Question

Find the points on the given surface at which the tangent plane is parallel to the indicated plane. x2 + y2 + z2 = 7; 2x + 4y + 6z = 1 (x, y, z) = (smaller x-value) (x, y, z)( (larger x-value)

Answer 1

To find the points on the given surface at which the tangent plane is parallel to the indicated plane x² + y² + z² = 7 and 2x + 4y + 6z = 1, we need to find the gradient vector of the surface and the normal vector of the plane. By setting the gradient vector equal to a scalar multiple of the normal vector, we can solve for the values of x, y, and z at which the two planes are parallel. The points are approximately (0.707, 1, 1.225) and (-0.707, -1, -1.225).

Step-by-step explanation:

To find the points on the given surface where the tangent plane is parallel to the indicated plane, we need to find the gradient vector of the surface and the normal vector of the indicated plane. Let's start with the given surface equation x² + y² + z² = 7. Taking the gradient of this equation, we get ∇(x² + y² + z²) = 2x∂x + 2y∂y + 2z∂z = 0.

The gradient vector of the surface is ∇(x² + y² + z²) = (2x, 2y, 2z). Next, let's find the normal vector of the indicated plane 2x + 4y + 6z = 1. The coefficients of x, y, and z in the plane equation give us the normal vector n = (2, 4, 6).

For the tangent plane to be parallel to the indicated plane, the gradient vector of the surface and the normal vector of the indicated plane must be parallel. This means that the two vectors must be scalar multiples of each other. Therefore, we can set up the following system of equations: 2x = λ(2), 2y = λ(4), 2z = λ(6), where λ is the scalar.

Solving this system of equations, we find that x = λ, y = 2λ, and z = 3λ. Substituting these values into the equation of the surface x² + y² + z² = 7, we have (λ)² + (2λ)² + (3λ)² = 7. Simplifying, we get 14λ² = 7. Solving for λ, we find λ = ±√(1/2).

So, the two sets of values for (x, y, z) at which the tangent plane is parallel to the indicated plane are:

(x, y, z) = (√(1/2), √(2/2), √(3/2)) ≈ (0.707, 1, 1.225)

(x, y, z) = (-√(1/2), -√(2/2), -√(3/2)) ≈ (-0.707, -1, -1.225)

Answer 2

The points on the surface
`x^2 + y^2 + z^2 = 7` at which the tangent plane is parallel to the plane 2x + 4y + 6z = 1 are (-√7, 0, 0) and (√7, 0, 0).

Step-by-step explanation:

The given surface equation
`x^2 + y^2 + z^2`= 7 represents a sphere centered at the origin with a radius of √7. The equation of the plane 2x + 4y + 6z = 1 can be rearranged into a normal vector form as (2, 4, 6). For the tangent plane to be parallel to this plane, the normal vector of the tangent plane should be parallel to (2, 4, 6).

The normal vector of the sphere at any point (x, y, z) is given by (2x, 2y, 2z). For the tangent plane to be parallel to the plane 2x + 4y + 6z = 1, the normal vector of the sphere should be a scalar multiple of the normal vector of the plane.

Comparing the components, we get:

2x = k * 2

2y = k * 4

2z = k * 6

Solving these equations gives x = k, y = 2k, and z = 3k. Substituting these values into
`x^2 + y^2 + z^2 = 7 yields 14k^2 = 7`, giving us k = ±1. Thus, the points on the surface are (-√7, 0, 0) and (√7, 0, 0), corresponding to the smaller and larger x-values, respectively.

This solution ensures that the tangent plane at these points on the sphere will be parallel to the plane 2x + 4y + 6z = 1.

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