The given problem involves finding the equations for:

1. The tangent plane to the surface defined by $y = x^2 - z^2$ at the point $(5, 21, 2)$.
2. The normal line to the surface at the same point.

Let's go step-by-step:

Step 1: Find the Tangent Plane

The equation of the tangent plane to the surface at a point can be written as: $f_x(x_0, z_0)(x - x_0) + f_y(x_0, z_0)(y - y_0) + f_z(x_0, z_0)(z - z_0) = 0$ where $f_x$, $f_y$, and $f_z$ are the partial derivatives of the function, and $(x_0, y_0, z_0)$ is the point on the surface.

The surface is defined by: $f(x, y, z) = y - (x^2 - z^2) = 0$

The partial derivatives are: $f_x = -2x, \quad f_y = 1, \quad f_z = 2z$

At the point $(x_0, y_0, z_0) = (5, 21, 2)$: $f_x(5, 2) = -2(5) = -10, \quad f_y(5, 21, 2) = 1, \quad f_z(5, 2) = 2(2) = 4$

Thus, the equation of the tangent plane is: $-10(x - 5) + 1(y - 21) + 4(z - 2) = 0$ Simplifying: $-10x + 50 + y - 21 + 4z - 8 = 0$ $-10x + y + 4z + 21 = 0$

Step 2: Find the Normal Line

The normal line is defined by the parametric equations: $x(t) = x_0 + f_x t, \quad y(t) = y_0 + f_y t, \quad z(t) = z_0 + f_z t$ Substituting the values of the partial derivatives and the point $(5, 21, 2)$: $x(t) = 5 - 10t, \quad y(t) = 21 + t, \quad z(t) = 2 + 4t$

So the parametric equation for the normal line is: $(x(t), y(t), z(t)) = (5 - 10t, 21 + t, 2 + 4t)$

Would you like further clarification or more details on any step?

Here are five related questions:

1. How do you find the normal vector to a surface at a point?
2. What is the general formula for the tangent plane of a surface?
3. Why do we use partial derivatives in tangent plane equations?
4. How would the tangent plane equation change if the surface had a more complex form?
5. How do you interpret the parametric equation of the normal line geometrically?

Tip: The normal vector is crucial when finding both the tangent plane and the normal line, as it directly influences the orientation of the plane and the direction of the line.