z2 = 6 and x2 + y2 − 2z = 3.

Sure, here is the step by step solution:

Step 1: Find the normal vectors of the surfaces at the given point.

The normal vector of a surface F(x, y, z) = 0 at a point (x0, y0, z0) is given by the gradient ∇F at that point.

For the first surface, F1(x, y, z) = x^2 + y^2 + z^2 - 6, the gradient ∇F1 = (2x, 2y, 2z). At the point (-2, 1, 1), ∇F1 = (-4, 2, 2).

For the second surface, F2(x, y, z) = x^2 + y^2 - 2z - 3, the gradient ∇F2 = (2x, 2y, -2). At the point (-2, 1, 1), ∇F2 = (-4, 2, -2).

Step 2: Find the angle between the normal vectors.

The angle θ between two vectors u and v is given by the formula cos(θ) = (u . v) / (||u|| ||v||), where "." denotes the dot product and "|| ||" denotes the norm (length) of a vector.

The dot product of ∇F1 and ∇F2 is (-4)(-4) + 22 + 2*(-2) = 12.

The norm of ∇F1 is sqrt((-4)^2 + 2^2 + 2^2) = sqrt(24), and the norm of ∇F2 is sqrt((-4)^2 + 2^2 + (-2)^2) = sqrt(24).

Therefore, cos(θ) = 12 / (sqrt(24) * sqrt(24)) = 12 / 24 = 0.5.

So, θ = arccos(0.5) = 60 degrees.

Therefore, the angle between the given surfaces at the point (-2, 1, 1) is 60 degrees.