Question

The tangent line to the curve of intersection of the surface \( x^2 + y^2 - z = 0 \) and the plane \( x + z = 3 \) at the point \( (1, 1, 2) \) passes through

• A. \( (-1, -2, 4) \)
• B. \( (-1, 4, 4) \)
• C. \( (3, 4, 4) \)
• D. \( (-1, 4, 0) \)

Hint:

To find the tangent line to the intersection of two surfaces, compute the gradients of the surfaces and take the cross product.

Answer 1

The Correct Option is B

Step 1: Find the equations of the surfaces.
The equation of the surface is \( x^2 + y^2 - z = 0 \), or equivalently, \( z = x^2 + y^2 \). The equation of the plane is \( x + z = 3 \), or equivalently, \( z = 3 - x \).
Step 2: Find the gradient vectors.
To find the tangent line, we compute the gradients of the surface and the plane to obtain the direction vector of the tangent line. The gradient of the surface is: \[ \nabla F(x, y, z) = (2x, 2y, -1), \] and the gradient of the plane is: \[ \nabla G(x, y, z) = (1, 0, 1). \]
Step 3: Compute the cross product.
The direction vector of the tangent line is given by the cross product of the gradients of the surface and the plane: \[ \nabla F \times \nabla G = (2x, 2y, -1) \times (1, 0, 1) = (2y, -2x, 2). \] At the point \( (1, 1, 2) \), the direction vector is \( (2, -2, 2) \).
Step 4: Equation of the tangent line.
The parametric equations of the tangent line are: \[ x = 1 + 2t, \quad y = 1 - 2t, \quad z = 2 + 2t. \] Substitute different values of \( t \) to find the point where the line intersects the given surface and plane.
Step 5: Conclusion.
Thus, the tangent line passes through \( (-1, 4, 4) \), so the correct answer is \( \boxed{(B)} \).