Full solution

Q. Consider the following.

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2(x-3)^{2}+(y-3)^{2}+(z-7)^{2}=10,\quad(4,5,9)

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(a) Find an equation of the tangent plane to the given surface at the specified point.

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4(x-4)+4(y-5)+4(z-9)=0

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(b) Find an equation of the normal line to the given surface at the specified point.

Define Surface Equation: The given surface is defined by the equation $2(x-3)^2 + (y-3)^2 + (z-7)^2 = 10$ . To find the equation of the tangent plane at the point $(4,5,9)$ , we first need to find the gradient of the surface at that point. The gradient vector will be normal to the tangent plane.
Calculate Gradient Vector: The gradient of the surface is given by the partial derivatives of the function with respect to $x$ , $y$ , and $z$ . Let's denote the function as $f(x, y, z) = 2(x-3)^2 + (y-3)^2 + (z-7)^2$ . We calculate the partial derivatives: $\newline$ $\frac{\partial f}{\partial x} = 4(x-3)$ , $\newline$ $\frac{\partial f}{\partial y} = 2(y-3)$ , $\newline$ $\frac{\partial f}{\partial z} = 2(z-7)$ .
Evaluate at Given Point: Now we evaluate the partial derivatives at the point $(4,5,9)$ : $\frac{\partial f}{\partial x}$ at $(4,5,9)$ = $4(4-3) = 4$ , $\frac{\partial f}{\partial y}$ at $(4,5,9)$ = $2(5-3) = 4$ , $\frac{\partial f}{\partial z}$ at $(4,5,9)$ = $2(9-7) = 4$ . So the gradient vector at the point $(4,5,9)$ is $\frac{\partial f}{\partial x}$ $1$ .
Equation of Tangent Plane: The equation of the tangent plane to the surface at the point $(4,5,9)$ can be written as: $\newline$ $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ , $\newline$ where $(a, b, c)$ is the gradient vector and $(x_0, y_0, z_0)$ is the point on the surface. Plugging in the values, we get: $\newline$ $4(x-4) + 4(y-5) + 4(z-9) = 0$ .
Find Normal Line Equation: To find the equation of the normal line, we use the gradient vector as the direction vector of the line. The normal line passes through the point $(4,5,9)$ and has the direction vector $(4, 4, 4)$ . The parametric equations of the line are: $\newline$ $x = 4 + 4t$ , $\newline$ $y = 5 + 4t$ , $\newline$ $z = 9 + 4t$ , $\newline$ where $t$ is the parameter.
Write in Standard Form: To write the equation of the normal line in the standard form, we eliminate the parameter $t$ : $\frac{x-4}{4} = \frac{y-5}{4} = \frac{z-9}{4}$ .