Question

The unit normal vector to the surface \( X^2 + Y^2 + Z^2 - 48 = 0 \) at the point (4, 4, 4) is

• A. \( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \)
• B. \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
• C. \( \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}} \)
• D. \( \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}} \)

Hint:

The unit normal vector is the normalized gradient of the surface equation.

Answer 1

The Correct Option is B

The surface equation is: \[ X^2 + Y^2 + Z^2 - 48 = 0. \] To find the unit normal vector, we take the gradient of the surface equation. The gradient \( \nabla f \) is: \[ \nabla f = \left( \frac{\partial}{\partial X}(X^2 + Y^2 + Z^2 - 48), \frac{\partial}{\partial Y}(X^2 + Y^2 + Z^2 - 48), \frac{\partial}{\partial Z}(X^2 + Y^2 + Z^2 - 48) \right) \] \[ \nabla f = (2X, 2Y, 2Z). \] At the point (4, 4, 4), the gradient is: \[ \nabla f = (8, 8, 8). \] The unit normal vector is given by normalizing the gradient: \[ \text{Unit normal vector} = \frac{\nabla f}{|\nabla f|} = \frac{(8, 8, 8)}{\sqrt{8^2 + 8^2 + 8^2}} = \frac{(8, 8, 8)}{\sqrt{192}} = \frac{(8, 8, 8)}{8\sqrt{3}} = \frac{1}{\sqrt{3}} (1, 1, 1). \] Final Answer: \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)