Question

In \( \mathbb{R}^3 \), the cosine of the acute angle between the surfaces \( x^2 + y^2 + z^2 - 9 = 0 \) and \( z - x^2 - y^2 + 3 = 0 \) at the point (2, 1, 2) is

• A. \( \frac{8}{5\sqrt{21}} \)
• B. \( \frac{10}{5\sqrt{21}} \)
• C. \( \frac{8}{3\sqrt{21}} \)
• D. \( \frac{10}{3\sqrt{21}} \)

Hint:

To find the cosine of the angle between two surfaces, use the gradients of the surfaces and apply the formula for the dot product and magnitudes of the gradients.

Answer 1

The Correct Option is C

Step 1: Understanding the problem. 
We are asked to find the cosine of the angle between two surfaces at a given point. The formula for the cosine of the angle between two surfaces is: \[ \cos \theta = \frac{\nabla f \cdot \nabla g}{|\nabla f| |\nabla g|} \] where \( \nabla f \) and \( \nabla g \) are the gradients of the functions defining the surfaces. 
 

Step 2: Computing the Gradients. 
The first surface is \( f(x, y, z) = x^2 + y^2 + z^2 - 9 \), and the second surface is \( g(x, y, z) = z - x^2 - y^2 + 3 \). We calculate their gradients: \[ \nabla f = (2x, 2y, 2z), \nabla g = (-2x, -2y, 1) \] At the point (2, 1, 2), we have: \[ \nabla f = (4, 2, 4), \nabla g = (-4, -2, 1) \]

Step 3: Computing the Dot Product and Magnitudes. 
The dot product \( \nabla f \cdot \nabla g \) is: \[ \nabla f \cdot \nabla g = (4)(-4) + (2)(-2) + (4)(1) = -16 - 4 + 4 = -16 \] The magnitudes are: \[ |\nabla f| = \sqrt{4^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \] \[ |\nabla g| = \sqrt{(-4)^2 + (-2)^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \]

Step 4: Finding the Cosine of the Angle. 
Now, we calculate the cosine of the angle: \[ \cos \theta = \frac{16}{6 \times \sqrt{21}} = \frac{8}{3\sqrt{21}} \]

Step 5: Conclusion. 
The correct answer is (C) \( \frac{8}{3\sqrt{21}} \).